%!TEX program = xelatex
\documentclass[a4paper]{report}
\usepackage[ngerman, english]{babel}
\usepackage[nottoc]{tocbibind}
\usepackage{csquotes}
\usepackage{amsmath,amsfonts,amssymb,amsthm,bbm,mathrsfs}
\usepackage{comment}
\usepackage{hyperref}
\usepackage{listings}
\usepackage{graphicx}
\usepackage{pgf,tikz,pgfplots}
\usetikzlibrary{arrows, positioning}
\pgfplotsset{compat=1.16}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem*{theorem*}{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{property}[theorem]{Property}
\newtheorem{properties}[theorem]{Properties}
\newtheorem{conjecture}[theorem]{Conjecture}

\newtheorem*{theoremmain*}{Theorem \ref{thm:DiscMainThm}}
\newtheorem*{propositionSafeDice}{Proposition \ref{prop:SafeDice}}
\newtheorem*{propositionBorelConFalse}{Proposition \ref{prop:BorelConFalse}}
\newtheorem*{propositionEnvelopeIndSafe}{Proposition \ref{prop:EnvelopeIndSafe}}

\theoremstyle{definition}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{fact}[theorem]{Fact}

\numberwithin{equation}{chapter}

\hypersetup{%
  colorlinks = true
}

%Afkortingen voor wiskundige symbolen
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\renewcommand{\C}{\mathbb{C}}

\let\P\relax
\DeclareMathOperator{\P}{\mathbb{P}}
\DeclareMathOperator{\V}{\mathbb{V}}
\DeclareMathOperator{\E}{\mathbb{E}}
\DeclareMathOperator{\1}{\mathbbm{1}}
\newcommand{\F}{\mathcal{F}}
\renewcommand{\G}{\mathcal{G}}
\renewcommand{\H}{\mathcal{H}}
\newcommand{\B}{\mathcal{B}}
\newcommand{\X}{\mathcal{X}}
\newcommand{\Y}{\mathcal{Y}}

\DeclareMathOperator{\supp}{supp}
\DeclareMathOperator{\range}{range}
\DeclareMathOperator{\acc}{acc}

\newcommand{\Pmod}{\mathcal{P}^*}
\newcommand{\Psafe}{\tilde{\P}}
\newcommand{\Uonesafe}{\tilde{\1}_U}
\newcommand{\Usafe}{\tilde{U}}

\newcommand{\EnvInd}{\1_{\{B=2A\}}}
\newcommand{\EnvIndSafe}{\tilde{\1}_{\{B=2A\}}}
\newcommand{\MontyInd}{\1_{\{U=a\}}}
\newcommand{\DieInd}{\1_{\{U=3\}}}
\newcommand{\DieIndSafe}{\tilde{\1}_{\{Y=3\}}}
\newcommand{\ChildInd}{\1_{\{U=bg\}}}
\newcommand{\ChildIndSafe}{\tilde{\1}_{\{U=bg\}}}
\newcommand{\ChildTwoInd}{\1_{\{U=bb\}}}
\newcommand{\ChildTwoIndSafe}{\tilde{\1}_{\{U=bb\}}}
\newcommand{\GeneralInd}{\1_{\{U=u\}}}
\newcommand{\GeneralGenInd}{\1_{\{U\in\X'\}}}
\newcommand{\GeneralGenIndSafe}{\tilde{\1}_{\{U\in\X'\}}}

\title{Thesis}
\author{Mathijs Kolkhuis Tanke}
\date{\today}

\begin{document}
\begin{titlepage}
\pagenumbering{gobble}
\input{titlepage.tex}
\end{titlepage}

\pagenumbering{roman}

\begin{abstract}
A vast number of questions and problems concerning probability theory need conditional probability for providing answers and solutions. From traditional games of dice to modern statistical applications and machine learning, all use conditional probability in some sense to obtain more insight in the problem. As fundamental conditional probability is, it is not without controversy. Problems and paradoxes like the Borel-Kolmogorov paradox, Monty Hall's three door problem and the two envelope problem have puzzled mathematicians, statisticians and psychologists for centuries, resulting into much debate and a vast amount of literature.

This thesis concerns some of the most well-known paradoxes in conditional probability. In all paradoxes, the paradoxical result arises from wrongly stating the probability space concerning the problem or wrongly applying conditional probability, like not giving the accompanying $\sigma$-algebra or not conditioning on a partition. All problems can be easily avoided by always stating the probability space with the $\sigma$-algebra when applying conditional probability.

The two most obvious examples are the Borel-Kolmogorov paradox and Monty Hall's problem. The Borel-Kolmogorov paradox is a good example of why conditioning on sets with zero measure is only possible with much care and why it is necessary to provide the accompanying $\sigma$-algebra with your solution.\\
Monty Hall's three door problem is a prime example of wrongly conditioning on a set of subsets that cannot form a partition of the sample space. The original problem asks for a single probability, however correctly applying conditional probability reveals that the probability of the car being behind the other door is dilated between the two values $\frac{1}{2}$ to $1$. In both cases the paradoxical results vanish when the whole probability space is considered and when conditional probability is applied correctly.

The dilation of the conditional probability like in Monty Hall's problem is investigated further in this thesis. Problems like Monty Hall and the boy or girl problem resemble each other in such a fundamental fashion that a generalization exists, encompassing them both. Furthermore, \emph{safe probability} introduced by Grünwald \cite{Grunwald18} can be applied to answer the following question: if one should pin a single probability on for example in Monty Hall's game the car being behind the other door, which probability should it be? This generalization can be applied to all problems with a countable space of outcomes with fixed probability measure and finite set of possible observations with sufficiently many possible probability measures, resolving several paradoxes in probability at once.
\end{abstract}

\pagenumbering{roman}
\setcounter{page}{2}
\tableofcontents

\chapter{Introduction}
\pagenumbering{arabic}
Probability theory is one of the most important and most researched fields in mathematics. It is in essence based on just three axioms first stated by Andrey Kolmogorov \cite{Kolmogorov33}, namely that the probability of an event is non-negative, the probability measure is a unit measure and the probability measure is countably additive on disjoint sets. These three axioms however do not prevent the existence of certain paradoxes, like the Borel-Kolmogorov paradox, Monty Hall's problem and the Sleeping Beauty problem. Some paradoxes arise from wrongly using probability theory, others are a misinterpretation of results.

This thesis focuses mainly on paradoxes arising from conditional probability, such as the Borel-Kolmogorov paradox, Monty Hall's problem and the two envelope problem. We will study these problems and address their paradoxical nature. Ultimately it will be shown that all paradoxes arise by incorrectly applying conditional probability.

Take a look at the Borel-Kolmogorov paradox first. A sphere is equipped with the uniform probability measure. Suppose a point exists on a given great circle of the sphere, but you do not know where that point is. Is there a probability distribution on this great circle for that point? If one models this problem using latitudes, the conditional distribution on the great circle is uniform. However, if one models this problem using meridians, the conditional distribution is a cosine.\\
Borel\index{Borel} \cite{Borel09} and Kolmogorov \cite{Kolmogorov33} both addressed this problem and Kolmogorov gave the following answer:
\foreignblockquote{ngerman}[Andrey Kolmogoroff, \cite{Kolmogorov33}]{Dieser Umstand zeigt, daß der Begriff der bedingten Wahrscheinlichkeit in bezug auf eine isoliert gegebene Hypothese, deren Wahrscheinlichkeit gleich Null ist, unzulässig ist: nur dann erhält man auf einem Meridiankreis eine Wahrscheinlichkeitsverteilung für [Breite] Theta, wenn dieser Meridiankreis als Element der Zerlegung der ganzen Kugelfläche in Meridiankreise mit den gegebenen Polen betrachtet wird.}
In summary, Kolmogorov states that the concept of conditional probability on a great circle is inadmissible, since the event that the point lies on a great circle of the sphere has zero measure. Furthermore, the sole reason of a cosine distribution on a great circle arising when considering meridians is that the meridian circle serves as an element of the decomposition of the whole spherical surface in meridian circles with the poles considered.\\
Despite Kolmogorov's explanation this problem is still under debate. Recently Gyenis, Hofer-Szabó and Rédei \cite{Gyenis17} studied this problem and provided more insight and an in my opinion satisfying discussion to the problem, eventually drawing the same conclusion as Kolmogorov already did. This problem is more elaborately discussed in Chapter~\ref{chap:BorelKolmogorov} where we will expand the analysis of \cite{Gyenis17} and not only consider latitudes and meridians, but their combination as well. Furthermore, we will look whether a conditional probability on a null set can be uniquely approached by traditional conditional probability on sets of decreasing measure as stated in Conjecture~\ref{con:BorelConjecture}. This conjecture turns out to be false, again affirming that conditional probabilities on sets of measure zero are not uniquely defined.

Another paradox arising from conditional probability is Monty Hall's problem. In Monty Hall's problem, a player is facing three doors called $a$, $b$ and $c$. One door has a car behind and the other two have goats. Suppose the player initially chooses door $a$. The game master then opens either door $b$ or $c$, but always a door with a goat. The player now faces two doors and is asked whether he wants to switch. One possible `solution' is that the player faces two doors without any preference for either door, thus the probability is 50\% of initially picking the correct door. Another possible solution is that first the player had a 33\% chance of correctly guessing the door with the car. If for example door $c$ is opened, door $b$ remains with a conditional probability of 67\% of having the car. Which solution is correct?\\
Let $\X=\{a,b,c\}$ be the set of doors and $U$ the $\X$-valued random variable denoting the location of the car. Conditional probability then provides
\begin{equation}\label{eq:IntMonty}
\P[U=a\mid U\in\{a,b\}]=\frac{\P[U\in\{a,b\}|U=a]\P[U=a]}{\P[U\in\{a,b\}]}=\frac{1\cdot\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}\tag{*},
\end{equation}
supporting the claim that a probability of 50\% of the car being behind the initially chosen door is the correct answer. However, the space conditioned on the event that door $c$ is opened is $\{a,b\}$ and the space conditioned on the event that door $b$ is opened is $\{a,c\}$. If door $a$ has the car, the game master can open either door and the set of events we must condition on is $\{\{a,b\},\{a,c\}\}$. These two events do not form a partition, preventing us from using traditional conditional probability: the answer \eqref{eq:IntMonty} is simply wrong, since traditional probability implicitly assumes one always conditions on the elements of a partition.

Thus, in addition to Kolmogorov's statement, we must not only be wary for conditioning on events with zero measure, we cannot condition on arbitrary events at all. I propose that when using conditional probability, one must provide a pair of a sub-$\sigma$-algebra and the event from that $\sigma$-algebra to condition on. Regarding Monty Hall's problem there is no $\sigma$-algebra on $\{a,b,c\}$ containing the set $\{\{a,b\},\{a,c\}\}$. In the case of the Borel-Kolmogorov paradox providing sub-$\sigma$-algebras immediately make clear why the conditional distribution on a great circle is not unique, as both calculations are supported by different sub-$\sigma$-algebras.

The crux of the Monty Hall problem is that the initial location of the car is unknown and the player does not know with which probability door $b$ is opened given the car is behind door $a$. Therefore, there is a set of different possible probability distributions where one is the correct distribution. Using the theory of \emph{safe probability} we can obtain a strategy that gives equal results for all distributions in a specified model. This theory is introduced by Grünwald \cite{Grunwald18} and summarized here in Chapter~\ref{chap:SafeProp}. Safe probability is then applied in Chapter~\ref{chap:DiscPara} to problems as the Monty Hall problem and to the two envelope problem in Chapter~\ref{chap:TwoEnvelope}.

In Chapter~\ref{chap:DiscPara} the results of the analysis of Monty Hall's problem are generalized to a theorem, which can be used to provide safe distributions for other problems like the boy or girl problem:
\begin{theoremmain*}
Let $\X$ be countable and $\Y$ be finite. Let $U$ be an $\X$-valued random variable and $V$ be a $\Y$-valued random variable. Let $\{p_u\}_{u\in\X}\subset[0,1]$ with $\sum_{u\in\X}p_u=1$. Let
\begin{equation*}
\Pmod\subseteq\{\P\mid\forall u\in\X:\P[U=u]=p_u\}
\end{equation*}
be our set of probability distributions on $\X\times\Y$ such that $|\Y|$ distributions $\P_1,\ldots,\P_{|\Y|}\in\Pmod$ exist imposing $|\Y|$ linearly independent vectors $(\P_i[V=v])_{v\in\Y}$ with $i\in\{1,\ldots,|\Y|\}$. Let $u\in\X$ be arbitrary and let $\Psafe$ be a distribution on $\X\times\Y$ with full support on $V$, then the following are equivalent:
\begin{enumerate}
    \item For all $v\in\Y$ we have $\Psafe[U=u|V=v]=p_{u}$.
    \item $\Psafe$ is safe for $\GeneralInd|[V]$.
    \item $\Psafe$ is safe for $\langle\GeneralInd\rangle|\langle V\rangle$.
\end{enumerate}
\end{theoremmain*}

This theorem and its notation will be explained, proven and applied in Chapter~\ref{chap:DiscPara}, but it essentially states the following: in the case of the Monty Hall problem if one assumes the car is initially distributed evenly between the doors, the probability of the car being behind the originally chosen door $a$ can be assumed to be $\frac{1}{3}$. Using this assumption one must always switch to the other door as the probability of the car being behind that door can be assumed to be $\frac{2}{3}$, resulting in a 67\% chance of winning the car. It is not at all clear whether this is the correct distribution and in most situations it definitely is not. However, this assumption always yields to a 67\% chance of \emph{winning} the car independent on the real probability of \emph{opening} door $b$ when the car is behind door $a$.

Another paradox we will treat is the two envelope problem. There are two envelopes filled with a certain value. The only knowledge to the player is that after filling the first envelope, a fair coin decided whether the second envelope has either half or twice the first envelope's value. An envelope is given to the player, with the player receiving either envelope with equal probability. Call this envelope~$A$ and suppose the player observes value $a$. Should he switch to envelope~$B$ or keep the contents of $A$?\\
At first glance, he should. Calling envelope $B$'s value $b$, one can reason that either $b=2a$ or $b=\frac{a}{2}$ holds with equal probability, giving \[\E[B|A=a]=\frac{1}{2}\cdot2a+\frac{1}{2}\cdot\frac{a}{2}=\frac{5}{4}a.\]
However, the player has received an envelope at random and only knows the contents of that particular envelope. So he could also have been given envelope~$B$ with value $b$, for which $\E[A|B=b]=\frac{5}{4}b$ holds. Therefore no matter what envelope the player receives, he should switch. This solution must clearly be wrong and it is. When conditioning on $A=a$, then either $b=2a$ or $a=2b$ holds with probability $1$ in the conditioned probability space. This renders the previous calculation of $\E[B|A=a]$ incorrect. However, we do not know whether $b=2a$ or $a=2b$ took place.\\
Now let $x=\min\{a,b\}$ be the lowest value of both envelopes, then either $a=x$ or $a=2x$ holds with equal probability. For both envelopes we now have
\[
\E[A|B=b]=\E[B|A=a]=\frac{1}{2}x+\frac{1}{2}\cdot2x=\frac{3}{2}x.
\]
Both envelopes thus have on average the same contents, as is expected when the envelopes are handed out uniformly. Unfortunately, we do not know the value of $x$. Furthermore, if $x$ is picked from an unbounded set and we assume that given an observation the other envelope must hold twice as much or half the value with equal probability, then $x$ must be taken from a uniform probability distribution on the unbounded set. Such a distribution does not exist. We conclude that the value of $\E[B|A=a]$ does depend on a prior distribution on the chosen value $x$.\\
There are two ways to address this problem. One way would be to replace the giant set of possible prior distributions by a single `safe distribution', as we could do for the Monty Hall problem. However, we will show that a useful `safe distribution' of the form presented in Theorem~\ref{thm:DiscMainThm} does not exist for this problem in general\footnote{It may exist though if we make specific assumptions about $a$ being taken from a finite set; also, safe probability comes in different types and a different `weaker' type may exist for general $a$. These questions are however for future work, as can be seen in Section \ref{sec:EnvelopeSafe}.}. Another method is using a switching strategy introduced by McDonell and Abbott~\cite{McDonnell09,Abbott10,McDonnell11}. Let $X$ be the random variable picking $x$ and let $f\colon (0,\infty)\to[0,1]$ be a function that takes an observation $a$ and equips it with a probability. The player must switch with probability $f(a)$ when observing value $a$. When $f$ is a decreasing function and strictly decreases on an interval where the distribution of $X$ has positive measure, then switching using $f$ will always return on average a higher value than $\frac{3}{2}\E[X]$. If $f$ is a threshold switch, for example $f=\1_{(0,a^*]}$, then for some distributions of $X$, in particular the ones with a strictly increasing cumulative distribution function, a strategy of this form is actually the best available. However, when playing the game, the player does not know how much strategy $f$ outperforms the trivial strategy of never switching. By any means, the distribution of $X$ gives $f$ an arbitrary small advantage over safe probability, making it an unnecessarily complicated method to marginally increase the average value won.\\
The two envelope problem will be discussed further in Chapter~\ref{chap:TwoEnvelope}.

For every problem there is a different reason for getting paradoxical results. Still, there is one recurring theme with all paradoxes, namely that when analysing these problems most of the times the underlying $\sigma$-algebra is not taken into account. This yields to various conflicting results, as conditioning on events that cannot be conditioned on to not recognizing that multiple probability distributions are possible. The common thread of this thesis is therefore that when doing probability theory, the underlying probability space and $\sigma$-algebra must never be ignored and not providing a sub-$\sigma$-algebra with a conditional distribution must become a bad habit instead of an accepted practice.

\chapter{The Borel-Kolmogorov paradox}\label{chap:BorelKolmogorov}
The first paradox we will study in more detail is the Borel-Kolmogorov paradox. Many accredit the origin of the problem to Borel in 1909 \cite{Borel09}, but the problem is originally conceived by Bertrand in 1889 \cite{Bertrand89}. Kolmogorov \cite{Kolmogorov33} first provided a meaningful answer to and discussion on the problem in 1933, which is why the problem is now called the Borel-Kolmogorov paradox. We will take a look at the following version of the paradox.

Suppose a random variable has a uniform probability distribution on the unit sphere. If one conditions on a great circle, what will the resulting conditional distribution be? If the great circle is viewed as a latitude, the conditional distribution will be uniform as well. However, if the great circle is modelled as a meridian, the conditional distribution has a cosine as probability density function. We have two different conditional distributions on the same set, thus which one is correct?

This problem seems a bit specific, however it can occur in many other more realistic cases. One example is given by Proschan and Presnell \cite{Proschan98}. There two random variables $X$ and $Y$ are considered, both distributed according to the standard normal distribution. A professor once asked the students on a test to describe the distribution of $Y$ given $X=Y$. In short, there are three different ways to attack this problem. Some students calculated the density of $Y$ given $Y-X=0$, others calculated the density of $Y$ given $\frac{Y}{X}=1$ and another student calculated the density of $Y$ given $\1_{\{Y=X\}}=1$. All three methods create different density functions, but the methods used by the students are all correct. The crux is that the original question posed by the professor is flawed: the event $\{X=Y\}$ has zero probability as $X$ and $Y$ both are continuous random variables. This is why this problem is related to the Borel-Kolmogorov paradox.

The basic solution to the paradox is to use measure-theoretic conditional probability, so that one is given not just a set like a great circle to condition on but rather a set and an accompanying sub-$\sigma$-algebra containing this set. The definition of measure-theoretic conditional probability now provides an answer that is unique and meaningful up to a set of zero measure.

When conditional probability is defined traditionally based on density functions, it is defined on every point even though each such point has measure zero. One may now ask the following question: can we generally extend the measure-theoretic conditional probability to be defined on all points once a sub-$\sigma$-algebra is provided, by treating the conditional probability at that point as a limit of traditional conditional probabilities where one conditions on smaller and smaller sets? In Section~\ref{sec:BorelCombining} we provide Conjecture~\ref{con:BorelConjecture} that would imply this, but we will then show this conjecture is wrong. If we condition on both a given meridian and a suitably rotated latitude, then two phenomena occur:
\begin{enumerate}
\item the only sub-$\sigma$-algebra that we can provide is the full Borel $\sigma$-algebra leading conditioning on the full $\sigma$-algebra to no effect and
\item the required limit is undefined as the limit of smaller and smaller meridians differs from the limit of smaller and smaller rotated latitudes.
\end{enumerate}

Kolmogorov \cite{Kolmogorov33} pointed out that the great circle has zero measure and according to him conditioning on a great circle must not be allowed. One will be tempted to disregard conditioning on null sets at all, but this brings many problems as argued by De Finetti \cite{Finetti72} and Howson \cite{Howson14}. One example from De Finetti is the field of statistical applications where it is realistic to regard the probability of making observation $E\in\R^d$ as $0$. Prediction models are highly based on these observations $E$, making conditioning on sets of zero measure necessary. Therefore, we cannot just disregard conditioning on probabilities with measure~$0$, as those conditional probabilities are needed for applying statistics.

Our solution, supported by \cite{Rao88,Billingsley95,Proschan98,Easwaran08,Myrvold15,Gyenis17}, is that when performing conditional probability, the $\sigma$-algebra of the sets conditioned on must always be taken into account. Otherwise, the event $E$ with zero measure has no meaning as there are many methods to approach $E$ by events with positive measures, giving different conditional probabilities.

Our analysis concerns conditioning on null sets using measure-theoretic conditional expectations as such conditioning is defined there. The article of Gyenis, Hofer-Szabó and Rédei \cite{Gyenis17} is followed to compute the conditional distributions on the great circle, viewing that circle both as latitude and as meridian. We will take it one step further by also considering the $\sigma$-algebra of both the meridians and latitudes and by considering conditional distributions on rotated latitudes making them coincide with meridians. We will then conclude that once again conditioning on only measure null sets must not be admissible, as it will lead to contradicting results.

After drawing these conclusions, we will rename the Borel-Kolmogorov paradox to the \emph{Borel-Kolmogorov phenomenon}, as it is more an example on why you should not carelessly condition on null sets than it is a paradox.

\section{Conditional expectation}
First we need to define conditional expectations. We use the standard measure-theoretic definitions as given by David Williams \cite{Williams91}.

Let $\Omega$ be an arbitrary sample space and let $\mathcal{F}$ be a $\sigma$-algebra on $\Omega$. The indicator function $\1_A$ on a set $A$ is defined as
\begin{equation}
\1_A(x)=\begin{cases}1,&x\in A,\\0,&x\not\in A.\end{cases}
\end{equation}
Let $\B(A)$ be the $\sigma$-algebra of all Borel-measurable sets on $A\subseteq\R^n$. We will start with the definition of a random variable.

\begin{definition}[Random variable]
Let $\X$ be a measurable space. A function $X\colon\Omega\to\X$ is an \emph{$\X$-valued random variable} if it is $\F$-measurable, thus if $X^{-1}(A)\in\F$ holds for all $A\in\G$ where $\G$ is a $\sigma$-algebra on $\X$.
\end{definition}

We can now define the conditional expectation as definition 9.2 of \cite{Williams91}.

\begin{definition}[Conditional expectation]\label{def:conexp}
Let $X$ be an $\F$-measurable random variable with finite $\E[|X|]$. Let $\G$ be a sub-$\sigma$-algebra of $\F$. There exists a random variable $Y$ such that
\begin{enumerate}
\item $Y$ is $\G$-measurable,
\item $\E[|Y|]$ is finite,
\item for every $G\in\G$ we have
\begin{equation}
\int_G Yd\P=\int_G Xd\P.
\end{equation}
\end{enumerate}
$Y$ is called a \emph{version of the conditional expectation} $\E[X|\G]$ of $X$ given $\G$, written as $Y=\E[X|\G]$ almost surely. Moreover, if $\tilde{Y}$ is another random variable with these properties, then $\tilde{Y}=Y$ equal almost surely. Since two versions of $\E[X|\G]$ coincide almost surely, $Y$ is also called \emph{the} conditional expectation $\E[X|\G]$.
\end{definition}

Note that when $\G=\sigma(Z)$ is the smallest $\sigma$-algebra generated by a set $Z\in\F$, we also write $\E[X|\G]=\E[X|Z]$. This generalizes to $\E[X|\G]=\E[X|Z_1,Z_2,\ldots]$ when $\G=\sigma(Z_1,Z_2,\ldots)$.

Definition~\ref{def:conexp} extends very nicely to the definition of conditional probability. Note that $\E[\1_F]=\P[F]$ holds for all $F\in\F$. This carries over to conditional probabilities.

\begin{definition}[Conditional probability]
If $\G$ is a sub-$\sigma$-algebra of $\F$ and $F\in\F$, then the \emph{conditional probability} $\P[F|\G]$ is a version of $\E[\1_F|\G]$.
\end{definition}

\subsection{Comparison to traditional definitions}
The traditional definition of traditional density-based conditional expectation and probability is the following: when $X$ and $Y$ are two random variables on $\R$, $f_{X,Y}$ is the joint density function of $X$ and $Y$ and $f_Y$ is the density function of $Y$, the conditional expectation $\E[X|Y=y]$ is defined as
\begin{equation}
\E[X|Y=y]=\int_\R x\frac{f_{X,Y}(x,y)}{f_Y(y)}dx.
\end{equation}
Definition~\ref{def:conexp} agrees on this traditional usage although it can be modified on a set of measure zero, as we will see now. Take $\G=\sigma(\{Y=y\})$, the smallest $\sigma$-algebra generated by all $\omega\in\R$ such that $Y(\omega)=y$, and define
\begin{equation}
g(y)=\int_\R x\frac{f_{X,Y}(x,y)}{f_Y(y)}dx,
\end{equation}
then $g(Y)$ is a version of $\E[X|Y]$. The proof and a more general statement can be found in section~9.6 of \cite{Williams91}.

The traditional conditional expectation can also be extended to traditional conditional probability.
\begin{definition}[Traditional conditional probability]
Let $(\Omega,\F, \P)$ be a probability space. Let $F\in\F$ be with positive measure and let $E\in\F$ be measurable. The \emph{traditional conditional probability} of $E$ given $F$ is given by
\begin{equation}
\P[E|F]=\frac{\P[E\cap F]}{\P[F]}.
\end{equation}
If $F$ has measure zero, the traditional conditional probability of $E$ given $F$ is undefined.
\end{definition}

\section{Formal description of the probability space}
\begin{figure}
\centering{
\input{figures/BorelTransform.tikz}
}
\caption{A visualization of coordinate transformation~\eqref{eq:BorelPolar} from Euclidean to spherical coordinates. The rectangle represents the set $S$. The horizontal lines in the rectangle like $C$ parametrize the latitudes. The vertical lines in the rectangle like $M$ parametrize the meridians. Note that the two blue vertical lines of $M$ are distance $\pi$ apart, this is needed to fully parametrize a meridian.}
\label{fig:BorelVis}
\end{figure}

To analyse the Borel-Kolmogorov paradox, we need to formalize our probability space. The following construction is visualized in Figure~\ref{fig:BorelVis}. Let $S$ be the unit sphere. We equip $S$ with polar coordinates to ease some calculations. Define $S=[0,2\pi)\times[0,\pi]$, then $S$ is described in the Euclidean space with the function
\begin{equation}\label{eq:BorelPolar}
S\to\R^3:(\phi,\psi)\mapsto(\cos\phi\sin\psi,\sin\phi\sin\psi,\cos\psi).
\end{equation}
Let $\B=\B(S)$ be the Borel-$\sigma$-algebra on $S$. The uniform distribution on $S$ is defined as
\begin{equation}
\P[B]=\frac{1}{4\pi}\iint_B\sin\psi d\psi d\phi
\end{equation}
for a $B\in\B$. The triple $(S,\B,\P)$ forms a probability space.

The set of latitudes is described by
\begin{equation}
\mathcal{C}=\{[0,2\pi)\times\{\psi\}\mid\psi\in[0,\pi]\}
\end{equation}
and will be horizontal lines in the rectangle of Figure~\ref{fig:BorelVis}. The set of meridians is described by
\begin{equation}
\mathcal{M}=\{\{\phi,\phi+\pi\}\times[0,\pi]\mid\phi\in[0,\pi)\}
\end{equation}
and will be the vertical lines in the rectangle of Figure~\ref{fig:BorelVis}.

Note that the function in \eqref{eq:BorelPolar} is not a bijection. The image of $S$ is not one-to-one on the north and south pole, but this is a null set and will not cause any problems in our case. The reason why \eqref{eq:BorelPolar} is not made a formal bijection is that it eases notation.

\section{Conditional distributions}
We will now explore various ways to describe the conditional distribution on a great circle.
\subsection{Traditional conditional probability}\label{sec:BorelNaive}
The first method is the naive method using traditional conditional probability. The probability space here is $(S,\B,\P)$. Let $F\in\B$ be a great circle. Note that $\P[F]=0$, as a great circle always has zero measure.

To be precise, let $f\colon S\to\R^3$ be the coordinate transformation of \eqref{eq:BorelPolar}. All great circles $F'\in\B$ have a rotation $O\colon\R^3\to\R^3$ such that $(f^{-1}\circ O\circ f)(F')=[0,2\pi)\times\left\{\frac{\pi}{2}\right\}$. Rotations are orthogonal and have determinant $1$, thus
\begin{align}
\P[F]&=\frac{1}{4\pi}\iint_F\sin\psi d\psi d\phi\\
&=\frac{1}{4\pi}\int_0^{2\pi}\int_{\frac{1}{2}\pi}^{\frac{1}{2}\pi}\sin(g(\psi)) \det(f^{-1}\circ O\circ f)d\psi d\phi\\
&=\frac{1}{4\pi}\int_0^{2\pi}\int_{\frac{1}{2}\pi}^{\frac{1}{2}\pi}\sin(g(\psi)) d\psi d\phi=0
\end{align}
where $g$ is a function needed to correctly perform the change of coordinates.
The inverse $f^{-1}$ is well-defined here since $f$ is a bijection locally around the circle $[0,2\pi)\times\left\{\frac{\pi}{2}\right\}$. This proves that any great circle has zero measure.

Let $E\subset F$ be a measurable subset of $F$, then the traditional conditional probability of $E$ given $F$ equals
\begin{equation}
\P[E|F]=\frac{\P[E\cap F]}{\P[F]},
\end{equation}
which is not defined as $\P[F]=0$.

Therefore, the traditional interpretation of conditional probability will not give an answer. Furthermore, a single great circle does not yield enough information to compute any conditional probability. In terms of Definition~\ref{def:conexp} the sub-$\sigma$-algebra considered here is $\G=\{\emptyset,S,B,S\setminus B\}$, which turns out to be too small.

\subsection{Conditional probability on latitudes}\label{sec:BorelLong}
Since four-element sub-$\sigma$-algebras are too small, we need to condition on larger sub-$\sigma$-algebras. One option is the $\sigma$-algebra of latitudes. Let
\begin{equation}
\mathfrak{C}=\sigma\left(\left\{[0,2\pi)\times A\mid A\in\B([0,\pi])\right\}\right)
\end{equation}
be the $\sigma$-algebra of all measurable subsets of latitudes. We can then calculate the conditional expectation of $\E[X|\mathfrak{C}]$.

\begin{proposition}
Let $X$ be $\B$-measurable. The conditional expectation of $\E[X|\mathfrak{C}]$ is given by
\begin{equation}
\E[X|\mathfrak{C}](\phi,\psi)=\frac{1}{2\pi}\int_0^{2\pi}X(\phi',\psi)d\phi'
\end{equation}
with $(\phi,\psi)\in S$.
\end{proposition}
\begin{proof}
The proof is taken from \cite{Gyenis17}. Take $A\in\B([0,2\pi))$ and consider $C=[0,2\pi)\times A\in\mathfrak{C}$. Since $\P$ is the uniform measure on the surface of the unit sphere, we have
\begin{equation}
\int_C Xd\P=\frac{1}{4\pi}\int_A\int_0^{2\pi}X(\phi,\psi)\sin\psi d\phi d\psi
\end{equation}
by the standard spherical to Euclidean coordinate transformation. We can now apply the same coordinate transformation on the integral of $\E[X|\mathfrak{C}]$:
\begin{equation}
\int_C\E[X|\mathfrak{C}]d\P=\frac{1}{4\pi}\int_A\int_0^{2\pi}\E[X|\mathfrak{C}](\phi,\psi)\sin\psi d\phi d\psi.
\end{equation}
Filling in $\E[X|\mathfrak{C}]$ and rewriting yields
\begin{align}
\int_C\E[X|\mathfrak{C}]d\P&=\frac{1}{4\pi}\int_A\int_0^{2\pi}\E[X|\mathfrak{C}](\phi,\psi)\sin\psi d\phi d\psi\\
&=\frac{1}{4\pi}\int_A\int_0^{2\pi}\frac{1}{2\pi}\left(\int_0^{2\pi}X(\phi',\psi)d\phi'\right)\sin\psi d\phi d\psi\\
&=\frac{1}{4\pi}\int_A\left(\int_0^{2\pi}\frac{1}{2\pi}d\phi\right)\int_0^{2\pi} X(\phi',\psi)\sin\psi d\phi' d\psi\\
&=\frac{1}{4\pi}\int_A\int_0^{2\pi}X(\phi,\psi)\sin\psi d\phi d\psi=\int_C Xd\P.
\end{align}
Note that $\mathfrak{C}$ is generated by sets like $C$, thus $\E[X|\mathfrak{C}]$ is a well-defined version of the $\mathfrak{C}$-conditional expectation of $X$.
\end{proof}

This derivation is slightly different from the one in \cite{Gyenis17}. The corrections on \cite{Gyenis17} are given in Appendix~\ref{app:CorLong}.

As we now have a $\mathfrak{C}$-conditional expectation, we can look at the measure space $(S,\mathfrak{C})$. Let $\psi'\in(0,\pi)$ be arbitrary, then from $\mathfrak{C}$ we can take a latitude $C=[0,2\pi)\times\{\psi'\}$ and a measurable arc $A=[\phi_1,\phi_2]\times\{\psi'\}\subset C$. Let $\P'$ be the probability measure taking $1$ on $C$ and $0$ on $S\setminus C$, then the conditional probability $\bar{\P}$ of points being on $A$ given there is a point on $C$ is
\begin{align}
\bar{\P}[A]&=\int_S\P[A|\mathfrak{C}]d\P'=\frac{1}{2\pi}\int_S\int_0^{2\pi} \1_A(\phi',\psi)d\phi' d\P'(\phi,\psi)\label{eq:BorelLongConProb}\\
&=\frac{1}{2\pi}\left(\int_{S\setminus C}\int_{\phi_1}^{\phi_2}0d\phi' d\P'+\int_{C}\int_{\phi_1}^{\phi_2}1d\phi' d\P' \right)\\
&=\frac{\phi_2-\phi_1}{2\pi}.
\end{align}
Thus the conditional probability distribution on $C$ is uniform, resulting in all latitudes having uniform conditional probability measure. Since the unit sphere itself has a uniform probability measure, this result is as expected.

\subsection{Conditional probability on meridians}\label{sec:BorelMer}
A great circle cannot only be described as a latitude, but also as a meridian. Therefore it is interesting whether describing great circles as meridians yield to the same conditional distribution. Let
\begin{equation}
\mathfrak{M}=\sigma(\{A\times[0,\pi]\mid A\in\B([0,2\pi))\})
\end{equation}
be the $\sigma$-algebra of measurable subsets of meridians. We can now calculate the conditional expectation $\E[X|\mathfrak{M}]$.
\begin{proposition}
Let $X$ be $\B$-measurable. The conditional expectation of $\E[X|\mathfrak{M}]$ is given by
\begin{equation}
\E[X|\mathfrak{M}](\phi,\psi)=\frac{1}{2}\int_0^\pi X(\phi,\psi')\sin\psi'd\psi'
\end{equation}
with $(\phi,\psi)\in S$.
\end{proposition}
\begin{proof}
The proof is largely taken from \cite{Gyenis17}. Let $A\in\B([0,2\pi))$ be measurable and consider $M=A\times[0,\pi]\in\mathfrak{M}$. The coordinate transformation between polar coordinates and the uniform measure on the circle $\P$ and further rewrites yield
\begin{align}
\int_M \E[X|\mathfrak{M}]d\P&=\frac{1}{4\pi}\int_0^\pi\int_A\E[X|\mathfrak{M}](\phi,\psi)\sin\psi d\phi d\psi\\
&=\frac{1}{4\pi}\int_0^\pi\int_A\left(\frac{1}{2}\int_0^{\pi}X(\phi,\psi')\sin\psi'd\psi'\right)\sin\psi d\phi d\psi\\
&=\frac{1}{4\pi}\int_A\int_0^{\pi}X(\phi,\psi')\sin\psi'd\psi'd\phi\\
&=\int_M Xd\P.
\end{align}

Note that $\mathfrak{M}$ is generated by sets like $M$, thus $\E[X|\mathfrak{M}]$ is a well-defined version of the $\mathfrak{M}$-conditional expectation of $X$.
\end{proof}

Note that this representation is different from equations (81) and (112) of \cite{Gyenis17}, where equation 81 is
\begin{equation}
\E[X|\mathfrak{M}](\phi,\psi)=\frac{1}{2}\int_0^{2\pi}X(\phi,\psi')|\sin\psi'|d\psi'
\end{equation}
and equation 112 is
\begin{equation}
\E[X|\mathfrak{M}](\phi,\psi)=\int_0^{\pi}X(\phi,\psi')\sin\psi'd\psi'.
\end{equation}
The corrections on \cite{Gyenis17} are given in Appendix~\ref{app:CorMer}. We can verify that our version is the correct one.\\
Take first the meridian $M=\{\phi',\phi'+\pi\}\times[0,\pi]\in\mathfrak{M}$ with $\phi'\in[0,\pi)$. Let $\psi_1^*,\psi_2^*\in\R$ be arbitrary with $\psi_2^*-2\pi\leq\psi_1^*\leq\psi_2^*$, define the angles $\psi_1=\psi_1^*\mod2\pi$ and $\psi_2=\psi_2^*\mod2\pi$ and define arc $A\subseteq M$ as
\begin{equation}
A=\begin{cases}
\{\phi'\}\times[\psi_1,\psi_2],&\psi_1,\psi_2\leq\pi,\\
\{\phi'\}\times[\psi_1,\pi]\cup\{\phi'+\pi\}\times[\psi_2-\pi,\pi],&\psi_1\leq \pi, \psi_2>\pi,\\
\{\phi'+\pi\}\times[\psi_1-\pi,\psi_2-\pi],&\psi_1,\psi_2>\pi,\\
\{\phi'\}\times[0,\psi_1]\cup\{\phi'+\pi\}\times[0,\psi_2-\pi],&\psi_2\leq\pi,\psi_1>\pi.
\end{cases}
\end{equation}
This definition is exhaustive, yet it provides all possible arcs on a meridian while restricting ourselves to the domain $[0,2\pi)\times[0,\pi]$. Now, analogous to the latitudes, let $\P'$ be the uniform measure taking $1$ on meridian $M$ and $0$ on $S\setminus M$. The conditional probability of $\hat{\P}$ of points being on $A$ with $\psi_1,\psi_2\leq\pi$ given there is a point on $M$ is given by
\begin{align}
\hat{\P}[A]&=\int_S\P[A|\mathfrak{M}]d\P'=\frac{1}{2}\int_S\int_0^\pi\1_A(\phi,\psi')\sin\psi'd\psi'd\P'(\phi,\psi)\\
&=\frac{1}{2}\left(\int_{S\setminus M}\int_{\phi_1}^{\phi_2}0d\psi'd\P'+\int_{M}\1_{\{\phi'\}\times[0,\pi]}\int_{\phi_1}^{\phi_2}\sin\psi'd\psi'd\P'\right)\\
&=\frac{1}{4}\int_{\phi_1}^{\phi_2}\sin\psi'd\psi'd\P'=\frac{\cos\psi_1-\cos\psi_2}{4}
\end{align}
since $\int_{\{\phi'\}\times[0,\pi]}d\P'=\frac{1}{2}$. On the other possible arcs of $M$ the probability $\hat{\P}[A]$ with $\psi_1,\psi_2$ as in the definition of $A$ becomes
\begin{equation}\label{eq:BorelMerConProb}
\hat{\P}[A]=\begin{cases}
\frac{1}{4}\left(\cos\psi_1-\cos\psi_2\right),&\psi_1,\psi_2\leq\pi,\\
\frac{1}{4}\left(2+\cos\psi_1-\cos\psi_2\right),&\psi_1\leq \pi, \psi_2>\pi,\\
\frac{1}{4}\left(\cos\psi_2-\cos\psi_1\right),&\psi_1,\psi_2>\pi,\\
\frac{1}{4}\left(2-\cos\psi_1+\cos\psi_2\right),&\psi_2\leq\pi,\psi_1>\pi.
\end{cases}
\end{equation}
Now one can immediately check that $\hat{\P}$ is well-defined on $M$ as
\begin{equation}
\hat{\P}[M]=\int_S\P[A|\mathfrak{M}]d\P'=\frac{1}{2}\int_M\int_0^\pi\sin\psi'd\psi'd\P'=-\frac{1}{2}\left(\cos\pi-\cos0\right)=1.
\end{equation}

Clearly this conditional distribution on meridians is not uniform. However, one should expect they are, as a meridian is just a rotation of a latitude and the points on the sphere are spread uniformly. An explanation of this difference is given in Section~\ref{sec:BorelExplained}.

\subsection{Combining latitudes and meridians}\label{sec:BorelCombining}
Another question one could ask is the following: if I define $\F=\sigma(\mathfrak{C},\mathfrak{M})$ as the smallest $\sigma$-algebra containing both measurable subsets of meridians and latitudes, what is then the conditional probability distribution on a great circle given $\F$? The answer is that in this approach the distributions in sections~\ref{sec:BorelLong} and~\ref{sec:BorelMer} can be recovered as limiting distributions of a sequence of traditional conditional probabilities defined in Section~\ref{sec:BorelNaive}.

First we'll further analyse the new $\sigma$-algebra $\F$. Consider an arbitrary rectangle $(a,b)\times(c,d)\subset[0,2\pi)\times[0,\pi]$. Since by definition $(a,b)\times[0,\pi]\in\mathfrak{M}$ and $[0,2\pi)\times(c,d)\in\mathfrak{C}$, we have
\begin{equation}
(a,b)\times(c,d)=\left((a,b)\times[0,\pi]\right)\cap\left([0,2\pi)\times(c,d)\right)\in\F.
\end{equation}
Thus all Borel-measurable sets on our sphere are contained in $\F$. Furthermore, as $\mathfrak{C}$ and $\mathfrak{M}$ contain only Borel-measurable sets, $\F=\sigma(\mathfrak{C},\mathfrak{M})$ can only have Borel-measurable sets. Therefore $\F=\B$ is the $\sigma$-algebra of Borel-measurable sets on the sphere.

Now take the following approach. Pick $x\in(0,\pi)$. Define the two rectangles $C_\epsilon=[0,2\pi)\times[x-\epsilon,x+\epsilon]$ and $R_\epsilon=[a,b]\times[x-\epsilon,x+\epsilon]\subseteq C_\epsilon$ for all ${\epsilon\in(0,\min\{\pi-x,x\})}$, then what is the limiting conditional probability of $R_0$ given $C_0$? Note that $C_0$ becomes a latitude and $R_0$ an arc on $C_0$. The conditional probability of $R_0$ given $C_0$ can be modelled as the limit of the sequence $\P[R_\epsilon|C_\epsilon]$ with $\epsilon\to0$, which will take on the uniform distribution on $C_0$ as calculated in Equation~\ref{eq:BorelLongConProb}.
\begin{proposition}\label{prop:BorelLongBayes}
Define the two rectangles $C_\epsilon=[0,2\pi)\times[x-\epsilon,x+\epsilon]$ and $R_\epsilon=[a,b]\times[x-\epsilon,x+\epsilon]\subset C_\epsilon$ for all $\epsilon\in(0,\min\{\pi-x,x\})$ with $x\in(0,\pi)$. The limiting conditional probability of $R_0$ given $C_0$ equals
\begin{equation}
\P[R_0|C_0]=\lim_{\epsilon\downarrow0}\P[R_\epsilon|C_\epsilon]=\frac{b-a}{2\pi}.
\end{equation}
\end{proposition}
\begin{proof}
Let $\epsilon\in(0,\min\{\pi-x,x\})$. Let $C_\epsilon=[0,2\pi)\times[x-\epsilon,x+\epsilon]\in\mathfrak{C}$. The probability $\P[R_\epsilon|C_\epsilon]$ equals
\begin{equation}
\P[R_\epsilon|C_\epsilon]=\frac{\P[R_\epsilon\cap C_\epsilon]}{\P[C_\epsilon]}=\frac{\int_{a}^{b}\int_{x-\epsilon}^{x+\epsilon}\sin\psi d\psi d\phi}{\int_{0}^{2\pi}\int_{x-\epsilon}^{x+\epsilon}\sin\psi d\psi d\phi}=\frac{b-a}{2\pi}.
\end{equation}
Thus $\P[R_\epsilon|C_\epsilon]=\frac{b-a}{2\pi}$ holds for all possible $\epsilon$, resulting in the limiting value
\begin{equation}
\P[R_0|C_0]=\lim_{\epsilon\downarrow0}\P[R_\epsilon|C_\epsilon]=\frac{b-a}{2\pi}.
\end{equation}
\end{proof}

A same proposition can be found for the meridians.

\begin{proposition}\label{prop:BorelMerBayes}
Let $x\in(0,2\pi)$ and $\epsilon\in(0,\min\{2\pi-x,x\})$. Consider $M_\epsilon=[x-\epsilon,x+\epsilon]\times[0,\pi]$ and $R_\epsilon=[x-\epsilon,x+\epsilon]\times[a,b]\subseteq M_\epsilon$, then the limiting conditional probability of $R_0$ given $M_0$ is
\begin{equation}
\P[R_0|M_0]=\lim_{\epsilon\downarrow0}\P[R_\epsilon|M_\epsilon]=\frac{1}{2}(\cos a-\cos b)
\end{equation}
\end{proposition}
\begin{proof}
The conditional probability $\P[R_\epsilon|M_\epsilon]$ equals
\begin{equation}
\P[R_\epsilon|M_\epsilon]=\frac{\P[R_\epsilon\cap M_\epsilon]}{\P[M_\epsilon]}=\frac{\int_{x-\epsilon}^{x+\epsilon}\int_a^b\sin\psi d\psi d\phi}{\int_{x-\epsilon}^{x+\epsilon}\int_0^{\pi}\sin\psi d\psi d\phi}=\frac{1}{2}\left(\cos a-\cos b\right).
\end{equation}
The limit of $\epsilon\downarrow0$ remains $\P[R_0|M_0]=\frac{1}{2}\left(\cos a-\cos b\right)$.
\end{proof}

Note that the probability in Proposition~\ref{prop:BorelMerBayes} has $\frac{1}{2}$ as normalization factor instead of $\frac{1}{4}$, as $M_\epsilon$ converges to only a half meridian. By symmetry the probability density function of the distribution on the whole meridian follows the absolute sine function.

The conditional probabilities on $\mathfrak{C}$ and $\mathfrak{M}$ therefore can be approached by traditional conditional probabilities on $\B$. This gives rise to the following conjecture, conjuring whether this is possible for all conditional probability distributions on sets of measure zero once a sub-$\sigma$-algebra is specified.
\begin{conjecture}\label{con:BorelConjecture}
Let $(X,\F,\P)$ be a probability space. Let $F\in\mathcal{F}$ have zero measure and let $E\subseteq F$ be measurable. Let $\G\subset\F$ be a sub-$\sigma$-algebra containing $E$ and $F$ with $\G\neq\F$. Let $\{F_n\}_{n\in\N}\subset\G$ be a sequence converging to $F$ and $\{E_n\}_{n\in\N}\subset\G$ be a sequence converging to $E$ with for all $n\in\N$ the sets $E_n$ and $F_n$ have positive measure, $E_n\subseteq F_n$ holds and the limit $\lim_{n\to\infty}\P[E_n|F_n]$ exists. Then
\begin{equation}
\P[E|F]:=\lim_{n\to\infty}\P[E_n|F_n]=\int_X\E[\1_F|\mathcal{G}]d\P'(E)
\end{equation}
where $\P'$ is the uniform probability measure taking $1$ on $F$ and $\E[\cdot|\G]$ is a version of the conditional expectation on $\G$.
\end{conjecture}

If Conjecture~\ref{con:BorelConjecture} is true, then all conditional probabilities on null sets can be computed by limiting traditional conditional probabilities. It seems to hold as for the Borel-Kolmogorov paradox when considering latitudes we can take $\F=\B$ and $\G=\mathfrak{C}$ and when considering meridians we can take $\F=\B$ and $\G=\mathfrak{M}$. It however is false as the following proposition will point out.

\begin{proposition}\label{prop:BorelConFalse}
Conjecture~\ref{con:BorelConjecture} is false.
\end{proposition}
\begin{proof}
The proof can be found in Appendix~\ref{app:ProofBorelConFalse}.
\end{proof}

A proposition like Proposition~\ref{prop:BorelConFalse} was already proven by Hájek \cite{Hajek03} and Jaynes \cite{Jaynes03}. Hájek argues that conditional probabilities on sets of measure zero cannot be approached by infinitesimals. This approach needs an entirely new model and the axiom of choice supports the existence of many models. Therefore there is no unique method to calculate conditional probabilities on a set of measure zero. Jaynes adds that a circle cannot be described uniquely and every method of describing a measure null set with a different limiting approach yields to a different result.

This demonstrates that when having an $F\in\F$ with zero measure, one cannot uniquely define $\P[E|F]$. A conditional expectation needs to be accompanied by a sufficiently large sub-$\sigma$-algebra, otherwise contradictory results can arise.

The idea underlying Conjecture~\ref{con:BorelConjecture} is however useful for guessing a version of the conditional expectation. The measure-theoretic definition of conditional expectation is only a list of properties, not a constructive definition. In the case of a $\mathfrak{C}$-conditional expectation, looking at Proposition~\ref{prop:BorelLongBayes} and its proof we observe that the resulting probability is $\frac{b-a}{2\pi}$, thus uniform. Furthermore, $\phi$ is the only variable that is integrated, thus one can guess $\phi$ is the only needed variable. Therefore one can suspect that
\begin{equation}
\E\left[\1_{[a,b]\times\left\{\frac{\pi}{2}\right\}}\middle|\mathfrak{C}\right](\phi,\psi)=\frac{1}{2\pi}\int_0^{2\pi}\1_{[a,b]\times\left\{\frac{\pi}{2}\right\}}(\phi',\psi)d\phi'=\frac{b-a}{2\pi}\1_{\left\{\frac{\pi}{2}\right\}}(\psi)
\end{equation}
holds and the guess can be generalized to
\begin{equation}
\E[X|\mathfrak{C}](\phi,\psi)=\frac{1}{2\pi}\int_0^{2\pi}X(\phi',\psi)d\phi'.
\end{equation}
This is in fact the same $\mathfrak{C}$-conditional expectation of $X$ by Proposition~\ref{prop:BorelLongBayes}. The same steps can be taken to guess $\E[X|\mathfrak{M}]$ using Proposition~\ref{prop:BorelMerBayes}.

Thus, we can conclude that it is wise to look at a converging sequence of traditional conditional probabilities for a potential conditional expectation when conditioning on null sets. However, it must be noted that the resulting conditional probability is not uniquely defined, must be checked whether it complies to the actual definition of conditional probability and depends on the chosen sub-$\sigma$-algebra.

\section{The Borel-Kolmogorov paradox explained}\label{sec:BorelExplained}
Following the arguments of \cite{Gyenis17} we will argue that the difference in conditional distribution is no paradox, but a misinterpretation of conditional probability.

A first explanation is quite intuitive. All meridians intersect at the north and south pole of the sphere, whereas latitudes do not intersect. Therefore the $\sigma$-algebras $\mathfrak{M}$ and $\mathfrak{C}$ are vastly different. If points are spread uniformly on the meridians, the distribution of mass on the whole sphere will not be uniform and the density of mass will be highest at the poles. This is simulated by \cite{Weisstein} and his simulations support this statement.

Let $\bar{\P}$ be the conditional probability measure on a latitude, defined by Equation~\ref{eq:BorelLongConProb} and let $\hat{\P}$ be the conditional probability measure on a meridian, defined by Equation~\ref{eq:BorelMerConProb}. If the spaces $(S, \mathfrak{C}, \bar{\P})$ and $(S, \mathfrak{M}, \hat{\P})$ are isomorphic with a measurable bijection, then $\hat{\P}$ and $\bar{\P}$ must have equal distribution and the results of sections~\ref{sec:BorelLong} and \ref{sec:BorelMer} must be paradoxical and contradictory. If there is a measurable bijection from a meridian to a latitude, their conditional probability distributions must agree. This is not the case here.
\begin{theorem}[\cite{Gyenis17}]
Let $f\colon S\to S$ be a measurable bijection with measurable inverse. There is no Boolean algebra isomorphism $h_f\colon\mathfrak{C}\to\mathfrak{M}$ that is determined by $f$.
\end{theorem}
\begin{proof}
The proof can be found in \cite{Gyenis17}, but we will repeat it here. Suppose such isomorphism $h_f$ exists. All latitudes $C$ in $\mathfrak{C}$ are the only atoms of $\mathfrak{C}$, therefore $h_f(C)$ are the only atoms of $\mathfrak{M}$ as well, making $h_f(C)$ meridians. We will now prove that $h_f(C)$ cannot be a meridian.

Let $m_0=\{(0,0), (0,\pi)\}$ be the set of north and south poles. Let $c_0\in\mathfrak{C}$ be such that $h_f(c_0)=m_0$. Then $c_0$ consists of two elements as well, thus we can choose latitude $C$ such that $C\cap c_0=\emptyset$ and $C$ not being the north or south pole. Note that $h_f(C)$ is a meridian, thus $m_0\subset h_f(C)$ as all meridians pass through the north and south poles. Since $h_f$ is a Boolean algebra isomorphism, we have $h_f(C\cap c_0)=h_f(C)\cap h_f(c_0)$. This all can be combined in the following line:
\begin{equation}
\emptyset=h_f(\emptyset)=h_f(C\cap c_0)=h_f(C)\cap h_f(c_0)=h_f(C)\cap m_0=m_0.
\end{equation}
This is clearly a contradiction.

Let $C$ be the north or south pole of the sphere. Since $h_f$ is a bijection, $h_f(C)$ is only a single point as well. Thus, $h_f(C)$ cannot be an atom of $\mathfrak{M}$. Therefore there is a contradiction here as well.

No single atom of $\mathfrak{C}$ becomes an atom of $\mathfrak{M}$ by applying $h_f$ and therefore function $h_f$ cannot exist.
\end{proof}

Therefore every measurable bijection between $(S,\mathfrak{C})$ and $(S,\mathfrak{M})$ has no Boolean algebra isomorphism $h_f\colon\mathfrak{C}\to\mathfrak{M}$ such that latitudes in $\mathfrak{C}$ are mapped to meridians on $\mathfrak{M}$. This theorem holds on all subalgebras of $\mathfrak{C}$ and $\mathfrak{M}$ as well, as is proven by \cite{Gyenis17}.

Now it should be clear why the conditional distribution on the latitudes and meridians of the sphere differ, since we are dealing with two entirely different structured spaces. Therefore, the question `why is the conditional distribution on the latitudes different from the conditional distribution on the meridians' is easy to answer: the probability spaces differ. To verify that two different methods of conditioning on the same set yield to the correct conditional probability, one must ask the following question: if one models a distribution on the sample space using conditional expectation on a sub-$\sigma$-algebra, does the resulting conditional probability distribution extend to the original distribution on the whole sample space? As earlier mentioned, Weisstein \cite{Weisstein} demonstrated that the answer to that question is yes. The uniform distribution on latitudes and the cosine on meridians both extend to the uniform distribution on the whole sphere. Thus, when dealing with different conditional distributions on measure zero sets, one should rather ask a question like the last one, as the answer to that question must be always yes.

Further reading and a more in-depth analysis can be found in \cite{Gyenis17}.

\section{Conclusion}
After analysing the Borel-Kolmogorov paradox, the first and most important conclusion we can take is that when conditioning, especially on a set $F$ of measure zero, one needs to provide the accompanying sub-$\sigma$-algebra of the event that is conditioned on. Otherwise, the Borel-Kolmogorov phenomenon appears, where one can give different sub-$\sigma$-algebras containing $F$ such that the resulting conditional probability distribution is different.

Secondly, we must accept that conditional probability on sets of measure zero are not uniquely defined. That uniqueness only appears when conditioning on sets of positive measure, as then the classical rule $\P[E|F]=\frac{\P[E\cap F]}{\P[F]}$ gives a version of the conditional probability. If a sub-$\sigma$-algebra can be used to model the entire sample space, for example the sub-$\sigma$-algebra of latitudes with the sphere as sample space, then the conditional probability must be able to extend to the original probability distribution. For example providing the uniform distribution on all latitudes gives back the uniform distribution on the sphere, as well as applying the cosine distribution on the meridians results into the uniform distribution on the sphere.

Thirdly, a conditional distribution on a null set must be calculated using the measure-theoretic conditional expectation. Conjecture~\ref{con:BorelConjecture} can help with finding a correct version of the conditional expectation, however Proposition~\ref{prop:BorelConFalse} proves that the resulting conditional probabilities do not need to be unique. Therefore Conjecture~\ref{con:BorelConjecture} cannot be used to calculate a conditional probability.

Lastly, as Kolmogorov \cite{Kolmogorov33} already stated in his work, there is nothing paradoxical going on in the Borel-Kolmogorov paradox. The space of the sphere with the latitudes as $\sigma$-algebra is not homeomorphic with the space of the sphere and the meridians as $\sigma$-algebra. The not-uniqueness of the conditional probability on a null set is thus as expected and we must give the sub-$\sigma$-algebra when conditioning on sets of measure zero.

\chapter{Safe probability}\label{chap:SafeProp}
Take a look at the following game of dice, devised by Grünwald \cite{Grunwald13}. A player and game master are present and the game master casts a fair die. He peeks at the value of the die and must either state that the value is 4 or lower or state that the value is 3 or higher. The game master is not allowed to lie. Upon hearing the game master's advice, what is the probability the die has landed on a 3?\\
To answer this question, suppose first that an alternative game is considered. Here the game master can only say that the value is either from 1 to and including 4 or that the value is 5 or 6. The question of the probability of the die landing on 3 becomes easy to answer. When 5 or 6 is revealed, the probability is 0 and when 1 to 4 is revealed, traditional conditional probability gives us that the probability is 0.25.\\
However, when either `4 or lower' or `3 or higher' is stated, the game master can choose his statement when the die lands on a 3 or 4. His strategy of choosing his statement heavily influences the probability of the die landing on 3 upon hearing the game master's statement, making that probability not uniquely defined.

What can we say about the probability of the die landing on 3? Traditional conditional probability will not suffice here. Unfortunately, measure-theoretic probability cannot be applied as well. We want to condition on sets with positive measure and the versions of conditional expectation then all almost surely coincide with traditional conditional probability, thus using measure-theoretic conditional probability will not give more insight than traditional conditional probability.

Thus, not even measure-theoretic probability gives a unique answer. Is a unique yet still meaningful answer possible nevertheless? For this, we can resort to \emph{safe probability}, introduced by Grünwald in 2018 \cite{Grunwald18}. Suppose there are two random variables, $U$ and $V$, and we want to use the distribution of $U|V$. Suppose however that there is no method of finding this distribution: you only know that it is a member of a set of joint distributions $\Pmod$. One can then create a new conditional distribution, not necessarily one from our set nor necessarily reflecting the true distribution at all, which gives a probability of $U$ given $V$ that on average performs equal for all distributions in our set $\Pmod$ in predicting $U$ given $V$. For example, one notion of safe probability is that we have a distribution $\Psafe$ such that $\E_{\P}[U]=\E_{\P}[\E_{\Psafe}[U|V]]$ holds for all $\P\in\Pmod$. Under this $\Psafe$ the expectation $\E_{\Psafe}[U|V]$ is an unbiased estimator of $U$ for all distributions $\P\in\Pmod$. Thus, no matter which strategy the game master has, distribution $\Psafe$ allows the player to estimate $U$ on average correctly. As a result, $\Psafe$ will behave like a `true' distribution and will be useful for making predictions for some, but not all prediction tasks that can be defined on a domain.

There are more notions of safety invoicing more stringent requirements on $\Psafe$. We will first introduce some notions of safe probability and how all notions are related to each other. The example of the dice game is then continued to illustrate how safe probability can be applied. The main idea is that such safe distributions can be used \emph{as if} they were true, and the world does behave as if they were true in some respects, even though they are not.

\section{Definitions}\label{sec:SafeDef}
Before we will define safe probability, we need to introduce some notation.

\begin{definition}[Range and support]
Let $S\colon\X\to\Y$ be a random variable. The \emph{range} of $S$ is
\begin{equation}
\range(S)=\{s\in\Y\mid s=S(x),x\in\X\}.
\end{equation}
Let $\P$ be a probability distribution on $\X$. The \emph{support} of $S$ under $\P$ is
\begin{equation}
\supp_{\P}(S)=\{s\in \Y\mid\P[S=s]>0\}.
\end{equation}
\end{definition}

Let $U$ and $U'$ be two random variables and suppose there is a function $f$ such that $f(U)=U'$ holds, then we will write $U\stackrel{f}{\rightsquigarrow}U'$. If a function $f$ exists such that $U\stackrel{f}{\rightsquigarrow}U'$, we will write $U\rightsquigarrow U'$. Random variable $U'$ is then called a \emph{coarsening} of $U$ or equivalently we state that $U$ determines $U'$.

Let $\Pmod$ be a set of probability measures, then $\Pmod$ is called a \emph{model}. All distributions in $\Pmod$ are candidate to be the true distribution of $U|V$.

Lastly, if we keep $U$ and $V$ as random variables, the distribution of $U$ under $\P$ is written as $\P[U]$. The distribution of $U$ given $V$ under $\P$ is written as $\P[U|V]$.

We are now able to define conditional probability. The following definitions and propositions are all taken from \cite{Grunwald18}.

\begin{definition}[Safe probabilities]\label{def:SafeProp}
Let $\Omega$ be a sample space and let $\Pmod$ be a set of distributions on $\Omega$. Let $U$ be a real-valued random variable with finite expectation and let $V$ be a random variable on $\Omega$. Let $\Psafe$ be a probability distribution on $\Omega$. Then $\Psafe$ is \emph{safe} with respect to $\Pmod$ for
\begin{itemize}
    \item $\langle U\rangle|\langle V\rangle$ if $\E_{\P}[U]=\E_{\P}[\E_{\Psafe}[U|V]]$ holds for all $\P\in\Pmod$. This $\Psafe$ is called \emph{unbiased} for $U|V$.
    \item $\langle U\rangle|[V]$ if $\E_{\P}[U]=\E_{\Psafe}[U|V=v]$ holds for all $v\in\supp_{\Psafe}(V)$. This $\Psafe$ is called \emph{marginally valid} for $\langle U\rangle|V$.
\end{itemize}
\end{definition}
Since in almost all situations we only consider a single set $\Pmod$, we will from now on omit the statement `with respect to $\Pmod$'. When we say that $\Psafe$ is safe for $\langle U\rangle|[V]$, we actually mean that `for all $\P\in\Pmod$ we have $\supp_{\P}(V)\subseteq\supp_{\Psafe}(V)$, $\E_{\P}[U]$ is well-defined and finite, $\E_{\Psafe}[U|V=v]$ is well-defined for all $v\in\supp_{\Psafe}(V)$ and both expectations are equal'. This statement is quite lengthy, therefore we will abbreviate it by stating that $\Psafe$ is safe for $\langle U\rangle |[V]$. Such abbreviations are performed for other notions of safety as well.

\begin{definition}[Stronger safety]\label{def:SafeStrongProp}
Let $\Omega,\Pmod,U,V$ and $\Psafe$ as above. Then $\Psafe$ is \emph{safe} for
\begin{itemize}
    \item $U|\langle V\rangle$ if for all real-valued random variables $U'$ on $\Omega$ with $U\rightsquigarrow U'$ distribution $\Psafe$ is safe for $\langle U'\rangle|\langle V\rangle$.
    \item $U|[V]$ if for all real-valued random variables $U'$ on $\Omega$ with $U\rightsquigarrow U'$ distribution $\Psafe$ is safe for $\langle U'\rangle|[V]$.
\end{itemize}
\end{definition}

Both definitions introduce a lot of new notation. In practice, $\cdot|\langle V\rangle$ implies that $\E_{\Psafe}[U|V]$ must be an unbiased estimator for $U$ for all $\P\in\Pmod$.\\
When $\cdot|[V]$ is considered, the expected value of $U|V=v$ under $\Psafe$ must equal the expected value of $U$ under $\P$ for all $v\in\supp_{\P}(V)$ and $\P\in\Pmod$. In other words, the outcome of $V$ will not influence the conditional expectation value of $U$ under $\Psafe$ and this expectation will equal the expectation of $U$ under all $\P\in\Pmod$. Note that when $\E_{\P}[U]$ does not remain constant for all $\P\in\Pmod$, safety for $\cdot|[V]$ is immediately impossible.\\
Lastly, the difference between $\langle U\rangle|\cdot$ and $U|\cdot$ is that for $\langle U\rangle|\cdot$ the requirement for safety does only have to be met for $U$. When $U|\cdot$ is considered, the requirement must be met for all coarsenings $U'$ of $U$.

Definition~\ref{def:SafeStrongProp} is not easy to work with as constructing a $\Psafe$ that is safe for $U|[V]$ requires checking $\Psafe$ against all coarsenings $U'$ of $U$. The following proposition from \cite{Grunwald18} gives us tools to actually create a safe $\Psafe$ for $U|\langle V\rangle$, $U|[V]$ and a few extra notions.

\begin{proposition}\label{prop:SafeProperties}
Let $\Omega$, $\Pmod$, $U$, $V$ and $\Psafe$ be as above. Then
\begin{enumerate}
    \item $\Psafe$ is safe for $U|\langle V\rangle$ if and only if for all $\P\in\Pmod$ there is a distribution $\P'$ on $\Omega$ with $\P'[U=u,V=v]=\Psafe[U=u|V=v]\P[V=v]$ for all $(u,v)\in\range((U,V))$ such that $\P'[U]=\P[U]$.
    \item $\Psafe$ is safe for $\langle U\rangle|V$ if and only if
    \begin{equation}
    \E_{\P}[U|V]=\E_{\Psafe}[U|V]
    \end{equation}
    holds with probability $1$ for all $\P\in\Pmod$. This $\Psafe$ is called \emph{squared error-optimal} for $U|V$.
    \item $\Psafe$ is safe for $U|V$ if and only if
    \begin{equation}
    \P[U|V]=\Psafe[U|V]
    \end{equation}
    holds with probability $1$ for all $\P\in\Pmod$. This $\Psafe$ is called \emph{valid} for $U|V$.
    \item $\Psafe$ is safe for $U|[V]$ if and only if $\P[U]=\Psafe[U|V=v]$ for all $\P\in\Pmod$ and $v\in\supp_{\Psafe}(V)$.
\end{enumerate}
\end{proposition}
\begin{proof}
The proof is in \cite{Grunwald16}, section A.2.
\end{proof}

When a $\Psafe$ is safe for $U|V$, then all probability measures in $\Pmod$ have an almost surely equal distribution on $U|V$. In that case we should not bother ourselves with safe probability and just use that one distribution from $\Pmod$ as our probability. Furthermore, if $\Pmod=\{\P\}$ consists of only a single probability measure, then $\P$ is automatically safe for $U|V$. Therefore, unless one is very sure that some specific $\P$ is true, it is wise to make the model $\Pmod$ large enough when considering safe probabilities.

It is easy to see that safety for $\langle U\rangle|[V]$ implies safety for $\langle U\rangle|\langle V\rangle$. If we have $\E_{\Psafe}[U|V=v]=\E_{\P}[U]$ for all $v\in\supp_{\Psafe}(V)$, then
\begin{equation}
\E_{\P}[\E_{\Psafe}[U|V]]=\E_{\P}[\E_{\P}[U]]=\E_{\P}[U]
\end{equation}
holds by the projective properties of the expectation value. All implications are stated in the following proposition.

\begin{figure}
\centering{
\input{figures/SafeDiagram.tikz}
}
\caption{A diagram of implications between safe probabilities. If $\Psafe$ is safe for X, then it is safe for all statements that can be reached from X as well.}
\label{fig:SafeDiagram}
\end{figure}

\begin{proposition}\label{prop:SafeImply}
Let $U$, $V$ and $\Psafe$ be as above.
\begin{enumerate}
\item If $\Psafe$ is safe for $U|V$, it is safe for $\langle U\rangle|V$ and $U|[V]$.
\item If $\Psafe$ is safe for $U|[V]$, it is safe for $U|\langle V\rangle$ and $\langle U\rangle|[V]$.
\item If $\Psafe$ is safe for $\langle U\rangle|V$, $U|\langle V\rangle$ or $\langle U\rangle|[V]$, it is safe for $\langle U\rangle|\langle V\rangle$.
\end{enumerate}
All relations are visualised in Figure~\ref{fig:SafeDiagram}.
\end{proposition}
\begin{proof}
All implications are found and proven in \cite{Grunwald18}.
\end{proof}

\section{The dice game}\label{sec:SafeDice}
Recall the dice game from the introduction of this chapter, introduced by \cite{Grunwald13}. The game master casts a fair die and observes a value. This value is either in the set $\{1,2,3,4\}$ or in the set $\{3,4,5,6\}$. The game master must reveal one of the two sets to the player and cannot lie. The player must then guess the probability that the die has landed on 3.

It is insufficient to take $\Omega=\{1,2,3,4,5,6\}$ as our sample space. Let $U$ be a random variable denoting the die's value and let $V$ be the statement of the game master, then $\P[U=3|V=\{1,2,3,4\}]$ cannot be calculated. When the die has rolled to a $3$, the game master can choose between $\{1,2,3,4\}$ and $\{3,4,5,6\}$ and this must be taken into account. The smallest $\sigma$-algebra containing both $\{1,2,3,4\}$ and $\{3,4,5,6\}$ is
\begin{equation}
\G=\sigma(\{1,2\},\{3,4\},\{5,6\}),
\end{equation}
thus $\{3\}$ is no member of $\G$ as $\G$ cannot have sets of odd size. We can add $\{3\}$ to $\G$, however we still do need the probability of $\{\{1,2,3,4\},\{3,4,5,6\}\}$. Therefore we cannot condition in $\Omega$ to compute $U=3$ given $V=\{1,2,3,4\}$.

We need to extend our sample space. We will use the extension introduced in \cite{Grunwald13}. Take $\X=\{1,2,3,4,5,6\}$ and $\Y=\{\{1,2,3,4\},\{3,4,5,6\}\}$ and let our sample space be $\Omega=\X\times\Y$. Let $U$ be an $\X$-valued random variable and let $V$ be a $\Y$-valued random variable. In this space conditioning on $\{1,2,3,4\}$ is valid, as $\G=\sigma(\{3\}\times Y)$ is a strict sub-$\sigma$-algebra of the power set $2^\Omega$ for all subsets $Y\subseteq\Y$.
\subsection{Using traditional and measure-theoretic conditional probability}
First we will try to solve the problem using traditional probability. Here our sample space is $\Omega'=\{1,2,3,4,5,6\}$ with random variable $U'$. We are interested in $\P[U'=3|U'\in\{1,2,3,4\}]$. Bayes' rule delivers the answer
\begin{align}
\P[U'=3|U'\in\{1,2,3,4\}]&=\frac{\P[U'\in\{1,2,3,4\}|U'=3]\P[U'=3]}{\sum_{u\in\Omega'}\P[U'\in\{1,2,3,4\}|U'=u]\P[U'=u]}\\
&=\frac{1\cdot\frac{1}{6}}{4\cdot1\cdot\frac{1}{6}+2\cdot0\cdot\frac{1}{6}}=\frac{1}{4},
\end{align}
which is the answer most people initially will give \cite{Grunwald13}. However, as stated earlier, $U'=3$ is also possible when $U'\in\{3,4,5,6\}$, thus the conditioning is not executed on a partition of sample space $\Omega'$. Therefore, traditional conditional probability is not allowed and the answer of $\frac{1}{4}$ is not correctly obtained.

Take now the viewpoint of measure-theoretic conditional probability with $\Omega$ as our sample space. Let $\P$ be a probability measure with the uniform distribution on $U$. To model the strategy of the game master, we need $p,q\in[0,1]$ such that
\begin{align}
\P[V=\{3,4,5,6\}\mid U=3]&=p,&\P[V=\{3,4,5,6\}\mid U=4]&=q.
\end{align}
We can now for example calculate the probability of the die landing on $6$ after the game master reveals $\{3,4,5,6\}$. This calculation can be found in equation~2 of \cite{Grunwald13} and the probability becomes
\begin{equation}\label{eq:SafeDiceCon6}
\P[U=6\mid V=\{3,4,5,6\}]=\frac{1}{p+q+2}.
\end{equation}
Therefore, the conditional probability of rolling $6$ ranges from $\frac{1}{4}$ to $\frac{1}{2}$ depending on the game master's strategy. This should not suffice as an answer, as the player does not know the game master's strategy. Measure-theoretic conditional probability will thus also not be of any help.

\subsection{Using safe probabilities}
Can we create a distribution $\Psafe$ such that the probability of $U=3$ given $V$ is independent of $p$ and $q$ and behaves on average like the dice? The answer is yes and we need to use safe probability. Consider as model
\begin{equation}
\Pmod=\left\{\P\middle|\forall u\in\X:\P[U=u]=\frac{1}{6},\forall v\in\Y:\P[U\in v\mid V=v]=1\right\}.
\end{equation}
We do not know the strategy of the game master when $3$ or $4$ is rolled. However, we do know that the die is fair and that the game master is not able to lie, thus we put this information in our model.

For shorthand purposes, we write $y_1=\{1,2,3,4\}$ and $y_2=\{3,4,5,6\}$ such that $\Y=\{y_1,y_2\}$ from now on.

\begin{proposition}\label{prop:SafeDice}
Let $\X$, $\Y$, $\Omega$, $U$, $V$ and $\Pmod$ be as before. Let
\begin{equation}
\tilde{\mathcal{P}}=\left\{\P\middle|
\begin{array}{l}
\P[U=1\mid V=y_1]=\P[U=2\mid V=y_1],\\
\P[U=3\mid V=y_1]=\frac{1}{2}-5\P[U=1\mid V=y_1],\\
\P[U=5\mid V=y_2]=\P[U=6\mid V=y_2],\\
\P[U=3\mid V=y_2]=3\P[U=5\mid V=y_2]+\frac{1}{2},\\
\P[U=1\mid V=y_1],\P[U=5\mid V=y_1]\in\left[0,\frac{1}{10}\right]
\end{array}
\right\}
\end{equation}
be a set of probability distributions on $\Omega$. Let $\Psafe$ be a probability distribution on $\Omega$ with $\Psafe[U=1|V=y_1]=\Psafe[U=2|V=y_1]$, $\Psafe[U=5|V=y_2]=\Psafe[U=6|V=y_2]$ and $\Psafe[U\in y|V=y]=1$ for all $y\in\Y$.

The following are equivalent:
\begin{enumerate}
    \item $\Psafe$ is a member of $\tilde{\mathcal{P}}$.
    \item $\Psafe$ is safe for $\langle U\rangle|[V]$.
    \item $\Psafe$ is safe for $\langle U\rangle|\langle V\rangle$.
\end{enumerate}

Furthermore, distribution $\Psafe$ is not safe for $U|[V]$.
\end{proposition}
\begin{proof}
The proof can be found in Appendix~\ref{app:ProofSafeDice}.
\end{proof}

The original question of the dice game did not concern an overall safe probability distribution for the whole die, but rather the probability of rolling $3$. Proposition~\ref{prop:DiscDiceSafe} states that a distribution $\Psafe$ is safe for $\langle\DieInd\rangle|\langle V\rangle$ as well as $\DieInd|[V]$ if and only if $\Psafe[U=3|V=v]=\frac{1}{6}$ holds for all $v\in\Y$. It is therefore safe to ignore the statement of the game master when guessing the probability of the die rolling to a $3$ and safe to state that the probability of rolling to a $3$ remains $\frac{1}{6}$. We will not prove Proposition~\ref{prop:DiscDiceSafe} now as it is a result of Theorem~\ref{thm:DiscMainThm}.

A safe distribution $\Psafe$ for $\langle\1_{\{U=1\}}\rangle|\langle V\rangle$ requiring that the game master still does not lie cannot exist, as the following proposition will prove.

\begin{proposition}
There is no probability distribution $\Psafe$ on $\Omega$ that is safe for $\langle\1_{\{U=u\}}\rangle|\langle V\rangle$ with $u\in\{1,2,5,6\}$ when $\Psafe[U\in v|V=v]=1$ is required for all $v\in\Y$.
\end{proposition}
\begin{proof}
Pick $u=1$ and suppose $\Psafe$ is safe for $\langle\1_{\{U=1\}}\rangle|\langle V\rangle$. Let $p,q\in[0,1]$ be as in the proof of Proposition~\ref{prop:SafeDice}. As $V=y_1$ is the only possible option when rolling $1$, we have
\begin{align}
\E_{\P}[\E_{\Psafe}[\1_{\{U=1\}}|V]]&=\Psafe[U=1|V=y_1]\P[V=y_1]\\
&=\Psafe[U=1|V=y_1]\left(\frac{2}{3}-\frac{p+q}{6}\right).
\end{align}
We need $\E_{\P}[\E_{\Psafe}[\1_{\{U=1\}}|V]]=\P[U=1]=\frac{1}{6}$, but there is no possible value for $\Psafe[U=1|V=y_1]$ such that
\begin{equation}
\Psafe[U=1|V=y_1]\left(\frac{2}{3}-\frac{p+q}{6}\right)=\frac{1}{6}
\end{equation}
holds for all $p,q\in[0,1]$. This only happens when $p+q=1$, so there is a contradiction. Therefore $\Psafe$ cannot be safe for $\langle\1_{\{U=1\}}\rangle|\langle V\rangle$.

This proof applies to all $v\in\{1,2,5,6\}$, where in the case of $v\in\{5,6\}$ we need to condition on $V=y_2$ instead of $V=y_1$.
\end{proof}

\subsection{Conclusion}
To conclude, in the example of the dice game, we cannot compute a unique conditional distribution for $U|V$. However, we were able to create a set of probability distributions that are all unbiased for $U|V$. When only considering the probability of the die rolling to a specific value, that is the random variable $\1_{\{U=u\}}$, a safe distribution $\Psafe$ for $\langle\1_{\{U=u\}}\rangle|\langle V\rangle$ with $u\in\{1,2,5,6\}$ can only exist if we accept that according to $\Psafe$ the game master is able to lie. This is due to these $u\in\{1,2,5,6\}$ not being in both conditioned outcome spaces $\{1,2,3,4\}$ and $\{3,4,5,6\}$.

When $u\in\{3,4\}$ is considered we can create a safe distribution $\Psafe$ for $\langle\1_{\{U=u\}}\rangle|\langle V\rangle$ as well as $\1_{\{U=u\}}|[V]$ while retaining the property that the game master cannot lie. This will be proven in Proposition~\ref{prop:DiscDiceSafe}, which is a direct application of Theorem~\ref{thm:DiscMainThm}. This is why safe probability for $\1_{\{U=3\}}|V$ and $\1_{\{U=4\}}|V$ has not been treated in this section.

\chapter{Discrete conditional paradoxes}\label{chap:DiscPara}
Consider the following three problems:

\paragraph{The Monty Hall problem}
Pretend there are three doors. Two doors have a goat behind and one door a car. The player initially chooses a door, say he chooses door $a$. The game master then opens either door $b$ or door $c$, however he never opens the door with the car. After opening a door, the player is asked if he wants to switch. He ultimately wins the contents behind the chosen door. Should the player switch to the other door?

\paragraph{The boy or girl problem}
Suppose you encounter a stranger in a bar and the two of you get in conversation. Suddenly, the stranger states that he has two children and at least one of them is a girl. What is the probability he has two daughters?

\paragraph{The dice game}
This game was already introduced in Section~\ref{sec:SafeDice}, but here is a repeat. A die is cast. Only the game master is able to observe the die's value. He then tells you whether the value of the die is in the set $\{1,2,3,4\}$ or in the set $\{3,4,5,6\}$. What is the probability of the die landing on a $3$ given the game master's statement?

\paragraph{}
All three problems essentially have the same structure. Firstly, there is a finite set of outcomes in each problem. Monty Hall's problem has three doors, the boy or girl problem has three distinct family compositions and there are six values on a die. Secondly, the game master must reveal some information where it is essential that at least one outcome remains possible no matter what the game master reveals. For that outcome the game master is free to choose his revelation. If in Monty Hall's problem door $a$ has the car, the game master can choose to either open door $b$ or door $c$. If for the boy or girl problem the stranger has one son and one daughter, he is free to choose whether he tells at least one child is a boy or at least one is a girl. If the die rolled 3 in the dice game, the game master is free to choose to reveal $\{1,2,3,4\}$ or $\{3,4,5,6\}$.\\
Since there is at least one outcome remaining possible no matter which revelation the game master makes, traditional conditional probability does not suffice. This is seen for example in the dice game in Section~\ref{sec:SafeDice}, where Equation~\ref{eq:SafeDiceCon6} states that the probability of rolling $6$ given $\{3,4,5,6\}$ is dilated from $\frac{1}{4}$ to $\frac{1}{2}$. Measure-theoretic conditional probability does not suffice as well, as the smallest sub-$\sigma$-algebra of $\{1,2,3,4,5,6\}$ containing the subset $\{\{1,2,3,4\},\{3,4,5,6\}\}$ does not contain the set $\{3\}$. When a larger sub-$\sigma$-algebra is chosen that does contain $\{3\}$, a version of the measure-theoretic conditional expectation cannot be found as the problem does not state the probability of for example revealing $\{1,2,3,4\}$ when a $3$ is rolled.

All three problems can be analysed using the same methods; they are essentially equal. The game master has a choice and therefore is able to employ a strategy. To counteract each strategy we need to first model all possible probability distributions. We can then create a pragmatic probability distribution using safe probability that the decision maker can act on. This distribution can be used as a `true' distribution and it will perform on average as well as all probability distributions in the model.

In this chapter we first present our main theorem, Theorem~\ref{thm:DiscMainThm}, which creates safe probability distributions in the setting of the three introduced problems. Then each problem is treated individually, applying Theorem~\ref{thm:DiscMainThm} in each case, and we will discuss the results. The dice game will be treated in Section~\ref{sec:DiscDice}, the Monty Hall problem will be treated in Section~\ref{sec:DiscMonty} and the boy or girl problem will be treated in Section~\ref{sec:DiscChildren}. We will conclude with Section~\ref{sec:DiscConcl} by stating that there is nothing paradoxical going on with the three problems. The paradoxical statements are just misapplications of conditional probability.

\section{The main theorem}\label{sec:DiscMain}
In the general setting we consider a set $\X$ of all possible outcomes called the \emph{outcome space} and a set $\Y$ of observations the player can make called the \emph{observation space}. Our sample space then becomes $\Omega=\X\times\Y$. Let $U$ be an $\X$-valued random variable denoting the outcome of the game, e.g.~the door containing the car in the Monty Hall problem or the family composition in the boy or girl problem. Let $V$ be a $\Y$-valued random variable denoting the statement of the game master, e.g.~the opening of door $b$ in the Monty Hall problem. Using safe probability we will for all $u\in\X$ create a safe distribution for $\1_{\{U=u\}}|[V]$.

\begin{theorem}\label{thm:DiscMainThm}
Let $\X$ be countable and $\Y$ be finite. Let $U$ be an $\X$-valued random variable and $V$ be a $\Y$-valued random variable. Let $\{p_u\}_{u\in\X}\subset[0,1]$ with $\sum_{u\in\X}p_u=1$. Let
\begin{equation}\label{eq:DiscMainMod}
\Pmod\subseteq\{\P\mid\forall u\in\X:\P[U=u]=p_u\}
\end{equation}
be our set of probability distributions on $\X\times\Y$ such that $|\Y|$ distributions $\P_1,\ldots,\P_{|\Y|}\in\Pmod$ exist imposing $|\Y|$ linearly independent vectors $(\P_i[V=v])_{v\in\Y}$ with $i\in\{1,\ldots,|\Y|\}$. Let $u\in\X$ be arbitrary and let $\Psafe$ be a distribution on $\X\times\Y$ with full support on $V$, then the following are equivalent:
\begin{enumerate}
    \item For all $v\in\Y$ we have $\Psafe[U=u|V=v]=p_{u}$.
    \item $\Psafe$ is safe for $\GeneralInd|[V]$.
    \item $\Psafe$ is safe for $\langle\GeneralInd\rangle|\langle V\rangle$.
\end{enumerate}
\end{theorem}
\begin{proof}
The proof can be found in Appendix~\ref{app:ProofDiscMainThm}.
\end{proof}

It looks like Theorem~\ref{thm:DiscMainThm} can only be applied in a few circumstances. However, the restriction on model $\Pmod$ is in many cases quickly satisfied, as in most cases $\Pmod$ is not a strict subset of the set in Equation~\ref{eq:DiscMainMod}. If $\Pmod$ is equal to the set in Equation~\ref{eq:DiscMainMod}, then the distributions $\P_i$ with $\P_i[V=v_j]=\delta_{ij}$ for all $i,j\in\{1,\ldots,|\Y|\}$, with $\delta$ the Kronecker delta, form a sequence of $|\Y|$ linearly independent unit vectors imposed from $|\Y|$ distributions of $\Pmod$. In other cases when excluding the unit vectors, if for all $v\in\Y$ a $\P\in\Pmod$ exists with $v\in\supp_{\P}(V)$ is in the support of $V$ under $\P$, then finding $|\Y|$ different distributions that impose $|\Y|$ linearly independent marginal probabilities on $V$ is not a hard task; otherwise the model $\Pmod$ is simply too small. This leads to the following lemma that is helpful when checking whether the model is large enough to apply Theorem~\ref{thm:DiscMainThm}.
\begin{lemma}\label{lem:DiscMainReq}
Let $\X$, $\Y$, $U$, $V$, $\{p_u\}_{u\in\X}$ be as in Theorem~\ref{thm:DiscMainThm}.  Let
\begin{equation}
\Pmod\subseteq\{\P\mid\forall u\in\X:\P[U=u]=p_u\}.
\end{equation}
It is necessary that $\bigcup_{\P\in\Pmod}\supp_{\P}(V)=\Y$ holds for $\Pmod$ to fulfil the requirements of Theorem~\ref{thm:DiscMainThm}.
\end{lemma}
\begin{proof}
Suppose $\bigcup_{\P\in\Pmod}\supp_{\P}(V)\neq\Y$, then there is a $v'\in\Y$ with $\P[V=v']=0$ for all $\P\in\Pmod$. Take this $v'$ and let $\P_1,\ldots,\P_{|\Y|}\in\Pmod$ be an arbitrary sequence of distributions. Then the vectors $(\P_1[V=v])_{v\in\Y},\ldots,(\P_{|\Y|}[V=v])_{v\in\Y}$ are not linearly independent as the vectors are of size $|\Y|$ as well and the entry of $v'$ is always zero.
\end{proof}

Furthermore, Theorem~\ref{thm:DiscMainThm} implies that a safe probability for $\GeneralInd|[V]$ can be created for all $u\in\X$. However, earlier in the introduction of Section~\ref{sec:DiscMain} we specifically discussed a special $u'\in\X$ that is supported by $U|V=v$ on all $\P\in\Pmod$ with $v\in\supp_{\P}(V)$.\\
Take for example the dice game. There either $3$ or $4$ can be picked as $u'$ as those values are still possible after the game master's statement that the value of the die is either in $\{1,2,3,4\}$ or in $\{3,4,5,6\}$. According to Theorem~\ref{thm:DiscMainThm} we can pick $u'=1$ as well to create a safe distribution $\Psafe$ for $\1_{\{U=1\}}|[V]$, however this implies
\begin{equation}
\Psafe[U=1|V=\{3,4,5,6\}]=\frac{1}{6}.
\end{equation}
Thus according to safe probability we need to assume that the game master lies with probability of at least $\frac{1}{6}$ when he tells the die has rolled three or higher, while in the problem we specifically assume the game master is not able to lie. This solution or distribution should therefore not be regarded as admissible, while it is still mathematically valid. When calculating a safe distribution $\Psafe$ for $\DieInd|[V]$, as is done in Proposition~\ref{prop:DiscDiceSafe}, we obtain $\Psafe[U=3|V=v]=\frac{1}{6}$ for all $v\in\Y$. Since for all $v\in\Y$ a $\P\in\Pmod$ exists with $\P[U=3|V=v]>0$, putting a positive probability on $U=3$ given any value of $V$ is a wise thing to do.\\
Therefore, while Theorem~\ref{thm:DiscMainThm} can be applied in many cases, it is wise to only apply it on $\GeneralInd|V$ when firstly $\P[U=u]=p_u>0$ holds for all $\P\in\Pmod$ and secondly when this $u$ is a member of the set
\begin{equation}\label{eq:DiscMainUReq}
\bigcap_{\P\in\Pmod}\bigcap_{v\in\supp_{\P}(V)}\range(U|V=v).
\end{equation}
Explaining the second requirement, since the event $\{U=u\}$ has positive and equal probability for all $\P\in\Pmod$ and since we have $\bigcup_{\P\in\Pmod}\supp_{\P}(V)=\Y$ by Lemma~\ref{lem:DiscMainReq}, for each $v\in\Y$ a $\P\in\Pmod$ must exist with $\P[U=u|V=v]>0$ when $u$ is a member of the set in Equation~\ref{eq:DiscMainUReq}. This makes it sensible to put positive safe probability $\Psafe[U=u|V=v]=\P[U=u]>0$ on the event $\{U=u\}$ given $V=v$.

The requirement $\Y=\bigcup_{\P\in\Pmod}\supp_{\P}(V)$ from Lemma~\ref{lem:DiscMainReq} is somewhat restrictive. The following proposition drops this requirement, but the equivalence in Theorem~\ref{thm:DiscMainThm} disappears as well leaving only the implications from 1 to 2 and from 2 to 3.

\begin{proposition}\label{prop:DiscSafeMargGen}
Let $\X$, $\Y$, $U$, $V$ and $\{p_{u}\}_{u\in\X}$ be as before. Let
\begin{equation}
\Pmod\subseteq\{\P|\forall u\in\X:\P[U=u]=p_u\}
\end{equation}
be non-empty and let $\Psafe$ be a distribution on $\X\times\Y$. Let $u\in\X$, then $\Psafe$ is safe for $\GeneralInd|[V]$ if $\Psafe[U=u|V=v]=p_{u}$ holds for all $v\in\Y$. This $\Psafe$ is safe for $\langle\GeneralInd\rangle|\langle V\rangle$ as well.
\end{proposition}
\begin{proof}
Let $\P\in\Pmod$ and $v\in\Y$ be arbitrary, then we have
\begin{equation}
\Psafe[U=u|V=v]=p_{u}=\P[U=u].
\end{equation}
Proposition~\ref{prop:SafeProperties} states that this $\Psafe$ is safe for $\GeneralInd|[V]$.

Safety for $\langle\GeneralInd\rangle|\langle V\rangle$ now immediately follows for $\Psafe$ by Proposition~\ref{prop:SafeImply}.
\end{proof}

Note that from the proof of Proposition~\ref{prop:DiscSafeMargGen} one can deduce that the elaborate set-up in Theorem~\ref{thm:DiscMainThm} is not necessarily needed and that Proposition~\ref{prop:DiscSafeMargGen} is nothing more than an applied restatement of the fourth property of Proposition~\ref{prop:SafeProperties}. However, in the case of Proposition~\ref{prop:DiscSafeMargGen}, you do not know whether distributions exist that are safe for $\langle \GeneralInd\rangle|\langle V\rangle$ but not for $\GeneralInd|[V]$. In the case of Theorem~\ref{thm:DiscMainThm}, safety for $\langle \GeneralInd\rangle|\langle V\rangle$ and $\GeneralInd|[V]$ are equivalent. Finding a safe distribution for $\GeneralInd|[V]$ is easy when the distribution on $U$ in $\Pmod$ is known, however for knowing these are the only safe distributions for $\langle \GeneralInd\rangle|\langle V\rangle$ we need Theorem~\ref{thm:DiscMainThm}.

Theorem~\ref{thm:DiscMainThm} only considers single $u\in\X$. We can construct safe probability distributions for $\GeneralGenInd|V$ for larger $\X'\subset\X$ as well in the following corollary.
\begin{corollary}\label{cor:DiscSafeGeneral}
Let $\X$, $\Y$, $U$, $V$, $\{p_u\}_{u\in\Y}$ and $\Pmod$ be as in Theorem~\ref{thm:DiscMainThm}. Let $\X'\subseteq\X$ be non-empty. Let $\Psafe$ be a distribution on $\X\times\Y$ with full support on $V$, then the following are equivalent:
\begin{enumerate}
    \item For all $u\in\X'$ and $v\in\Y$ we have $\Psafe[U=u|V=v]=p_{u}$.
    \item $\Psafe$ is safe for $\GeneralGenInd|[V]$.
    \item $\Psafe$ is safe for $\langle\GeneralGenInd\rangle|\langle V\rangle$.
\end{enumerate}
\end{corollary}
\begin{proof}
The proof is in essence a repetition of the proof of Theorem~\ref{thm:DiscMainThm} where $\GeneralGenInd=\max_{u\in\X'}\GeneralInd=\sum_{u\in\X'}\GeneralInd$ must be used.
\end{proof}

\subsection{Accuracy}
Suppose you are in a situation where you need to guess $\GeneralGenInd$ for an $\X'\subset\X$. One example is the Monty Hall game where on live television you need to guess whether your door $a$ has the car behind, thus whether $\1_{\{U=a\}}=1$ holds given the opened door $V$. This guess can be modelled as a \emph{randomized action} $\alpha$, which is a probability distribution on sample space $\Omega$.

\begin{definition}[Randomized action]
Let $\Omega$ be a sample space with $\sigma$-algebra $\F$. Let $\alpha\colon\F\to[0,1]$ be a probability measure, then $\alpha$ is called a \emph{randomized action} or \emph{action} in short.
\end{definition}

The key point here is that the distribution of random variable $\1_{\{U=a\}}$ is fixed by an unknown $\P\in\Pmod$. We want to impose an action $\alpha$ on predicting $\1_{\{U=a\}}$, where $\alpha$ needs to be `as equal to $\P$ as possible' with the challenge that $\P$ is unknown. The probability of action $\alpha$ correctly guessing a random variable $X$ is called the \emph{accuracy}, which can also be defined as the minus expected 0/1-loss of $\alpha$. For further reading in the subject of 0/1-losses the reader is referred to the article `Game theory, maximum entropy, minimum discrepancy and robust bayesian decision theory' by Grünwald and Dawid \cite{Grunwald04}.

\begin{definition}[Accuracy]\label{def:DiscAccuracy}
Let $\mathcal{Z}$ be a countable set and let $\Omega$ be the sample space. Let $X\colon\Omega\to\mathcal{Z}$ be a random variable with $\sigma$-algebra $\F$. Let $\Pmod$ be a set of probability measures on $\Omega$. Let $\alpha\colon\F\to[0,1]$ be a randomized action. The \emph{accuracy of $\alpha$ for guessing $X$ distributed by $\P$} is defined by
\begin{equation}
\acc_X^{\P}(\alpha)=\sum_{k\in\mathcal{Z}}\alpha[X=k]\P[X=k].
\end{equation}
When $\acc_X^{\P}(\alpha)$ is constant for all $\P\in\Pmod$, this value is called the \emph{accuracy of $\alpha$ for guessing $X$} and is denoted by $\acc_X(\alpha)$. If from context it is clear we want to guess $X$, we write \emph{the accuracy of $\alpha$} and denote this by $\acc(\alpha)$.
\end{definition}

The next theorem optimizes the accuracy of an action $\alpha^*$ for predicting $\GeneralGenInd$ against all $\P\in\Pmod$.
\begin{theorem}\label{thm:DiscAccOpt}
Let $\X$, $\Y$, $U$, $V$, $\{p_u\}_{u\in\Y}$ and $\Pmod$ be as in Proposition~\ref{prop:DiscSafeMargGen} and let $\X'\subseteq\X$ be non-empty. Let $\alpha^*$ be an action on $\X\times\Y$ such that
\begin{equation}\
\alpha^*[U\in\X']=\begin{cases}
1,&\sum_{u\in\X'}p_u>\frac{1}{2},\\
0,&\sum_{u\in\X'}p_u<\frac{1}{2},\\
p,&\sum_{u\in\X'}p_u=\frac{1}{2},p\in[0,1].
\end{cases}
\end{equation}
This action guesses $\GeneralGenInd$ correctly with probability
\begin{equation}
\acc^{\P}\left(\alpha^*\right)=\max\left\{\sum_{u\in\X'}p_u,1-\sum_{u\in\X'}p_u\right\}
\end{equation}
for all $\P\in\Pmod$. Any other action $\alpha$ has a lower accuracy than $\alpha^*$.
\end{theorem}
\begin{proof}
Let $\P\in\Pmod$ be arbitrary and let $\alpha\colon2^{\X\times\Y}\to[0,1]$ be an arbitrary action. We will first prove that $\acc^{\P}(\alpha)$ is constant for all $\P\in\Pmod$ and then prove that $\alpha=\alpha^*$ maximizes the accuracy of $\alpha$.

By Definition~\ref{def:DiscAccuracy} and the fact that $\P\in\Pmod$ we have
\begin{align}
\acc_{\P}(\alpha)&=\sum_{k\in\{0,1\}}\alpha[\GeneralGenInd=k]\P[\GeneralGenInd=k]\\
&=\alpha[U\not\in\X']\sum_{u\in\X\setminus\X'}\P[U=u]+\alpha[U\in\X']\sum_{u\in\X'}\P[U=u]\\
&=\alpha[U\not\in\X']\left(1-\sum_{u\in\X'}p_u\right)+\alpha[U\in\X']\sum_{u\in\X'}p_u,
\end{align}
thus $\acc^{\P}(\alpha)$ is constant for all $\P\in\Pmod$. Abbreviate the terms $\sum_{u\in\X'}p_u=x$ and $\alpha[U\in\X']=p$. We now need to maximize
\begin{equation}
f(p):=px+(1-p)(1-x)=p(2x-1)+1-x.
\end{equation}
Note that both $x,p\in[0,1]$. Consider the following different cases for $x$.
\begin{itemize}
    \item[$x>\frac{1}{2}$:] When $x>\frac{1}{2}$ holds, the function $f$ is strictly increasing in $p$. Its maximum is then achieved at $p=1$, which is $p$. Thus the action putting all mass on $\{U\in\X'\}$ maximizes its accuracy.
    \item[$x<\frac{1}{2}$:] When $x<\frac{1}{2}$ holds, the function $f$ is strictly decreasing in $p$. Its maximum is then achieved at $p=0$, which is $1-x$. Thus the action putting all mass on $\{U\not\in\X'\}$ maximizes its accuracy.
    \item[$x=\frac{1}{2}$:] When $x=\frac{1}{2}$ holds, the function $f$ is constant. Its value is $f(p)=\frac{1}{2}=x$ for all $p\in[0,1]$. Therefore the mass put on $\{U\in\X'\}$ by $\alpha$ can be of arbitrary value, the accuracy is still maximized.
\end{itemize}
Therefore $\alpha$ has maximal accuracy when $\alpha=\alpha^*$ and the accuracy of $\alpha^*$ is
\begin{equation}
\acc\left(\alpha^*\right)=\max\left\{\sum_{u\in\X'}p_u,1-\sum_{u\in\X'}p_u\right\}.
\end{equation}
In the case of $\sum_{u\in\X'}p_u\neq\frac{1}{2}$, any other action $\alpha$ putting different mass on $\{U\in\X'\}$ than $\alpha^*$ has a lower accuracy as the maxima of $f$ are uniquely achieved. When $\sum_{u\in\X'}p_u=\frac{1}{2}$ holds, any action is optimal with accuracy~$\frac{1}{2}$.
\end{proof}

Suppose the player uses a safe distribution $\Psafe$ to determine his action $\alpha$ for guessing $\GeneralGenInd$, what will the accuracy of $\alpha$ then be? The following proposition answers this question.

\begin{proposition}\label{prop:DiscAccSafe}
Let $\X$, $\X'$, $\Y$, $U$, $V$, $\{p_u\}_{u\in\Y}$ and $\Pmod$ be as in Corollary~\ref{cor:DiscSafeGeneral}. Let $\Psafe$ be a safe distribution for $\GeneralGenInd|\langle V\rangle$ and define action $\tilde{\alpha}$ as
\begin{equation}
\tilde{\alpha}[U\in\X']=\Psafe[U\in\X'|V=v]=\sum_{u\in\X'}p_u.
\end{equation}
The accuracy of $\tilde{\alpha}$ for all $\P\in\Pmod$ becomes
\begin{equation}
\acc^{\P}\left(\tilde{\alpha}\right)=\left(\sum_{u\in\X'}p_{u}\right)^2+\left(1-\sum_{u\in\X'}p_{u}\right)^2.
\end{equation}
\end{proposition}
\begin{proof}
Let $\P\in\Pmod$ be arbitrary. By Corollary~\ref{cor:DiscSafeGeneral} and Definition~\ref{def:DiscAccuracy} we have
\begin{align}
\acc^{\P}\left(\tilde{\alpha}\right)&=\sum_{k=0}^1\tilde{\alpha}\left[\GeneralGenInd=k\right]\P[\GeneralGenInd=k]\\
&=\left(1-\sum_{u\in\X'}p_u\right)\P[U\not\in\X']+\left(\sum_{u\in\X'}p_u\right)\P[U\in\X'].
\end{align}
Since $\P[U\in\X']=\sum_{u\in\X'}p_u$ holds for all $\P\in\Pmod$, we can conclude that
\begin{equation}
\acc\left(\tilde{\alpha}\right)=\left(\sum_{u\in\X'}p_{u}\right)^2+\left(1-\sum_{u\in\X'}p_{u}\right)^2.
\end{equation}
\end{proof}

Now we know when a safe distribution $\Psafe$ for $\GeneralGenInd|[V]$ results into being the most accurate guess for $\GeneralGenInd$ as well.

\begin{corollary}
Let $\X$, $\X'$, $\Y$, $U$, $V$, $\{p_u\}_{u\in\Y}$ and $\Pmod$ be as in Corollary~\ref{cor:DiscSafeGeneral}. Let $\alpha^*$ be as in Theorem~\ref{thm:DiscAccOpt}. Let $\tilde{\alpha}$ be as in Proposition~\ref{prop:DiscAccSafe}. We only have $\acc(\alpha^*)=\acc(\tilde{\alpha})$ when
\begin{equation}
\sum_{u\in\X'}p_u\in\left\{0,\frac{1}{2},1\right\}.
\end{equation}
\end{corollary}
\begin{proof}
This is already proven by Grunwald and Dawid \cite{Grunwald04}, albeit in the different context of 0/1-loss. The proof can in our case be easily recreated by comparing $\acc(\alpha^*)$ and $\acc(\tilde{\alpha})$.
\end{proof}

We see that the action $\alpha^*$ putting all mass on $\X'$ or no mass on $\X'$ is always optimal. When the decision maker wants to apply safe probability, it is rarely the case that the safe distribution puts all or no mass on $\X'$ as well. In this case for all $\P\in\Pmod$ we have $\P[U\in\X'|V=v]=\P[U\in\X']$ for all $v\in\supp_{\P}(V)$ as the events $\{U\in\X'\}$ and $\{V=v\}$ become independent, thus the conditional distribution of $\GeneralGenInd|V=v$ is known. The application of safe probability becomes trivial and unnecessary.

Knowing this, one must keep in mind that safe probability for $\GeneralGenInd|[V]$ gives in almost no case a prediction for $\GeneralGenInd|V$ with optimal accuracy. Safe probability merely simulates the distribution of $\GeneralGenInd$, where in the setting of Theorem~\ref{thm:DiscMainThm} the revelation of the game master is in no case of influence. A safe probability distribution must therefore in this setting not directly be used to make predictions for $\GeneralGenInd$, but it can be used to create actions that does predict $\GeneralGenInd$ with optimal accuracy. For example, in the Monty Hall game you should not switch with probability $\frac{2}{3}$, as then the probability of winning the car is $\frac{5}{9}$. If the player observes that the safe distribution in the Monty Hall game puts a probability of $\frac{1}{3}$ on door $a$, then never choosing door $a$ yields to a probability of winning the car of $\frac{2}{3}$, which is optimal by Theorem~\ref{thm:DiscAccOpt}.

\section{Dice game}\label{sec:DiscDice}
With the main theorems stated, we can apply them to some well-known paradoxes. We will first continue our example of the dice game, started in Section~\ref{sec:SafeDice}. This game is a prime example to demonstrate the main theorems first as it removes all context that is present in problems like Monty Hall and boy or girl, which is why Grünwald devised it in the first place \cite{Grunwald13}.

Recall that a die is cast with values in $\X=\{1,2,3,4,5,6\}$ and the game master is, after observing the die's value, only able to reveal a member of the set $\Y=\{\{1,2,3,4\},\{3,4,5,6\}\}=:\{y_1,y_2\}$. Our sample space is $\Omega=\X\times\Y$ with its power set $\F=2^\Omega$ as $\sigma$-algebra. Let $U$ be an $\X$-valued random variable and $V$ be a $\Y$-valued random variable. The game master is not allowed to lie and the die is fair, thus any probability distribution for this game must be a member of \begin{equation}
\Pmod=\left\{\P\middle|\forall u\in\X:\P[U=u]=\frac{1}{6},\forall y\in\Y:\P[U\in y|Y=y]=1\right\}.
\end{equation}

\subsubsection{Safe probability}
Proposition~\ref{prop:SafeDice} gives a set $\tilde{\mathcal{P}}$ of distributions that are all safe for $\langle U\rangle|[V]$. However, in the original problem we are not interested in the whole distribution of $U$ given $V$, but only want to know the probability of rolling $3$ after the game master's statement. Therefore finding safe distributions for $\DieInd|[V]$ is sufficient.

\begin{proposition}\label{prop:DiscDiceSafe}
Let $\X$, $\Y$, $\Omega$, $U$, $V$ and $\Pmod$ be as in the dice game. Let $\Psafe$ be a distribution on $\Omega$ with full support on $V$. Let $u\in\X$, then the following statements are equivalent:
\begin{enumerate}
\item For all $v\in\Y$ we have $\Psafe[U=u|V=v]=\frac{1}{6}$.
\item $\Psafe$ is safe for $\langle \1_{\{U=u\}}\rangle|\langle V\rangle$.
\item $\Psafe$ is safe for $\1_{\{U=u\}}|[V]$.
\end{enumerate}
\end{proposition}
\begin{proof}
This proposition is a direct application of Theorem~\ref{thm:DiscMainThm}. The set $\X$ is of size $6$ and therefore countable and the set $\Y$ is of size $2$, which makes it finite. $U$ is a random variable on $\X$ and $V$ is a random variable on $\Y$. In this dice game we have $p_u=\frac{1}{6}$ as $\P\in\Pmod$ implies $\P[U=u]=\frac{1}{6}$ for all $u\in\X$. Lastly, the set $\Pmod$ is a subset of $\{\P|\forall u\in\X:\P[U=u]=p_u\}$.

Take $\P_1,\P_2\in\Pmod$ with $\P_i[V=y_i|U=3]=\P_i[V=y_i|U=4]=1$. Let $i\in\{1,2\}$ be arbitrary, then
\begin{align}
\P_i[V=y_i]&=\sum_{u=1}^6\P_i[V=y_i|U=u]\P_i[U=u]\\
&=\frac{1}{6}\left(4\cdot1+2\cdot0\right)=\frac{2}{3}.
\end{align}
Thus the vectors $\left(\P_i[V=v]\right)_{v\in\Y}$ become $\left(\frac{1}{3},\frac{2}{3}\right)$ and $\left(\frac{2}{3},\frac{1}{3}\right)$, which are linearly independent.

Now we can directly apply Theorem~\ref{thm:DiscMainThm} to conclude the proof.
\end{proof}

From Proposition~\ref{prop:DiscDiceSafe} it immediately follows that the distribution with $\Psafe[U=3|V=v]=\Psafe[U=4|V=v]=\frac{1}{6}$ for all $v\in\Y$ and the requirement $\Psafe[U\in\{5,6\}|V=y_1]=\Psafe[U\in\{1,2\}|V=y_2]=0$ is safe for $\1_{\{U=3\}}|[V]$ and $\1_{\{U=4\}}|[V]$. We do have a certain freedom when putting a probability mass the other probabilities like $\Psafe[U=1|V=y_1]$. The only requirement in order for $\Psafe$ to be safe for both $\1_{\{U=3\}}|[V]$ and $\1_{\{U=4\}}|[V]$ is that
\begin{align}
\Psafe[U=1|V=y_1]+\Psafe[U=2|V=y_1]&=\\
\Psafe[U=5|V=y_2]+\Psafe[U=6|V=y_2]&=\frac{2}{3}.
\end{align}
When one acts assumes the random variable $\DieInd$ behaves like $\Psafe$, thus assuming that the probability of rolling $3$ and the probability of rolling $4$ stay $\frac{1}{6}$ after hearing the game master's statement and still applying the assumption that the game master does not lie, he gets a distribution that performs as well as any distribution from $\Pmod$ on guessing the probability of rolling $3$ or $4$. This $\Psafe$ does not necessarily need to be the correct distribution, in fact there are numerous distributions $\Psafe$ fulfilling the requirements of Proposition~\ref{prop:DiscDiceSafe}, but it is useful to act as if $\Psafe$  is a `true' distribution.

Therefore if one asks what the probability of rolling $3$ is given the game master's statement, then `not unique, but pretend like it is always $\frac{1}{6}$' must be a satisfying answer.

\subsubsection{Accuracy}
Suppose you want to bet on this game and want to know whether you must put money on rolling $3$. As the following proposition will state, after observing the game master's statement, always stating that the die has not rolled to a $3$ has a $\frac{5}{6}$ chance of being correct independently on the game master's strategy and is the optimal accuracy.

\begin{proposition}
Let $\X$, $\Y$, $U$, $V$ and $\Pmod$ be as in the dice game. Consider the action $\alpha^*$ with $\alpha^*[U=3]=0$, then to predict $\DieInd$ this $\alpha^*$ has accuracy
\begin{equation}
\acc^{\P}\left(\alpha^*\right)=\frac{5}{6}
\end{equation}
for all $\P\in\Pmod$. All other actions $\alpha\neq\alpha^*$ have $\acc^{\P}(\alpha)<\acc^{\P}(\alpha^*)$ for all $\P\in\Pmod$.
\end{proposition}
\begin{proof}
We want to apply Theorem~\ref{thm:DiscAccOpt}. Firstly in this case $\X'=\{3\}\subset\X$ is non-empty. Secondly, we have $\sum_{u\in\X'}p_u=p_3=\frac{1}{6}<\frac{1}{2}$. Then Theorem~\ref{thm:DiscAccOpt} states that $\alpha^*$ with $\alpha^*[U=3]=0$ optimizes the accuracy for predicting $\DieInd$ with
\begin{equation}
\acc\left(\alpha^*\right)=\max\left\{\frac{1}{6},\frac{5}{6}\right\}=\frac{5}{6}.
\end{equation}
\end{proof}

We now observe that using the information given by the game master does not lead to a higher probability of guessing $\DieInd$ correct. When always stating that the die has not rolled $3$ you maximize the probability of having a correct guess, which is $5$ out of $6$. The information revealed by the game master can therefore be ignored as it will not yield to a higher probability of guessing correctly.

\subsubsection{Final remarks}
We have seen that the conditional probability of rolling $3$ given the game master's statement is not uniquely defined; it is dilated from $0$ to $\frac{1}{4}$ depending on the to the player unknown strategy of the game master. However, when playing like the probability of rolling $3$ is always $\frac{1}{6}$ disregarding the game master's statement, you are able to simulate the probability distribution of $\DieInd$. Furthermore, when always stating that the die has not rolled $3$, you are correct in on average five out of six games, which is the maximal fraction of games you can be correct on.

The paradoxical nature of this problem is that most people treat this problem with sample space $\Omega'=\{1,2,3,4,5,6\}$ and condition on for example the information $\{3,4,5,6\}$. However, as we have seen in Section~\ref{sec:SafeDice}, we also need to condition on the possibility $\{1,2,3,4\}$ and those two sets cannot form a partition of $\Omega'$. Therefore the sample space $\Omega'$ is insufficient for solving this problem, resulting in the paradoxical statements.

This underlines the argument that when conditioning the accompanying sub-$\sigma$-algebra must always be given, as then it is easily seen that conditioning on a sub-$\sigma$-algebra containing $\sigma(\{1,2,3,4\},\{3,4,5,6\})\subset2^{\Omega'}$ is not possible.

\section{Monty Hall}\label{sec:DiscMonty}
Consider now Monty Hall's three door problem. The problem was already explained in the introduction of this chapter, but we will repeat it here.

There are three doors called $a$, $b$ and $c$.  Two doors have a goat behind and one door has a car. The player chooses one door, where after the game master opens one of the remaining doors. The game master does not open the door with the car. The player is then asked whether he wants to switch to the remaining door. The player wins the contents of the door he ultimately chooses.

A logical question, and the centre of this paradox, is whether the player should switch when a door is opened. This question can be answered with multiple viewpoints. Throughout the whole analysis we assume that initially the car is distributed between the doors with equal probability. Without loss of generality we now assume the player has chosen door $a$, as the initial choice of door has become independent of the question whether the player should switch.

Most initially resort to the following viewpoint. Say door $c$ is opened by the game master, then either door $a$ or door $b$ has a car with equal probability $\frac{1}{2}$. Since there is a 50-50 chance, switching does not increase your chances. This viewpoint is widely disputed.

Another viewpoint using classic conditional probability states the following. In the beginning there is a $\frac{1}{3}$-chance that door $a$ has a car. Assume after choosing door $a$ the game master reveals door $c$. There was a chance of $\frac{2}{3}$ that the car was behind door $b$ or $c$, leaving a $\frac{2}{3}$-chance that the car is behind door $b$. Therefore switching is advised, doubling your chances of winning the car. This viewpoint is however disputed as well.

\subsection{History and discussion of the viewpoints}\label{sec:MontyHist}
The Monty Hall problem is first stated by Selvin in 1975 \cite{Selvin75a} with a table of all possible outcomes, concluding that switching is preferable in six out of nine cases. This article sparked little debate and later that year Selvin \cite{Selvin75b} posted a letter arguing further why the answer of $\frac{1}{3}$ is correct. The problem gained popularity when a reader of Parade Magazine posted a letter to Vos~Savant \cite{Marilyn90a}, asking her view on the problem. She argued that $\frac{1}{3}$ is correct as well. Now the problem sparked much more controversy, forcing Vos~Savant to treat the problem three more times in her column \cite{Marilyn90b,Marilyn91a,Marilyn91b}. Following this debate a vast amount of literature has been written on the problem. A thorough overview of all different viewpoints and literature written on the problem can be found on the Wikipedia page of the Monty Hall problem \cite{WikiMonty}.

After Vos~Savant's original column \cite{Marilyn90a}, she received approximately 10000 letters concerning the problem with a majority disagreeing with her \cite{Tierney91}. Most letters come from mathematicians and statisticians. Following this, Tierney \cite{Tierney91} describes that Monty Hall himself simulated this problem in his home with 30 independent attempts. Monty Hall drew two conclusions, namely that Vos~Savant and Selvin originally were correct but also that Vos~Savant was not meticulous in writing down the problem. Vos~Savant found it charming that Monty Hall himself was consulted, but not a scientifically sound way of proceeding \cite{Savant91c}. She argues that `the television game show host is without relevance and consulting one erroneously introduces personality'.

What is most extraordinary on the answer of the probability being $\frac{1}{2}$ is that humans are surprisingly persistent on their belief that the answer of $\frac{1}{2}$ is correct. Granberg and Brown \cite{Granberg95} found out that in their experiment only 13\% of the contestants wanted to switch and in a study on more variants of the Monty Hall problem by Mueser and Granberg \cite{Mueser99} only 19\% of the contestants wanted to switch. When two slightly altered versions of the problem are posed, only 11\% and 18\% of the contestants wanted to switch. It can be assumed that the contestants who did not switch believed that the probability of winning after switching is $\frac{1}{2}$ or lower, as otherwise they would have believed the answer of $\frac{2}{3}$ and did want to switch. The in my opinion most remarkable result is achieved by Herbranson and Schroeder \cite{Herbranson10} who tested the Monty Hall problem on pigeons. The result is that after 30 days of playing the experiment, almost all of the pigeons learned that switching yields to a higher probability of winning, whereas humans stubbornly stay with their initial beliefs that $\frac{1}{2}$ is the correct answer.

The first to support the answer of $\frac{1}{3}$ with rigorous mathematical proofs was Morgan et al.~in 1991 \cite{Morgan91a}. They do point out that Vos~Savant's methods and proofs are incorrect, however her conclusion is not. Morgan et al.~explicitly drop the assumption that the game master chooses a door with equal probability after initially the player choosing the door with the car, proving that the probability of winning after switching is in the interval $\left[\frac{1}{2},1\right]$ and not a single probability. This approach received many comments \cite{Seymann91,Morgan91b,Savant91c,Bell92,Rao92,Hogbin10}. Seymann \cite{Seymann91} commented that Morgen et al.~correctly pointed out that the Monty Hall problem is stated with many different versions, all having different solutions. Morgan et al.~solve the `vos~Savant scenario', which is the scenario we treat here as well. Vos~Savant \cite{Savant91c} defended that the problem could not be stated explicitly and without doubt in her column; the column would otherwise be too long. She noted after reading nearly 3000 letters that nearly all of her critics understood the scenario intended without raising questions on ambiguity, where most concluded that the answer is $\frac{1}{2}$ as only two options are left after opening a door. Bell \cite{Bell92} pointed out that in the article of Morgan the events `car behind door $a$' and `door $c$ is opened' are independent, but `door $c$ is opened' and `car behind door $b$' are not. It is incorrect to state that the independence of some events imply independence of the random variables indicating them. Rao \cite{Rao92} provides a four-door problem giving another insight to the Monty Hall problem. Lastly, Hogbin and Nijdam \cite{Hogbin10} corrected a mistake of Morgan et al.~\cite{Morgan91a}. Morgan et al.~stated that the probability of the car being behind door $a$ given door $c$ is opened is $\frac{q}{1+q}$, where $q\in[0,1]$ is the probability of opening door $c$ given the car is behind door $a$. Morgan et al.~then stated that the average probability of winning after switching given door $c$ is opened is $\int_0^1(1+q)^{-1}dq=\ln2$. Hogbin and Nijdam comment that this analysis is wrong as Morgen et al.~integrate over the prior density instead of the posterior density. They are complimented by Morgan et al.~\cite{Hogbin10} for finding the mistake.

We could cover more literature on this problem, however discussing all literature on Monty Hall's problem can be a thesis on its own. A reader wanting more sources on this problem is once more kindly referred to the Wikipedia page \cite{WikiMonty} as it provides an exhaustive list of sources with relevant discussions.

\subsection{Our thoughts on the problem}
Let us start with a formal statement of the problem. Let $\X=\{a,b,c\}$ be our outcome space and $\Y=\{b,c\}$ be our observation space. Let $\Omega=\X\times\Y$ be our sample space with $\F=2^\Omega$ as $\sigma$-algebra. Let $U$ be an $\X$-valued random variable denoting the placement of the car and let $V$ be a $\Y$-valued random variable denoting the opening of a door.\\
Note that the doors are called $a$, $b$ and $c$ and not $1$, $2$ and $3$, as in the latter case the expectation $\E[U]$ suddenly can be given a misleading meaning. There is no such thing as the `average door', as the labels on the doors are nothing more than names, not aspects indicating any value. Therefore we choose to name the doors $a$, $b$ and $c$ so as not to give rise to any confusion.\\
We assume that the car is initially distributed with equal probability and the game master cannot open a door with a car. A probability distribution must therefore be a member of the set
\begin{equation}
\Pmod=\left\{\P\middle|\forall u\in\X:\P[U=u]=\frac{1}{3},\forall v\in\Y:\P[V=v|U=v]=0\right\}.
\end{equation}

In this section we first give a repetition of the arguments of \cite{Grunwald13} as they view the Monty Hall problem in the viewpoint of traditional probability. Then we apply safe probability and Theorem~\ref{thm:DiscMainThm} to the Monty Hall problem, with as result that it is safe to state that the car is behind door $a$ with probability~$\frac{1}{3}$.

\subsubsection{Traditional and measure-theoretic conditional probability}
Let us start with the viewpoint that results in a probability of $\frac{1}{2}$ of obtaining the car after switching. This viewpoint takes $\Omega'=\{a,b,c\}$ as sample space. Let $U'$ be an $\Omega'$-valued random variable denoting the cars location. Equip $\Omega'$ with the uniform distribution $\P$ such that $\P[U'=u]=\frac{1}{3}$. Traditional conditional probability and Bayes' rule then give
\begin{align}
\P[U'=a|U'\in\{a,b\}]&=\frac{\P[U'\in\{a,b\}|U'=a]\P[U'=a]}{\sum_{u\in\Omega}\P[U'\in\{a,b\}|U'=u]\P[U'=u]}\label{eq:DiscMontyWrong}\\
&=\frac{1\cdot\frac{1}{3}}{\frac{1}{3}\left(1+0+1\right)}=\frac{1}{2}.
\end{align}

What is wrong with the above reasoning from a measure-theoretic point of view? Suppose door $a$ has the car. The game master is now able to choose between doors $b$ and $c$. If door $b$ is opened, the resulting set of possibilities is $\{a,c\}$ and when door $c$ is opened, the resulting set of possibilities is $\{a,b\}$. We need to condition on both the possibilities $\{a,b\}$ and $\{a,c\}$, however those sets cannot form a partition of $\Omega'$. Furthermore, the smallest $\sigma$-algebra containing both sets is
\begin{equation}
\G=\sigma(\{a,b\},\{a,c\})=2^{\{a,b,c\}}=2^{\Omega'},
\end{equation}
our original $\sigma$-algebra. As stated by Williams \cite{Williams91} and also easy to prove, we have $\P[F|\G]=\P[F]$ for all $F\in\G$ when $\G=2^{\Omega'}$ is our original $\sigma$-algebra. We are therefore simply not able to use measure-theoretic conditional probability.

Can we say nothing sensible at all about the probability of winning the car after switching? Consider the $\Omega$, $U$ and $V$ from the formal statement in the introduction of this section. Now we have $\Omega=\X\times\Y$, thus $\Omega$ is split up in outcome space $\X=\{a,b,c\}$ and observation space $\Y=\{b,c\}$. We can condition on all possibilities of open doors, as $\sigma(\{a\}\times\{b\},\{a\}\times\{c\})$ is a strict sub-$\sigma$-algebra of $2^\Omega$.\\
We can now calculate the resulting conditional probabilities. Let $\P\in\Pmod$, thus such that $\P$ is uniform distributed on $U$ and a door with a car cannot be opened. Then a $p\in[0,1]$ exists such that $\P[V=b|U=a]=p$. This results into
\begin{align}
\P[U=a|V=b]=\frac{\P[V=b|U=a]\P[U=a]}{\sum_{u\in\Omega}\P[V=b|U=u]\P[U=u]}=\frac{p}{p+1},\label{eq:DiscMontyDilate}\\
\P[U=a|V=c]=\frac{\P[V=c|U=a]\P[U=a]}{\sum_{u\in\Omega}\P[V=c|U=u]\P[U=u]}=\frac{1-p}{2-p}.
\end{align}
In both cases the probability of door $a$ having the car is dilated from~$0$ to~$\frac{1}{2}$. Furthermore, the answer of $\frac{1}{2}$ disregarding which door is opened cannot be true, as the probability of $\frac{1}{2}$ after opening door $v\in\Y$ is only achieved when $p=\1_{\{b\}}(v)$ which is dependent on door $v$. This is a strategy that is not possible to deploy.\\
The player is advised to switch no matter the strategy of the game master. However, we cannot pin down a single probability of door $a$ having the car given the game master's statement. We can put a prior on $\P[V=b|U=a]$, however we do need more information to apply such priors wisely.

This subsection shows another example of why when doing conditional probability one always needs to provide the accompanying sub-$\sigma$-algebra. Unlike the Borel-Kolmogorov paradox in Chapter~\ref{chap:BorelKolmogorov}, here the measure-theoretic conditional expectation defines a unique probability measure. However, most people use $\Omega'=\{a,b,c\}$ as sample space and without taking the accompanying $\sigma$-algebra in account when conditioning, one could not find a reason not to do this. Only when correctly applying conditional probability and noticing that both $\{a,b\}$ and $\{a,c\}$ are possible options, one finds out that $\Omega'=\{a,b,c\}$ is insufficient as sample space and paradoxical results will rise when continuing to use $\Omega'$.

\begin{remark}[Three prisoners problem]
Consider again the measure-theoretic approach taken at the Monty Hall problem in Equation~\ref{eq:DiscMontyDilate} with now $p=\frac{1}{2}$. This case is called the \emph{three prisoners problem} which is first described by Martin Gardner in 1959 \cite{Gardner59}. We will take a brief look at this paradox using the version posed on the problem's Wikipedia page \cite{WikiPrisoners}.

There are three prisoners conveniently called $a$, $b$ and $c$ on death row and located in separate cells. One prisoner however will be pardoned and the warden knows which one. Prisoner~$a$ is able to gain some information from the warden. If $a$ will not be pardoned, he hears from the warden the name of either prisoner~$b$ or $c$ and that prisoner will not be pardoned as well. If $a$ will be pardoned, the warden flips a fair coin to decide whether to tell $b$ or $c$'s name.\\
After hearing prisoner~$b$'s name, prisoner $a$ believes his chance of survival rises from $\frac{1}{3}$ to $\frac{1}{2}$. However, when $a$ tells this news to $c$, prisoner~$c$ argues that $a$'s chance of survival is still $\frac{1}{3}$ but his is risen to $\frac{2}{3}$. Prisoner $c$'s reasoning is correct as we will soon see.

The only difference between this problem and Monty Hall's is that we now have full information on how the player, in this case prisoner~$a$, obtains his observation. In the Monty Hall problem the game master could deploy any strategy of either opening door $b$ or $c$ when the car is behind door $a$. Here prisoner~$b$ or $c$'s name is chosen by the flip of a fair coin if prisoner~$a$ will be pardoned. Therefore, adopting the Monty Hall problem's mathematical terminology, ${\P[V=b|U=a]=\frac{1}{2}}$ must hold here instead of ${\P[V=b|U=a]\in[0,1]}$. If $U$ is distributed uniformly on $\{a,b,c\}$ and the warden is not allowed to lie, then now our model for the three prisoners problem $\Pmod=\{\P\}$ only contains one single distribution.

The resulting probability of prisoner~$a$ being pardoned remains $\frac{1}{3}$ regardless of which prisoner's name is called:
\begin{align}
\P[U=a|V=b]=\frac{\P[V=b|U=a]\P[U=a]}{\sum_{u\in\Omega}\P[V=b|U=u]\P[U=u]}=\frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{3}\left(\frac{1}{2}+0+1\right)}=\frac{1}{3},\\
\P[U=a|V=c]=\frac{\P[V=c|U=a]\P[U=a]}{\sum_{u\in\Omega}\P[V=c|U=u]\P[U=u]}=\frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{3}\left(\frac{1}{2}+1+0\right)}=\frac{1}{3}.
\end{align}
The statement of prisoner $c$ is therefore correct. The reason why people first claim that $\frac{1}{2}$ is correct is the same as in the Monty Hall problem. They first take $\Omega'=\{a,b,c\}$ as sample space and then condition only on the event $\{U\in\{a,c\}\}$, which will result in the answer of $\frac{1}{2}$ after following the logic of Equation~\ref{eq:DiscMontyWrong}. The fact that the warden could also have flipped a coin before declaring a name is ignored when $\frac{1}{2}$ is given as an answer. This mistake can easily be avoided by applying conditional probability with more care, taking all information stated in the problem into account.
\end{remark}

\subsubsection{Safe probability}
Using safe probability it is possible to find a pragmatic distribution $\Psafe$ on $\Omega$ that is safe against all $\P\in\Pmod$. We only need to apply Theorem~\ref{thm:DiscMainThm}.

\begin{proposition}\label{prop:DiscMontySafe}
Let $\X$, $\Y$, $\Omega$, $U$, $V$ and $\Pmod$ be as in the Monty Hall problem. Let $\Psafe$ be a distribution on $\Omega$ with full support on $V$. Let $u\in\X$, then the following statements are equivalent:
\begin{enumerate}
\item For all $v\in\Y$ we have $\Psafe[U=u|V=v]=\frac{1}{3}$.
\item $\Psafe$ is safe for $\langle \1_{\{U=u\}}\rangle|\langle V\rangle$.
\item $\Psafe$ is safe for $\1_{\{U=u\}}|[V]$.
\end{enumerate}
\end{proposition}
\begin{proof}
This proposition is a direct application of Theorem~\ref{thm:DiscMainThm} and its proof is equal to the proof of Proposition~\ref{prop:DiscDiceSafe}. The set $\X$ is of size $3$ and therefore countable and the set $\Y$ is of size $2$, which makes it finite. $U$ is a random variable on $\X$ and $V$ is a random variable on $\Y$. In this Monty Hall problem we have $p_u=\frac{1}{3}$ as $\P\in\Pmod$ implies $\P[U=u]=\frac{1}{3}$ for all $u\in\X$. Lastly, the set $\Pmod$ is a subset of $\{\P|\forall u\in\X:\P[U=u]=p_u\}$.

Take $\P_b,\P_c\in\Pmod$ with $\P_i[V=i|U=a]=1$. Let $i\in\{b,c\}$ be arbitrary, then
\begin{align}
\P_i[V=i]&=\sum_{u=1}^3\P_i[V=i|U=u]\P_i[U=u]\\
&=\frac{1}{3}\left(2\cdot1+0\right)=\frac{2}{3}.
\end{align}
Thus the vectors $\left(\P_i[V=v]\right)_{v\in\Y}$ become $\left(\frac{1}{3},\frac{2}{3}\right)$ and $\left(\frac{2}{3},\frac{1}{3}\right)$, which are linearly independent.

Now we can directly apply Theorem~\ref{thm:DiscMainThm} to conclude the proof.
\end{proof}

Note that in Proposition~\ref{prop:DiscMontySafe} first a $u\in\{1,2,3\}$ is chosen and then a safe distribution for $\1_{\{U=u\}}|[V]$ is created. The distribution stating that door $a$ has probability $\frac{1}{3}$ of having the car is safe for $\MontyInd|[V]$. When applying the assumption that the game master never opens the door with the car, only one safe distribution $\Psafe$ exists: $\Psafe[U=b|V=c]=\Psafe[U=c|V=b]=\frac{2}{3}$. When the variant with Monty Hall being able to open the door with the car and the player not being allowed to switch to the opened door is considered, then any distribution $\Psafe$ with $\Psafe[U=a|V=v]=\frac{1}{3}$ for $v\in\Y$ is safe for $\MontyInd|[V]$ and plausible as a true distribution.

Proposition~\ref{prop:DiscMontySafe} also states that $\Psafe[U=b|V=v]=\frac{1}{3}$ for all $v\in\{b,c\}$ is safe for $\1_{\{U=b\}}|[V]$, which seems counter-intuitive when considering the probability $\Psafe[U=b|V=b]=\frac{1}{3}$. However, this arises from the fact that the event $\{U=b\}$ is not possible in all conditioned outcome spaces. Distribution $\Psafe$ must be safe against all $\P\in\Pmod$, thus also against the $\P$ that never produce conditioned outcome space $\{a,b\}$ in the event $\{U=a\}$. By acting like $\Psafe[U=b|V=b]=\frac{1}{3}$ is true, one is able to perform as well as all $\P\in\Pmod$, even though this $\Psafe$ is clearly not the correct distribution.

Returning to the discussion presented in Section~\ref{sec:MontyHist}, stating that the probability of the car being behind either door is fifty-fifty is not safe. When one acts like this distribution is true, the game master is easily able to pick a strategy that minimizes your chances of winning the car. Furthermore, the distribution $\Psafe_{\text{vS}}$ of Vos~Savant and Selvin  \cite{Marilyn90a,Selvin75a} that the probability of the car being behind door $a$ is always $\frac{1}{3}$ is not the correct measure-theoretic probability as well, unless one knows the quiz master uses a fair coin to decide which door to open when presented with a choice. The distribution $\Psafe_{\text{vS}}$ is however a safe distribution for the random variable $\MontyInd$ of interest, therefore Vos~Savant can act as if this distribution is true.

\subsubsection{Accuracy}
Suppose you are playing the Monty Hall game, which action optimizes the probability of winning the car? The answer is that you must always switch for a probability of $\frac{2}{3}$ of winning the car, which follows from applying Theorem~\ref{thm:DiscAccOpt}.

\begin{proposition}
Let $\X$, $\Y$, $U$, $V$ and $\Pmod$ be as in the Monty Hall problem. Consider the action $\alpha^*$ with $\alpha^*[U=a]=0$, then this $\alpha^*$ has accuracy
\begin{equation}
\acc^{\P}\left(\alpha^*\right)=\frac{2}{3}
\end{equation}
for all $\P\in\Pmod$. All other actions $\alpha\neq\alpha^*$ have $\acc^{\P}(\alpha)<\acc^{\P}(\alpha^*)$ for all $\P\in\Pmod$.
\end{proposition}
\begin{proof}
We want to apply Theorem~\ref{thm:DiscAccOpt}. Firstly in this case $\X'=\{a\}\subset\X$ is non-empty. Secondly, we have $\sum_{u\in\X'}p_u=p_a=\frac{1}{3}<\frac{1}{2}$. Theorem~\ref{thm:DiscAccOpt} states that when $\alpha^*$ puts zero mass on $\{U=a\}$, the accuracy for guessing $\MontyInd$ correctly is optimized with value
\begin{equation}
\acc\left(\alpha^*\right)=\max\left\{\frac{1}{3},\frac{2}{3}\right\}=\frac{2}{3}.
\end{equation}
\end{proof}

As in the dice game, the information given by the game master does not lead to a higher probability of winning the car. When always switching doors, the probability of winning the car becomes $\frac{2}{3}$ and this is the optimal probability you can have.

One can argue that Selvin and Vos Savant \cite{Selvin75a,Marilyn90a} initially answered this question on how to optimize the probability of winning the car and if they did, they gave the correct answer of $\frac{2}{3}$ as optimal probability obtained by always switching. However, as measure-theoretic conditional probability already has pointed out, it is not correct to state that the probability of winning the car when switching is $\frac{2}{3}$ thus the probability of the car being behind the other door is $\frac{2}{3}$.

\subsubsection{Final remarks}
The Monty Hall problem applied Theorems~\ref{thm:DiscMainThm} and \ref{thm:DiscAccOpt} in the same manner as the dice game. In the setting of the Monty Hall problem, it is safe to say that the car has probability $\frac{1}{3}$ of being behind the originally chosen door. Always switching therefore yields to an optimal probability of $\frac{2}{3}$ for winning the car.

The Monty Hall problem is essentially equal to the dice game. Here the paradoxical nature of the Monty Hall problem stems from the fact that most people use $\Omega'=\{a,b,c\}$ when first trying to solve the problem. However, in that case we cannot create a partition to apply traditional conditional probability or a $\sigma$-algebra to apply measure-theoretic conditional probability. This problem can only be spotted when performing conditional probability correctly, thus with stating the accompanying sub-$\sigma$-algebra. Then it becomes clear that $\Omega'$ must be extended with an observation space $\Y=\{b,c\}$. This makes the Monty Hall problem yet another argument for why the $\sigma$-algebra must be given when performing conditional probability.

\section{Boy or girl problem}\label{sec:DiscChildren}
The last problem we will study in this chapter is the boy or girl problem. Suppose you are approached by a stranger and you two get in conversation. The stranger tells you that he has two children and at least one of them is a girl. You want to know the probability of him having two daughters.

The first mistake made by many people is misinterpreting the setting. Suppose you ask the stranger if he has at least one girl. If he says `yes', the probability of him having two girls is $\frac{1}{3}$, as he can also have a boy and a girl and you do not know whether the earlier mentioned girl is his oldest child. This calculation is not controversial, as his answer `yes' implies the set of possibilities $\{bg, gb, gg\}$ and is answer `no' implies the set of possibilities $\{bb\}$. These sets do not overlap, thus traditional probability theory can be applied without any problems:
\begin{equation}
\P[\text{two girls}|\text{at least one girl}]=\frac{\P[\{gg\}]}{\P[\{bg, gb, gg\}]}=\frac{1}{3}.
\end{equation}
Suppose now the stranger tells you out of the blue that he has two children and at least one of them is a girl. Now the setting changes, as he could also have told you that at least one of his children is a boy. If he said that he has a girl, the set of possibilities is $\{bg, gb, gg\}$ and if he said that he has a boy, the set of possibilities is $\{bg, gb, bb\}$. Now there is an overlap of $\{bg, gb\}$ and traditional probability theory cannot be applied any-more. The probability of the stranger having two girls given he has at least one is now dilated between~$\frac{1}{3}$ and $1$, depending on the willingness of the stranger to reveal he has a girl. This dilation can be calculated in the same manner as is done for the Monty Hall problem in Equation~\ref{eq:DiscMontyDilate}.

\subsection{History of the problem}
The problem originates from 1959 and is conceived by Martin Gardner \cite{WikiChildren}. The problem was phrased with the following questions:
\begin{enumerate}
	\item Mr.~Jones has two children. The older child is a girl. What is the probability that both children are girls?
	\item Mr.~Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
\end{enumerate}
The first question is unambiguous and traditional conditional probability can be applied to calculate the probability being $\frac{1}{2}$, assuming boys and girls are born with equal rate. In 1982 Bar-Hillel and Falk \cite{BarHillel82} discussed both questions in a version posed by Falk in 1978. For the ambiguous second question, their setting is that the stranger is walking around with his son and after meeting him, he tells you he has two children. You then ask yourself what the probability of his second child being a boy is. This question is equal to the Gardner's second and the problem posed in the introduction of this section. Bar-Hillel and Falk first argue that the event given must not be `the stranger has at least one son'. It must be phrased as `the child we meet is a boy' as the latter event has a probability of $\frac{1}{2}$ instead of $\frac{3}{4}$ for the first event. Furthermore, they state that the probability space $\{bg,gb,bb,gg\}$ is insufficient for solving this problem, since it is not known how the boy the stranger walks with is picked. Using an argument of Jeffrey \cite{Jeffrey68} the sample space of Bar-Hillel and Falk's boy or girl problem is relabelled as $\{b_mb_h,b_mg_h,g_mb_h,g_mg_h\}$ where subscript $m$ indicates the child met and the subscript $h$ indicates the child at home. They argue that all four events have equal probability and meeting a boy leaves $\{b_mb_h,b_mg_h\}$, giving a probability of $\frac{1}{2}$ of the stranger having two boys. However, as they argue as well, the preference of the stranger taking his boy for a walk is not given, rendering the answer of $\frac{1}{2}$ still as incomplete. The importance of the phrasing of the question is also touched upon by Freund in 1965 \cite{Freund65}, who regards many versions of a slightly different card game and shows how each slight variation needs a different approach.

Most of the confusion on the problem originates from the statement of the problem and the inability of humans to correctly interpret such statement. Johnson-Laird et al.~\cite{Laird99} discussed the second of Gardner's questions and argued that most responders initially assume they have to calculate the probability of the second child being a girl, which they reason to be $\frac{1}{2}$. The responders however calculate an absolute probability where a conditional one is asked. Fox and Levav \cite{Fox04} argue that the initial answer of $\frac{1}{2}$ stems from a naive partitioning of the sample space. When stating Gardner's version of the problem to MBA students finishing their probability course, 85\% gave the answer of $\frac{1}{2}$ and 3.3\% gave the answer of $\frac{1}{3}$. Fox and Levav then rephrased the problem to `a stranger has two children and both are not a girl', where now only 39\% gave the answer of $\frac{1}{2}$ and 31\% gave the answer of $\frac{1}{3}$. They concluded that in Gardner's question the conditioned sample space $\{bb,bg\}$ is created by the participants, where $bg$ means the stranger has one boy and one girl. The students employ uniform probability on both outcomes, not taking the birth order of the children in the option $bg$ into account. When stating that no two children are a girl, the sample space created is more likely to be $\{bb,bg,gb\}$ with uniform probability. The statement that both children are not girls increases the likelihood of a responder taking the birth order of the children into account when thinking of all possible outcomes.

Mlodinow \cite{Mlodinow09} posed a different boy or girl problem analysed by Marks and Smith \cite{Marks11}. This different version is treated in Section~\ref{sec:DiscChild2}. The sample space $\{bb,gb,bg,gg\}$ is pruned by Mlodinow to be $\{bb,bg,gb\}$ after learning at least one child is a boy. Mark and Smith pointed out that this is a wrong application of conditional probability. The general mistake Mlodinow made is applying conditional probability on the boy or girl problem by pruning the sample space and analysing the resulting conditional space. You should rather ask yourself what the probability of learning at least one child being a boy is as in the most extreme case $bb$ cannot be an outcome. This taking into account the probability of observing information, also touched upon by Morgan et al.~\cite{Morgan91a}, is the key point of our analysis as well.

\subsection{Our thoughts on the problem}
We will now construct our model. Let $\X=\{bb,bg,gg\}$ be our outcome space and let $\Y=\{b,g\}$ be our observation space. Then $\Omega=\X\times\Y$ becomes our sample space with its power set $2^\Omega$ as its accompanying $\sigma$-algebra. We do not care about whether the girl is older than the boy in the event that the stranger has a boy and a girl, thus we remove the outcome $gb$. Let $U$ be a random variable on $\X$ denoting the children of the stranger and $V$ be a random variable on $\Y$ denoting whether the stranger told you he as at least one boy, event $\{V=b\}$, or at least one girl, event $\{V=g\}$. All probabilities $\P$ on $\Omega$ must be a member of
\begin{equation}
\Pmod=\left\{\P\middle|\begin{array}{l}
\P[U=bb]=\frac{1}{4},\P[U=bg]=\frac{1}{2},\\
\P[V=g|U=bb]=\P[V=b|U=gg]=0
\end{array}
\right\}.
\end{equation}

\subsubsection{Traditional and measure-theoretic conditional probability}
The analysis of the problem using traditional conditional probability will be very short here, as it is essentially the same analysis as in sections~\ref{sec:SafeDice} and \ref{sec:DiscMonty}.

First take a look at the naive approach using traditional conditional probability. Let $\Omega'=\{bg,bb,gg\}$ be the sample space with random variable $U'$. To calculate the probability of having a boy and a girl given there is a girl, most try to calculate $\P[U'=bg|U'\in\{bg,gg\}]$. They then conclude using Bayes' rule that
\begin{align}
\P[U'=bg|U'\in\{bg,gg\}]&=\frac{\P[U'\in\{bg,gg\}|U'=bg]\P[U'=bg]}{\sum_{u\in\Omega'}\P[U'\in\{bg,gg\}|U'=u]\P[U'=u]}\\
&=\frac{1\cdot\frac{1}{2}}{1\cdot\frac{1}{2}+0\cdot\frac{1}{4}+1\frac{1}{4}}=\frac{2}{3}.
\end{align}
This is the argument used by most that are convinced the probability is $\frac{2}{3}$. However, similar to the dice game and Monty Hall's problem, the conditioned sample space of the family composition given at least one girl is $\{bg,gg\}$ and the other conditioned sample space given at least one boy is $\{bg,bb\}$. These sets cannot form a partition of $\Omega'$ and traditional conditional probability can therefore not be applied.

Consider now the viewpoint of measure-theoretic probability. Let $\P\in\Pmod$ be arbitrary, then there is a $p\in[0,1]$ with $\P[V=g|U=bg]=p$. Using Bayes' rule we get
\begin{align}
\P[U=bg|V=g]&=\frac{\P[V=g|U=bg]\P[U=bg]}{\sum_{u\in\X}\P[V=g|U=u]\P[U=u]}\\
&=\frac{p\cdot\frac{1}{2}}{0\cdot\frac{1}{4}+p\cdot\frac{1}{2}+1\cdot\frac{1}{4}}=\frac{2p}{2p+1},
\end{align}
thus the probability of him having two girls given the stranger told you he has at least one daughter is dilated from $\frac{1}{3}$ to $1$. Furthermore, the probability of a boy and a girl given at least one girl being $\frac{2}{3}$ is only possible when $p=1$, that is the stranger always states he has a girl whenever possible. Therefore measure-theoretic conditional probability does not give a unique answer.

\subsubsection{Safe probability}

Since $U$ is not real-valued, we cannot calculate $\E_{\P}[U]$ for any $\P\in\Pmod$. Thus there is no notion of safety for $U|V$. We want to speak of safety for $\ChildInd|V$ as the event $\{U=bg\}$ arises both in conditioning on $\{V=b\}$ and on $\{V=g\}$.
\begin{proposition}\label{prop:DiscChildSafe}
Let $\X$, $\Y$, $\Omega$, $U$, $V$ and $\Pmod$ be as in the boy or girl problem. Let $\Psafe$ be a distribution on $\Omega$ with full support on $V$. Let $u\in\X$, then the following are equivalent:
\begin{enumerate}
    \item For all $v\in\Y$ we have $\Psafe[U=u|V=v]=p_u$.
    \item $\Psafe$ is safe for $\langle\1_{\{U=u\}}\rangle|\langle V\rangle$.
    \item $\Psafe$ is safe for $\1_{\{U=u\}}|[V]$.
\end{enumerate}
\end{proposition}
\begin{proof}
The proof is a direct application of Theorem~\ref{thm:DiscMainThm} and an almost direct copy of the proof of Proposition~\ref{prop:DiscMontySafe}. We will nevertheless state the full proof here.

The set $\X$ is of size $3$ and therefore countable and the set $\Y$ is of size $2$, which makes it finite. $U$ is a random variable on $\X$ and $V$ is a random variable on $\Y$. In this boy or girl problem we have $p_{bg}=\frac{1}{2}$ and $p_{bb}=p_{gg}=\frac{1}{4}$ by $\P\in\Pmod$. Lastly, the set $\Pmod$ is a subset of $\{\P|\forall u\in\X:\P[U=u]=p_u\}$.

Take $\P_{b},\P_{g}\in\Pmod$ with $\P_i[V=i|U=bg]=\frac{3}{4}$. Let $i\in\{b,g\}$ be arbitrary, then
\begin{align}
\P_i[V=i]&=\sum_{u\in\X}\P_i[V=i|U=u]\P_i[U=u]\\
&=\frac{1}{2}\cdot\frac{3}{4}+\frac{1}{4}\cdot\frac{1}{4}+0\cdot\frac{1}{4}=\frac{7}{16}.
\end{align}
Thus the vectors $\left(\P_i[V=v]\right)_{v\in\Y}$ become $\left(\frac{7}{16},\frac{9}{16}\right)$ and $\left(\frac{9}{16},\frac{7}{16}\right)$, which are linearly independent.

Now we can directly apply Theorem~\ref{thm:DiscMainThm} to conclude the proof.
\end{proof}

One example of a safe distribution for $\ChildInd|[V]$ is $\Psafe[U=bg|V=v]=\frac{1}{2}$ for all $v\in\Y$ and $\Psafe[U=gg|V=b]=\Psafe[U=bb|V=g]=0$. This is also the most `natural' safe distribution, as it is the only one where we still ensure that the stranger does not lie when stating he has at least one boy or at least one girl. It is important to note that only $\Psafe[U=bg|V=v]=\frac{1}{2}$ is required for $\Psafe$ to be safe for $\ChildInd|[V]$, all other conditional probabilities under $\Psafe$ can be chosen at will. For example, when $\Psafe[U=gg|V=g]=\frac{1}{6}$ and $\Psafe[U=bb|V=g]=\frac{1}{3}$ are chosen, then this $\Psafe$ is still safe for $\ChildInd|[V]$. It is however not natural to choose this $\Psafe$ as here we assume the stranger lies with probability $\frac{1}{3}$ after his statement he has at least one girl.

We now see that the probability distribution $\Psafe[U=bg|V=g]=\frac{2}{3}$ initially given by most people who answer this problem for the first time is not safe for $\ChildInd|[V]$. By pretending like the probability of the stranger having one boy and one girl disregarding his statement, the player obtains a distribution that seems to be true.

\subsubsection{Accuracy}
In this problem the optimal accuracy for guessing $\ChildInd$ correctly becomes interesting. When playing this game multiple times, it does not matter in which frequency the player guesses that the stranger has two children of different gender against two children of equal gender. According to Proposition~\ref{prop:DiscChildSafe} distribution $\Psafe$ with $\Psafe[U=bg|V=v]=\frac{1}{2}$ for all $v\in\Y$ is safe for $\ChildInd|[V]$, thus Theorem~\ref{thm:DiscAccOpt} gives us that all actions $\alpha$ optimize the accuracy of guessing $\ChildInd$ correctly.
\begin{proposition}
Let $\X$, $\Y$, $U$, $V$ and $\Pmod$ be as in the boy or girl problem. Let $\alpha$ be an arbitrary action, then this $\alpha$ has accuracy
\begin{equation}
\acc^{\P}(\alpha)=\frac{1}{2}
\end{equation}
for all $\P\in\Pmod$ on predicting $\ChildInd$.
\end{proposition}
\begin{proof}
We want to apply Theorem~\ref{thm:DiscAccOpt}. Firstly in this case $\X'=\{bg\}\subset\X$ is non-empty. Secondly, we have $\sum_{u\in\X'}p_u=p_{bg}=\frac{1}{2}$. Let $\P\in\Pmod$ be arbitrary, then Theorem~\ref{thm:DiscAccOpt} states that an action $\alpha^*$ with $\alpha^*[U=bg]\in[0,1]$ optimizes the accuracy for guessing $\ChildInd$ with
\begin{equation}
\acc^{\P}\left(\alpha^*\right)=\max\left\{\frac{1}{2},\frac{1}{2}\right\}=\frac{1}{2}.
\end{equation}
Since any action $\alpha$ has $\alpha[U=bg]\in[0,1]$, all actions have equal accuracy.
\end{proof}

We can therefore just guess randomly whether the stranger has two children of different gender or not. On average we will be correct 50\% of the times we find ourselves in this situation, which is the best we can do.

\subsection{Boy or girl problem 2.0}\label{sec:DiscChild2}
There is a second variant of the boy or girl problem stated in a lecture by Peter Grünwald \cite{Grunwald19} and is similar to a version posed by Mlodinow \cite{Mlodinow09} and analysed by Marks and Smith \cite{Marks11}. Grünwald stated a revised boy or girl problem as such:
\begin{enumerate}
    \item A person is picked uniformly at random from a population and a conversation is started.
    \item You ask if at least one of his children is a boy. Suppose his answer is `yes', then the probability of two boys is $\frac{1}{3}$. As stated in section~\ref{sec:DiscChildren}, this is not controversial.
    \item You ask if he can give you the name of his son. He can choose which son if he has two. The answer is `Martin'.
    \item You ask if Martin is his youngest child.
    \begin{itemize}
        \item[Yes:] If he answered `yes', then his youngest child is a boy and the probability of two boys is now $\frac{1}{2}$.
        \item[No:] If he answered `no', then his oldest child is a boy and the probability of two boys is now $\frac{1}{2}$.
    \end{itemize}
\end{enumerate}

The paradoxical nature of this problem is that after learning the name of a son of the stranger, the probability of him having two boys shifts from $\frac{1}{3}$ to $\frac{1}{2}$. However, as the stranger can state any string as name, just learning the name after knowing he has at least a boy must not influence the probability of the stranger having two boys.

Marks and Smith \cite{Marks11} use a Bayesian analysis to derive the probability to be $\frac{1}{2}$ without paradoxical results when the stranger mentions a child and name with no preference. Furthermore, when families with at least one boy considered, Mlodinow's analysis initially states that the probability of two boys is $\frac{1}{2}$ as well \cite{Marks11,Mlodinow09}. Marks and Smith however argue that the actual probability is $\frac{2-p}{4-p}$ with $p$ the probability of a boy being called Martin. Here Marks and Smith point out that the probability of obtaining the information that a boy is called Martin influences the probability of the stranger having two boys. In reality, such probability is small enough to approximate $\frac{2-p}{4-p}$ to $\frac{1}{2}$, taking $p\approx0$.

\subsubsection{Our thoughts on the problem}

This problem is stated more elaborately, but the underlying foundation is, as we will soon see, exactly equal to the Monty Hall problem. The outcome space after the stranger answering `yes' is $\{gb,bb\}$ and the outcome space after the stranger answering `no' is $\{bg, bb\}$. Now there is an overlap of $bb$. Furthermore, asking a boy's name and then asking if he is the youngest is essentially the same as asking whether his youngest child is a boy.

Therefore we can adapt our model to this situation. Let $\X=\{bg,gb,bb\}$ be the outcome space, let $\Y=\{y,o\}$ be the observation space and take $\Omega=\X\times\Y$ as sample space. Now we have both $bg$ and $gb$ in $\X$ as we now do care about whether the boy is the oldest of the children. The outcome $gg$ is not included in $\X$ as we assume he answered `yes' on the question whether he has at least one boy. Let $U$ be a random variable on $\X$ denoting the family composition of the stranger, where $\{U=bg\}$ is the event that his oldest child is a boy. Let $V$ be a random variable on $\Y$ denoting which of the children is a boy for sure, where $\{V=y\}$ is the event that the youngest child is a boy and $\{V=o\}$ is the event that the oldest child is a boy. Assume lastly that the stranger does not lie. All probability distributions $\P$ on $\X\times\Y$ must lie in
\begin{equation}
\Pmod=\left\{\P\middle|\forall u\in\X:\P[U=u]=\frac{1}{3},\P[V=bg|Y=y]=\P[V=gb|Y=o]=0\right\}.
\end{equation}
We now have the exact same model as for the Monty Hall problem in Section~\ref{sec:DiscMonty}. We will therefore only state the result of applying safe probability.

For safe probability we can now look at safety for $\ChildTwoInd|V$ as $bb$ is an element of the intersection of the outcome spaces conditioned on $\{V=y\}$ or $\{V=o\}$.
\begin{proposition}
Let $\X$, $\Y$, $\Omega$, $U$, $V$ and $\Pmod$ be as in the boy or girl problem 2.0. Let $\Psafe$ be a distribution on $\Omega$. Let $u\in\X$, then the following are equivalent:
\begin{enumerate}
    \item For all $v\in\Y$ we have $\Psafe[U=u|V=v]=\frac{1}{3}$.
    \item $\Psafe$ is safe for $\langle\1_{\{U=u\}}\rangle|\langle V\rangle$.
    \item $\Psafe$ is safe for $\1_{\{U=u\}}|[V]$.
\end{enumerate}
\end{proposition}
\begin{proof}
This proof is equal to the proof of Proposition~\ref{prop:DiscMontySafe} as the boy or girl problem 2.0 is equal to the Monty Hall problem.
\end{proof}

For this problem we can talk about traditional and measure-theoretic conditional probability and accuracy, but as the proof above already stated, this revised boy or girl problem is in essence equal to the Monty Hall problem. Therefore if the reader wants to dive further into this subject, he is kindly referred to Section~\ref{sec:DiscMonty}.

\subsection{Final remarks on both problems}
Both versions of the boy or girl problem are in essence or just exactly equal to the dice game and the Monty Hall problem. When in both versions of the boy or girl problem $\Omega'=\{bb,bg,gb,gg\}$ is chosen as sample space, then in the first version the conditioned sample spaces become $\{bg, gb, bb\}$ and $\{bg, gb, gg\}$ and in the second version the conditioned sample spaces become $\{bg, bb\}$ and $\{gb, bb\}$. In both versions the conditioned sample spaces cannot form a partition as there is an overlap in the sets, thus traditional conditional probability cannot be applied. The sample space must be extended, which can only be noticed when performing conditional probability correctly, thus with taking the accompanying $\sigma$-algebras into account.

In both versions there is no unique probability of the stranger either having two girls or having two boys. In the first version the probability of having two girls is dilated from $\frac{1}{3}$ to $1$ and in the second version the probability of him having two boys is dilated from $0$ to $\frac{1}{2}$. The latter can be proven by following the reasoning in Equation~\ref{eq:DiscMontyDilate}.

When Theorem~\ref{thm:DiscMainThm} is applied, we find that the first version has a safe distribution stating the probability of having a boy and a girl is $\frac{1}{2}$. In the second version, we found out that the safe distribution puts a probability of $\frac{1}{3}$ on him having two boys.

Most importantly, the second version of the paradox is actually equal to the Monty Hall problem. The first version can also be solved using the same techniques as the only difference between the problems is the set of probability distributions.

\section{Conclusion}\label{sec:DiscConcl}
After going through the dice game, the Monty Hall problem and the boy or girl problem, we see that Theorem~\ref{thm:DiscMainThm} can be applied nicely in each case. When a problem is founded upon a sample space $\Omega=\X\times\Y$ where $\X$ is the countable space of all possible outcomes, $\Y$ is the finite space of all statements the game master can make to the player and $\Omega$ is equipped with a large enough set of probabilities $\Pmod$ with fixed distribution on $\X$, we can apply Theorem~\ref{thm:DiscMainThm} to find safe distributions for single probabilities. The set of probability distributions must be large enough and fulfil an important requirement, however this requirement is almost always fulfilled when $\Pmod$ has $|\Y|$ distinct distributions with at least one having full support on $\Y$. When $\Pmod$ has fewer than $|\Y|$ distributions, then you should at first ask yourself whether $\Pmod$ is not too restrictive. Taking $\Pmod=\{\P\}$ as a single distribution makes this $\P\in\Pmod$ even safe for $U|V$, however this result is not quite helpful as in that case one would never think of using safe probability and just applies traditional conditional probability.

Furthermore, seemly distinct problems like the boy or girl problem and the Monty Hall problem are in essence equal. Therefore the studies on both of these problems can be combined, but with care as both problems do have different probability models.

The last and most important observation of this chapter is that the paradoxical results people find in the Monty Hall problem or the boy or girl problem all arise when $\Omega=\X$, thus when the outcome space is taken as sample space. In this case traditional conditional probability cannot be applied as the resulting outcome spaces after conditioning on the observed events do not form a partition.\\
This can be seen when the $\sigma$-algebra $2^{\Omega}$ is taken into account. Using loose notation, take $y_1,y_2\in\Y$ and suppose $x\in\X$ is present in both conditioned spaces $\X|y_1$ and $\X|y_2$. Then for computing the conditional probability of $x$ you need to condition on $\{\{y_1\},\{y_2\}\}$, which is neither an element of $2^\Omega$ nor does it have a known probability. Thus a sub-$\sigma$-algebra sufficient for conditioning cannot be found.\\
This mistake is easily spotted when taking the $\sigma$-algebra into account when performing conditional probability. Therefore these problems are examples as well of why when applying conditional probability the accompanying $\sigma$-algebras must always be provided.

\chapter{The two envelopes problem}\label{chap:TwoEnvelope}
The last paradox we will consider is the two envelope problem. This problem is first stated in 1953 by Kraitchik \cite{Kraitchik53} as the necktie problem. In this problem each person claims to have the finer necktie. A judge must make a decision, where the winner has to give his necktie to the loser. The argument of each contestant is: `I know what my tie is worth. I may lose it, but I may also win a better one, so the game is to my advantage.' However, a game that is favourable for both players cannot exist. Furthermore, the game is symmetrical in terms of the contestants, thus Kraitchik states that the chance of winning is actually fifty-fifty.

However, \cite{Kraitchik53} only provides a discussion and no rigorous proofs. In 1989 Nalebuff \cite{Nalebuff89} states the two envelope problem with many variants and provides a wide discussion from various sources. We will study one of the variants of the paradox, in my opinion the one with the most interesting results, and show how all paradoxical statements are incorrect. The reasoning will stay valid for various extensions to the problem. Those extensions will be treated briefly in Section~\ref{sec:EnvelopeAli}.

Pick the following version of the two envelope problem from \cite{Nalebuff89}. There are two envelopes where one contains a fixed amount of money $y$ and the other contains either value $\frac{1}{2}y$ or value $2y$ determined by the flip of a fair coin. The player can choose between both envelopes and call the chosen envelope $A$. The other envelope is then named envelope $B$. After choosing, the player is given the choice whether he wants to switch. Reasoning that envelope~$B$ contains $\frac{1}{2}a$ with probability $\frac{1}{2}$ and contains $2a$ with probability $\frac{1}{2}$ with $a$ the observed value, he computes the expected value of envelope~$B$ to be
\begin{equation}\label{eq:EnvelopeIntroWrong}
\E[B]=\frac{1}{2}\cdot\frac{1}{2}a+\frac{1}{2}\cdot2a=\frac{5}{4}a.
\end{equation}
Thus the player always wants to switch when initially picking envelope~$A$. However, by symmetry, if the player has chosen envelope~$B$ and observed $b$, then to his knowledge $A$ contains either $\frac{1}{2}b$ or $2b$, thus he wants to switch as well. It is however not possible to have a greater expected value in the other envelope in all cases, as the total amount of money in the whole game is fixed. This result is called the \emph{two envelope paradox}. Some even reason that if the player possesses no memory, he wants to switch infinitely many times, believing to gain infinite money.

Most people try to solve the problem by fixing the minimal value of the envelopes to be $x=\min\{A,B\}$ and then prove that $\E[A]=\E[B]$, concluding there is nothing paradoxical going on. This `solution' will be discussed in Section~\ref{sec:EnvelopeNaive} and corrects the calculation in Equation~\ref{eq:EnvelopeIntroWrong}. It is however incomplete, as firstly the player does not know the minimal value $x$ when observing the contents of envelope $A$ and secondly $x$ can be drawn from a probability distribution on $(0,\infty)$. This will be formalized in Section~\ref{sec:EnvelopeFormal}, where we will create a probability space that enables $x$ to take on all real positive values. In Section~\ref{sec:EnvelopePrior} we will investigate certain prior distributions on the lowest value $x$ of both envelopes and their influence on the problem. Since the correct probability distribution on $x$ is not known, safe probability will be applied in Section~\ref{sec:EnvelopeSafe}. The result however is that a safe distribution in most cases cannot exist. Section~\ref{sec:EnvelopeSwitching} introduces Cover switching, which provides a switching strategy that does on average improve the amount of money won. Lastly, an alternative version called the Ali~Baba problem will be introduced in Section~\ref{sec:EnvelopeAli} and shortly discussed with many other variants of the two envelope problem. Eventually we will wrap up in Section~\ref{sec:EnvelopeConcl} giving concluding remarks.

\section{Fixing the minimal value of the envelopes}\label{sec:EnvelopeNaive}
First we need to address the flaw in the reasoning of the two envelope paradox. The biggest flaw is in Equation~\ref{eq:EnvelopeIntroWrong}, where $\E[B]$ is calculated with the following method:
\begin{equation}
\E[B]=\frac{A}{2}\P\left[B=\frac{A}{2}\right]+2A\P[B=2A].
\end{equation}
This equation is however a misapplication of notation. Let $A$ and $B$ be the random variables denoting the values of respectively envelopes $A$ and $B$.  In correct notation, the calculation becomes
\begin{align}\label{eq:EnvelopeIntroCorrect}
\E[B]&=\E[B|A<B]\P[A<B]+\E[B|A>B]\P[A>B]\\
&=\frac{1}{2}\E[2A|A<B]+\frac{1}{2}\E\left[\frac{1}{2}A|A>B\right]\\
&=\E[A|A<B]+\frac{1}{4}\E[A|A>B].
\end{align}
The problem now is that both expectation values are unknown. Unless we know the expectation values, no conclusions can be drawn from the above calculation. For further reading on the above calculation and an in-depth discussion on all the mistakes made when performing this calculation, the reader is referred to the article `The two envelope problem: there is no conundrum' by O'Brien and Mitchell \cite{Brien14}.

An initial proposal for a solution fixes the minimal value put in envelopes~$A$ and $B$ to be $\min\{A,B\}=x$. Many do not accept this as a satisfying solution as for the player the value of $x$ is unknown. However, this proposal does remove the paradoxical nature of the results of Equation~\ref{eq:EnvelopeIntroWrong}, which is why we will discuss it shortly.

Let $x=\min\{A,B\}$ be the lowest value of the envelopes. The envelopes can only obtain two values: $x$ and $2x$. The probability of $A$ having $x$ is $\P[A=x]=\frac{1}{2}$, thus the probability space is
\begin{equation}
\left(\Omega=\{(x,2x),(2x,x)\},\F=2^\Omega,\P\right)
\end{equation}
with $\P$  the uniform probability measure on $\Omega$. As suggested by \cite{Schwitzgebel08,Brien14}, this probability space can now be applied to Equation~\ref{eq:EnvelopeIntroCorrect} which gives
\begin{align}
\E[B]&=\E[B|A<B]\P[A<B]+\E[B|A>B]\P[A>B]\\
&=2x\cdot\frac{1}{2}+x\cdot\frac{1}{2}=\frac{3}{2}x.
\end{align}
We can calculate using the same method that $\E[A]=\frac{3}{2}x$, thus the expected amount of money in either envelope is equal. According to this approach switching does not increase one's chances of getting more money.

Note again that the paradoxical statement of $\E[B]=\frac{5}{4}\E[A]$ is proven to be false as when fixing the minimal value of the envelopes $\E[B]=\E[A]$ is quickly achieved. However, the minimal value $x=\min\{A,B\}$ is not known. Even though the above calculations are mathematically correct, they do not provide a satisfying solution to the two envelopes problem.

\section{The problem formalized}\label{sec:EnvelopeFormal}
Actually we are interested in calculating $\E[B|A=a]$, the expectation value of envelope $B$ given $a$ is observed in envelope $A$, and this calculation must be valid for all $a\in(0,\infty)$. For this, we need to create a probability space for which the event $\{A=a\}$ is measurable in the union of the supports of $(X,2X)$ and $(2X,X)$ with $X=\min\{A,B\}$ as random variable. If the event $\{A=a\}$ is not measurable, we cannot condition on it.

Therefore we need to create a different probability space where we can always condition on the event $\{A=a\}$. Let $\X=(0,\infty)$ be the space containing all possible lowest values $x$ of the envelopes. Equip $\X$ with $\F_{\X}=\B((0,\infty))$, the Borel $\sigma$-algebra on $(0,\infty)$. Let $\Y=\left\{(x,2x),(2x,x)\in(0,\infty)^2\middle|x\in\X\right\}$ be the lines $\xi\mapsto2\xi$ and $\xi\mapsto\frac{1}{2}\xi$ where all possible values in the envelopes lie. The set $\Y$ is equipped with $\sigma$-algebra $\F_{\Y}=\B\left(\Y\right)$, the Borel-$\sigma$-algebra of the lines $\xi\mapsto2\xi$ and $\xi\mapsto\frac{1}{2}\xi$. Let $\Omega=\X\times\Y$ be our sample space equipped with $\F=\F_{\X}\times\F_{\Y}$ as $\sigma$-algebra.\\
Let $X$ be an $\X$-valued continuous random variable denoting the lowest value of the envelopes and $Y$ be a $\Y$-valued random variable denoting the actual values. All probability distributions $\P$ on $(\Omega,\F)$ must be a member of
\begin{equation}
\Pmod=\left\{\P\middle|\P[Y=(x,2x)|X=x]=\P[Y=(2x,x)|X=x]=\frac{1}{2},\E_{\P}[X]<\infty\right\}.
\end{equation}
The finiteness of $\E_{\P}[X]$ is addressed in Remark~\ref{rem:EnvelopePriorFinite}. All probability spaces we view as valid for the two envelope problem are now of the form
\begin{equation}
(\Omega=\X\times\Y,\F=\F_{\X}\times\F_{\Y},\P)
\end{equation}
with $\P\in\Pmod$. Let $\pi_i\colon\Y\to(0,\infty)$ with $i\in\{1,2\}$ be the projection on the $i$-th coordinate. Let $A=\pi_1(Y)$ and $B=\pi_2(Y)$ be the random variables denoting the values in respectively envelope~$A$ and $B$. Note that all $\P\in\Pmod$ are fixed on $Y$ when the value of $X$ is known. Thus $\Pmod$ is essentially implied by all the probability distributions on $\X$ where $X$ has finite expectation.

Assume that without loss of generality the player initially chooses envelope~$A$, then we would like to know the distribution of $B|A$. Suppose a value $a$ is observed in envelope $A$. The reasoning in Equation \ref{eq:EnvelopeIntroWrong} would be valid if
\begin{equation}\label{eq:EnvelopeProbHalve}
\P[B=2A|A=a]=\P\left[B=\frac{A}{2}\middle|A=a\right]=\frac{1}{2}
\end{equation}
holds for all $a\in(0,\infty)$. However, is this $\P$ still an element of $\Pmod$? As already pointed out by many \cite{Christensen92,Christensen93b,Navara17,Tzur18}, this is not the case. Equation~\ref{eq:EnvelopeProbHalve} implies that firstly $X$ is distributed uniformly on its support and secondly that the support of $X$ is unbounded both below and above. Therefore, $X$ is distributed uniformly on an infinite set, which is not possible.\\
Now we have seen that we are not allowed to state the probability of the other envelope having twice the observed value being equal to the probability of the other envelope having half the observed value.\footnote{If the support of $X$ is of finite size, this statement can be possible for all observed $a$ apart from the minimal and maximal value of $\supp(A)$. This is however a specific case of the two envelope problem, which we will not look into.}

\begin{remark}\label{rem:EnvelopePriorFinite}
There are multiple reasons why we should insist that $\E_{\P}[X]<\infty$ is finite for all $\P\in\Pmod$. Firstly when $x$ is drawn from a distribution $\P$ with $\E_{\P}[B]=\E_{\P}[A]=\infty$, then the law of total expectation $\E_{\P}[B]=\E_{\P}[\E_{\P}[B|A]]$ is in some cases not valid any-more \cite{Tzur18}. Secondly, in the case of infinite expectation the two envelope problem starts to behave like the St.~Petersburg problem\footnote{This problem concerns a game where a fair coin is flipped until it gives heads. Suppose the coin is flipped $n$ times, then $2^n$ is paid out. The expected value of the amount of money paid out is infinite while there is a probability of 87,5\% of a payout of at most 8, which was viewed as paradoxical when the problem was conceived in 1713. In present times the problem is merely an example of what can occur when dealing with infinite expectation values.} \cite{Brams95,Broome95,Tzur18}. This problem does not concern conditional probability and will therefore not be treated in this thesis. Interested readers are referred to William Feller, who described in 1950 the problem with an accessible discussion and solution \cite{Feller50} and provided in 1971 a general theorem for similar problems \cite{Feller71}. Thirdly, in our case, a prior with infinite expectation value on $X$ will lead to the paradoxical result of always having to switch, as will be shown in Section~\ref{sec:EnvelopePriorInfluence}.
\end{remark}

\section{Prior on envelope values}\label{sec:EnvelopePrior}
Recall that all probability distributions usable in the two envelope problem must be of the set
\begin{equation}
\Pmod=\left\{\P\middle|\P[Y=(x,2x)|X=x]=\P[Y=(2x,x)|X=x]=\frac{1}{2},\E_{\P}[X]<\infty\right\}.
\end{equation}
As we now do not have more information on the problem, any distribution from $\Pmod$ can be the correct one. However, a player playing the two envelope game probably has a suspicion on how $X$ is distributed. This suspicion can be modelled as either a single distribution $\P\in\Pmod$ or a subset $\mathcal{P}\subset\Pmod$ of distributions. Therefore, we can introduce the following definition of a prior.

\begin{definition}[Prior]
A distribution $\P\in\Pmod$ is called a \emph{prior} on the envelope values and set $P\subset\Pmod$ is called a \emph{set of priors}. The density function of prior $\P\in\Pmod$ on $\X$, the lowest value of the envelopes, will be denoted by $f\colon\X\to[0,\infty)$.
\end{definition}

Note that, as mentioned in Section~\ref{sec:EnvelopeFormal}, there are no two $\P_1,\P_2\in\Pmod$ that almost surely are equally distributed on $\X$ but differently on $\Y$, as the conditional distribution $Y|X$ is fixed in $\Pmod$. Thus the distribution on $\Omega$ of two $\P$'s in $\Pmod$ are almost surely equal if and only if those $\P$'s are almost surely equal on~$\X$.

\subsection{Prior with no information}
First consider the setting where no information on $\Pmod$ is known. An exhaustive analysis of this case is done by Albers, Kooi and Schaafsma \cite{Albers05}. They give three noteworthy examples for trying to solve the two envelope problem while nothing is known about prior $f$, however they conclude that the paradox is just hopeless to tackle without being giving more information.

\begin{example}[Conditional expectation]
Let $\P\in\Pmod$ be arbitrary and suppose $\P[A>B|A=a]=p\in[0,1]$ holds, thus the probability of receiving the higher envelope given you observed $A=a$ equals $p$. Here we assume the player knows of value $p$ as well. The expected value of $B$ given $A=a$ becomes
\begin{align}
\E_{\P}[B|A=a]&=\frac{a}{2}\P\left[B=\frac{a}{2}\middle|A=a\right]+2a\P[B=2a|A=a]\\
&=\frac{a}{2}p+2a(1-p)=a\left(2-\frac{3}{2}p\right).
\end{align}
Therefore $\E_{\P}[B|A=a]<a$ holds when $p>\frac{2}{3}$. Thus when it is likely that you have the higher envelope, switching is not advised.

However, note that
\begin{equation}
\P[A>B|A=a]=\E\left[\1_{\{A>B\}}\middle|A=a\right]
\end{equation}
holds. Following \cite{Albers05}, the estimation of $\P[A>B|A=a]$ is replaced by the prediction of $\1_{\{A>B\}}$, which is not possible without having more information on $\P$.
\end{example}
\begin{example}[Entropy analysis]
Another route is to estimate the entropy of the distribution on the other envelope. If we can maximize this entropy, an estimated prior will follow giving us advice on whether we want to switch.

Let $\mu\in(0,\infty)$ be arbitrary and let $\mathcal{P}\subset\Pmod$ be such that $\E_{\P}[B]=\mu$ holds for all $\P\in \mathcal{P}$. This will be the only assumption we will make on the prior. We can now construct an estimate $f'$ for the prior $f$ such that the entropy
\begin{equation}
H(f')=\begin{cases}
-\sum_{y=1}^\infty f'(y)\log f'(y),&X\text{ discrete},\\
-\int_{y=0}^\infty f'(y)\log f'(y),&X\text{ continuous},
\end{cases}
\end{equation}
is maximized with $\E[B]=\mu$ as restriction.

Following this route \cite{Albers05} found out that in the discrete case the envelope must be swapped for odd $a\in\N\setminus2\N$ and not swapped for even $a\in2\N$. In the continuous case, the player must always swap.

This answer raises a lot of questions, such as why the player should in the discrete case favour values $1001$ and $1003$ and discard value $1002$. No one playing the two envelope problem would be convinced that this is a working strategy. Furthermore, as we will see in Section~\ref{sec:EnvelopeSwitching}, in the continuous case (randomized) switching strategies exist yielding higher gain for all priors $f$ than always switching the envelope. Thus the continuous strategy using entropy can be called insufficient, as strategies exist that outperform `always switching'.
\end{example}

There is a third game-theoretic example given by \cite{Albers05}, however even Albers et al.~state multiple times that this perspective is unnecessarily complicated and will not be helpful.

The overall conclusion of \cite{Albers05} is that if insufficient information on the prior is known, a mathematician must be satisfied with the fact that he just cannot solve the problem. There is not enough relevant information to make justified decisions or create useful strategies.

\subsection{Investigating different priors}\label{sec:EnvelopePriorInfluence}
Take a distribution $\P$ on $\Omega$ and let $f$ be its accompanying prior on $X$. We will take a look at various variations of $f$. We will observe that for certain priors switching will always be advised, but the expectation $\E_{\P}[X]$ will be infinite as well. When $\E_{\P}[X]$ is finite, then useful switching strategies will arise.

\begin{example}[Navara and Šindelár (2017)]\label{ex:EnvelopePriorGain}
The following example is taken from \cite{Navara17}. Assume that $f$ is a discrete prior on $X$ and $\P$ its accompanying distribution on $\Omega$. We want to know how much we expect to gain by switching given value $a$ is in envelope $A$. Define the gain $G=B-A$. It is easy to see that
\begin{equation}
\E_{\P}[G]=\E_{\P}[B]-\E_{\P}[A]=\frac{3}{2}\E_{\P}[X]-\frac{3}{2}\E_{\P}[X]=0.
\end{equation}
The possible values of the gain given $A=a$ are $G=a$ or $G=-\frac{a}{2}$. The probability distribution of $G|A=a$ is according to \cite{Navara17} given by
\begin{equation}
\P[G=g|A=a]=\begin{cases}
\frac{f(a)}{f(a)+f\left(\frac{a}{2}\right)},&g=a,\\
\frac{f\left(\frac{a}{2}\right)}{f(a)+f\left(\frac{a}{2}\right)},&g=-\frac{a}{2}.
\end{cases}
\end{equation}
The expectation value of $G|A=a$ is then given by
\begin{equation}
\E_{\P}[G|A=a]=\frac{af(a)-\frac{a}{2}f\left(\frac{a}{2}\right)}{f(a)+f\left(\frac{a}{2}\right)}.
\end{equation}
Consider now the prior $f(2^t)=\frac{q^t}{1-q}$, then three different cases can be distinguished:
\begin{enumerate}
    \item[$q<\frac{1}{2}$:] Take $q<\frac{1}{2}$, then $\E_{\P}[X]$ exists and is equal to $\E_{\P}[X]=(1-q)^{-1}(1-2q)^{-1}$. In this case we have
    \begin{equation}
    \E_{\P}\left[G\middle|A=2^t\right]=\begin{cases}2^{t-1}\frac{2q-1}{q+1},&t\in\N_{\geq 1}\\1,&t=0.\end{cases}
    \end{equation}
    Therefore with this prior switching is only recommended when one observes a value of $1$ in his envelope, as $\E_{\P}[G|A=2^t]<0$ is negative for all $t\in\N_{\geq 1}$.
    \item[$q=\frac{1}{2}$:] Take $q=\frac{1}{2}$, then $\E_{\P}[X]$ does not exist any-more. The expected gain now simplifies to $\E_{\P}[G|A=2^t]=\1_{\{0\}}(t)$ for all $t\in\N_{\geq0}$, thus there is only gain on switching when $A=1$.
    \item[$q>\frac{1}{2}$:] Take $q>\frac{1}{2}$, then $\E_{\P}[X]$ does not exist as well. The conditional gain $G|A=2^t$ is still valid and equals
    \begin{equation}
    \E_{\P}\left[G\middle|A=2^t\right]=\begin{cases}2^{t-1}\frac{2q-1}{q+1},&t\in\N_{\geq 1},\\1,&t=0,\end{cases}
    \end{equation}
    which is strictly positive for all $t\in\N_{\geq0}$. Thus the player is motivated to always switch, which is a paradoxical result.
\end{enumerate}

The $q$ controlling the prior $f$ can lead to various results, from non-paradoxical advise to only switch when obtaining the smallest value $1$ knowing the other envelope must contain $2$, to the original paradox of always having to switch no matter the contents of $A$. The difference with the original paradox is that the paradoxical result of this example is supported by a prior on $X$, not by wrongly performing the calculations involved with the problem.
\end{example}

\begin{example}[Broome (1995)]
This example is taken from \cite{Broome95} and is given by many as a lighting example of when a prior on $X$ can have infinite expectation. Consider first the discrete case. Let $f(2^n)=\frac{2^n}{3^{n+1}}$ with $n\in\N_{\geq 0}$ be a prior. We now want to know the expected value in envelope $B$. The trivial case is $\E[B|A=1]=2$ as the only possible value in envelope $B$ is $2$ when observing $1$.\\
Suppose we observe a value $a\neq1$. Then $B=2a$ is possible if and only if $X=A=a$. When observing envelope $A=a$, then either $X=a$ or $X=\frac{a}{2}$ can hold. Therefore, following \cite{Broome95},
\begin{align}
\P[B=2a|A=a]&=\P\left[X=a\middle|X=a\lor X=\frac{a}{2}\right]\\
&=\frac{\frac{2^n}{3^{n+1}}}{\frac{2^n}{3^{n+1}}+\frac{2^{n-1}}{3^{n}}}=\frac{2}{5}.
\end{align}
We have $\P\left[B=\frac{a}{2}\middle|A=a\right]=1-\P[B=2a|A=a]=\frac{3}{5}$ as well, giving an expected value of envelope $B$ of
\begin{equation}
\E[B|A=a]=\frac{2}{5}\cdot2a+\frac{3}{5}\cdot\frac{a}{2}=\frac{11}{10}a.
\end{equation}
The paradoxical conclusion is that the player must always switch, no matter the contents of his envelope. The reason why $\E[B|A=a]>a$ for all $a=2^n$ with $n\in\N_{\geq0}$ is possible, is that $\E[X]=\frac{1}{3}\sum_{n=0}^\infty\left(\frac{4}{3}\right)^n=\infty$ is infinite.

Until now we have only considered discrete distributions, but paradoxical conclusions can result from continuous distributions as well. Let $f$ be a continuous prior on $X$, then the conditional expectation on $B|A=a$ is
\begin{equation}
\E[B|A=a]=a\frac{8f(a)+f\left(\frac{a}{2}\right)}{4f(a)+2f\left(\frac{a}{2}\right)}
\end{equation}
as is derived by \cite{Broome95}. Take density function $f(x)=(x+1)^{-2}$ with $x\in(0,\infty)$, then the conditional expectation of $B|A=a$ becomes
\begin{equation}
\E[B|A=a]=a\frac{2(a+2)^2+(a+1)^2}{(a+2)^2+2(a+1)^2}>a
\end{equation}
for all $a\in(0,\infty)$. Here the situation that switching is always advised arises as well. Note that the expectation value of the prior $\E[X]=\int_0^\infty \frac{x}{(x+1)^2}dx=\infty$ is infinite, explaining why $\E[B|A=a]>a$ for all $a\in(0,\infty)$ is again possible.
\end{example}

Take another look at the gain $G$ given $A=a$, defined in Example~\ref{ex:EnvelopePriorGain}. When considering the conditional expectation $\E[G|A=a]$, one can quantify when a distribution yields to paradoxical results. For discrete priors, $\E[G|A=a]$ equals \cite{Navara17,Tzur18}
\begin{equation}
\E[G|A=a]=\frac{af(a)-\frac{a}{2}f\left(\frac{a}{2}\right)}{f(a)+f\left(\frac{a}{2}\right)}
\end{equation}
and where it gives rise to the following definition.
\begin{definition}[Discrete paradoxical distribution]
A discrete prior $f$ on $X$ with accompanying $\P\in\Pmod$ is called \emph{paradoxical} if $f(x)>\frac{1}{2}f\left(\frac{x}{2}\right)$ holds for all $x\in\supp_{\P}(X)$.
\end{definition}
\begin{lemma}
Let $f$ be a discrete paradoxical prior on $X$, then the expectation value $\E[X]=\infty$ is infinite.
\end{lemma}
\begin{proof}
This proof is taken from \cite{Tzur18}. Let $x$ be a possible realization of $X$, then $2^k x$ with $k\in\N$ are possible realizations of $X$ as well by the setting of the two envelope problem. Note that $2^{k+1}xf(2^{k+1} x)>2^k xf(2^kx)$ holds for all $x\in\N$, implying an unbounded $\E[X]=\sum_{k=0}^\infty k f(k)>\sum_{k=0}^\infty 2^k xf(2^k x)$ as all terms $2^kxf(2^kx)$ will increase in $k$.
\end{proof}

The requirement for a continuous paradoxical distribution is slightly different. Brams, Kilgour and Blachman \cite{Brams95,Christensen96} calculated when the conditional expectation $\E[B|A=a]>a$ holds for continuous priors.

\begin{definition}[Continuous paradoxical distribution]
A continuous prior $f$ on $X$ with accompanying $\P\in\Pmod$ is called \emph{paradoxical} if $f(x)>\frac{1}{4}f\left(\frac{x}{2}\right)$ holds for all $x\in\supp_{\P}(X)$.
\end{definition}

As mentioned in Section~\ref{sec:EnvelopeFormal}, we will only consider non-paradoxical distributions from now on. It is now clear that, when having full information, paradoxical distributions still yield unexpected results. This is contributed to the expectation value of the prior being undefined, not to a wrong interpretation of the problem. When having full information non-paradoxical priors yield to well-defined and useful switching strategies.

\subsection{Optimal solution}
Suppose a non-paradoxical prior on $X$ is known, does an optimal switching strategy exist? This question is asked and answered by \cite{Christensen92} and further discussed by \cite{Christensen93a,Christensen93b,Christensen94,Christensen96}. Originally, \cite{Christensen92} proposed a solution for all general priors, however \cite{Christensen96} pointed out this is incorrect for continuous distributions. Here we will combine the discussions and provide a resolution.

\subsubsection{Discrete priors}
Let $f$ be a discrete prior on $X$ with distribution $\P\in\Pmod$, then \cite{Christensen92} states that
\begin{align}
\P\left[X=a\middle|A=a\right]&=\frac{\P[A=a|X=a]f(a)}{\P[A=a|X=a]f(a)+\P\left[A=a\middle|X=\frac{a}{2}\right]f\left(\frac{a}{2}\right)}\\
&=\frac{f(a)}{f(a)+f\left(\frac{a}{2}\right)},\\
\P\left[X=\frac{a}{2}\middle|A=a\right]&=\frac{\P\left[A=a\middle|X=\frac{a}{2}\right]f\left(\frac{a}{2}\right)}{\P[A=a|X=a]f(a)+\P[A=a|X=2a]f(2a)}\\
&=\frac{f\left(\frac{a}{2}\right)}{f(a)+f\left(\frac{a}{2}\right)}.
\end{align}
We can now compute $\E[B|A=a]$ to be
\begin{align}
\E[B|A=a]&=\frac{a}{2}\P\left[X=\frac{a}{2}\middle|A=a\right]+2a\P\left[X=a\middle|A=a\right]\\
&=\frac{af\left(\frac{a}{2}\right)+4af(a)}{2f(a)+2f\left(\frac{a}{2}\right)}.
\end{align}
Obviously when $\E[B|A=a]>a$ holds, switching is advised. This happens when $f\left(\frac{a}{2}\right)<2f(a)$.

Does a discrete prior $f$ exist for which the requirement $\E[B|A=a]\leq a$ holds for all $a\in\supp_{\P}(A)$, thus where the player must never switch? The answer is no, as the following proposition from \cite{Brams95} points out.
\begin{proposition}[Brams and Kilgour (1995)]
There is no discrete prior $f$ on $X$ for which $\E[B|A=a]\leq a$ holds for all $a\in\supp_{\P}(A)$.
\end{proposition}
\begin{proof}
The proof is taken from \cite{Brams95}. Assume $f$ is a prior with $\E[B|A=a]\leq a$ for all $a\in\supp_{\P}(A)$. In order for $\E[B|A=a]\leq a$ we must have $f\left(\frac{a}{2}\right)\geq2f(a)$. Let $a\in\supp_{\P}(A)$ be such that $f(a)>0$, then $f\left(a\right)\geq 2^{n}f(2^na)$ holds for all $n\in\Z$, leading to $f(a)\to0$ as $n\to-\infty$. We cannot have $f(a)$ being arbitrarily close to zero and strictly positive at the same time, rendering the assumption of the existence of $f$ false.
\end{proof}

On the contrary, does a discrete prior $f$ exist with $\E[B|A=a]>a$ for all $a\in\supp_{\P}(A)$? Then $f\left(\frac{a}{2}\right)>f(a)$ must hold for all $a\in\supp_{\P}(A)$, which is achieved for all paradoxical distributions since $\supp_{\P}(X)\subset\supp_{\P}(A)$. Various examples can be found in \cite{Christensen93a,Linzer94,Broome95} and previously in Section~\ref{sec:EnvelopePriorInfluence}, with $f(2^n)=\frac{2^n}{3^{n+1}}$ as the one most used. In this case a new paradox arises, where switching is always advised. Note that Christensen and Utts do not agree with this occurrence as being a paradox as stated in \cite{Christensen93a}, as $\E[B|A=a]$ is still finite and can be compared with $a$.

\subsubsection{Continuous case}
Let $f$ be a continuous prior on $X$ with distribution $\P\in\Pmod$. First \cite{Christensen92} states that
\begin{equation}
\E[B|A=a]=\frac{af\left(\frac{a}{2}\right)+4af(a)}{2f(a)+2f\left(\frac{a}{2}\right)}
\end{equation}
holds for continuous priors as well, however \cite{Christensen96} points out this is not correct.

Take a look at the following analysis from \cite{Christensen96}. Let $Y=\frac{X}{2}$ and let $g$ be a probability density function on $Y$. Then we have $\P[X\leq x]=\P\left[Y\leq \frac{x}{2}\right]$, thus $F(x)=G\left(\frac{x}{2}\right)$ where $F$ and $G$ are the cumulative density functions of $X$ and $Y$ respectively. Differentiating for the density functions yields $f(x)=\frac{1}{2}g\left(\frac{x}{2}\right)$. Now, analogous to the discrete case, we have
\begin{align}
\P[X=a|A=a]&=\frac{f(x)}{f(x)+g(x)}=\frac{f(x)}{f(x)+\frac{1}{2}f\left(\frac{x}{2}\right)},\\
\P\left[X=\frac{a}{2}\middle|A=a\right]&=\frac{g(x)}{f(x)+g(x)}=\frac{\frac{1}{2}f\left(\frac{x}{2}\right)}{f(x)+\frac{1}{2}f\left(\frac{x}{2}\right)}.
\end{align}
The expectation of $B|A=a$ now becomes
\begin{align}
\E[B|A=a]&=\frac{a}{2}\P\left[X=\frac{a}{2}\middle|A=a\right]+2a\P\left[X=a\middle|A=a\right]\\
&=\frac{af\left(\frac{a}{2}\right)+8af(a)}{4f(a)+2f\left(\frac{a}{2}\right)}.
\end{align}
This expectation value is different from the version in \cite{Christensen96}, as they made a mistake mixing up the probabilities while calculating the expectation value. Nevertheless their conclusion that now $4f(a)>f\left(\frac{a}{2}\right)$ must hold in order to get $\E[B|A=a]>a$ is still correct.

The same proposition as in the discrete case that $\E[B|A=a]>a$ must happen for at least one $a\in\supp_{\P}(A)$ is valid here as well.
\begin{proposition}[Brams and Kilgour (1995)]
There is no continuous prior $f$ on $X$ with accompanying prior $\P\in\Pmod$ for which $\E[B|A=a]\leq a$ holds for all $a\in\supp_{\P}(A)$.
\end{proposition}
\begin{proof}
This proposition is stated and proven as Theorem 2 in \cite{Brams95}.
\end{proof}

Examples for which $\E[B|A=a]>a$ holds for all $a\in\supp_{\P}(A)$ can be found in \cite{Christensen96,Broome95,Brams95} and previously in Section~\ref{sec:EnvelopePriorInfluence}, with $f(a)=(a+1)^{-2}$ for $a\in(0,\infty)$ being one of the most common example \cite{Broome95}.

\subsubsection{Summarizing switching strategies}
To summarize, let $f$ be a non-paradoxical prior on $X$ with distribution $\P\in\Pmod$. Suppose the player observes $a\in\supp_{\P}(A)$, then the following strategies optimize the values won by the player:
\begin{itemize}
\item[Discrete:] When $f$ is discrete, the player must switch his envelope when \begin{equation}
f(a)>\frac{1}{2}f\left(\frac{a}{2}\right).
\end{equation}
\item[Continuous:] When $f$ is continuous, the player must switch his envelope when
\begin{equation}
f(a)>\frac{1}{4}f\left(\frac{a}{2}\right).
\end{equation}
\end{itemize}

\section{Safe probability}\label{sec:EnvelopeSafe}
The problem with the priors is that we do not know which prior is correct. As stated in Section~\ref{sec:EnvelopeFormal}, the set $\Pmod$ essentially is the set of all probability distributions on $(0,\infty)$ with finite expectation value. There is a vast amount of possible distributions and we are not able to reduce the size of $\Pmod$ without making unfounded assumptions.  Therefore it would be helpful if a safe probability distribution exists for $\EnvInd|A$, which we can use as a `true' distribution for determining whether we want to switch the envelopes. The following proposition states that this is not possible.
\begin{proposition}\label{prop:EnvelopeIndSafe}
Let $\Omega$, $A$, $B$ and $\Pmod$ be as in the two envelope problem. There is no safe distribution $\Psafe$ on $\Omega$ for $\langle\EnvInd\rangle|\langle A\rangle$.
\end{proposition}
\begin{proof}
Assume a safe distribution $\Psafe$ for $\langle \EnvInd\rangle|\langle A\rangle$ exists. Let $\P\in\Pmod$ be arbitrary and let $F_X$ be the cumulative distribution function of the accompanying prior on $X$. Firstly we have
\begin{align}
\E_{\P}\left[\EnvInd\right]&=\P[B=2A]=\int_0^\infty\P[B=2X|X=x]dF_X(x)\\
&=\frac{1}{2}\int_0^\infty dF_X(x)=\frac{1}{2},
\end{align}
as $B=2A$ only happens when $A=X$ and $\P[B=2X|X=x]=\frac{1}{2}$ is dictated by our model $\Pmod$. Let $a\in\supp_{\Psafe}(A)$ be arbitrary, then
\begin{equation}
\E_{\Psafe}\left[\EnvInd\middle|A=a\right]=\Psafe[B=2A|A=a]
\end{equation}
must hold. Therefore we have
\begin{equation}
\E_{\P}\left[\E_{\Psafe}\left[\EnvInd\middle|A=a\right]\right]=\int_0^\infty\Psafe[B=2A|A=a]dF_A(a)
\end{equation}
with $F_A$ being the cumulative density function of $A$, resulting to the requirement
\begin{equation}\label{eq:EnvelopeSafeReq}
\int_0^\infty\Psafe[B=2A|A=a]dF_A(a)=\frac{1}{2}
\end{equation}
by safety for $\langle \EnvInd\rangle|\langle A\rangle$.

Note that $\P\in\Pmod$ is taken arbitrarily, thus \eqref{eq:EnvelopeSafeReq} must hold against all cumulative density functions $F_A$ on $(0,\infty)$. As those all integrate to $1$, the equation is only valid against all $\P\in\Pmod$ when
\begin{equation}
\Psafe[B=2A|A=a]=\frac{1}{2}
\end{equation}
holds for all $a\in(0,\infty)$. As earlier pointed out by \cite{Christensen92,Christensen93b,Navara17,Tzur18} and in Section~\ref{sec:EnvelopeFormal}, this is not possible.
\end{proof}

Note that in this case Theorem~\ref{thm:DiscMainThm} cannot be applied, as this theorem dictates that the sample space $\Omega$ is split up in $\Omega=\X'\times\Y'$, where $\X'$ is countable and $\Y'$ is finite. In our case we do have $\Omega=\X\times\Y$, however both $\X$ and $\Y$ are of uncountably infinite size.

Take a quick look at safety for $\langle B\rangle|\langle A\rangle$. After picking a $\P\in\Pmod$ the expectation values $\E_{\P}[A]=\E_{\P}[B]$ are equal. Therefore we can pick a distribution $\Psafe$ on $\Omega$ with $\E_{\Psafe}[B|A=a]=a$ for all $a\in(0,\infty)$, which will be safe for $\langle B\rangle|\langle A\rangle$:
\begin{align}
\E_{\P}[\E_{\Psafe}[B|A]]=\int_0^\infty \E_{\Psafe}[B|A=a]dF_A(a)=\int_0^\infty adF_A(a)=\E_{\P}[A]=\E_{\P}[B].
\end{align}
The existence and appearance of this $\Psafe$ is unknown and open for further research.

The problems with safe probability are that it cannot take into account the probability of observing $A=a$ and that it uses a model $\Pmod$ that is too large. The only reasonable suggestion for a safe distribution $\Psafe$ from the proof of Proposition~\ref{prop:EnvelopeIndSafe} is $\Psafe[B=2A|A=a]=\frac{1}{2}$ for all $a\in(0,\infty)$. However when playing this game, one should quickly dismiss the idea that all values $a\in(0,\infty)$ can be observed. Only that much money exists in the world and a minimal value one is able to put into an envelope must exist. When $\Pmod$ is reduced to only allow distributions with a finite support on the minimal value $x$, we do not know whether a safe distribution $\Psafe$ does exist. However, even if it does exist, it is not very informative. A player is not able to use this safe distribution to devise a strategy where he wins more than $\frac{3}{2}\E_{\P^*}[X]$ on a structural basis with $\P^*$ being the true distribution. This average value $\frac{3}{2}\E_{\P^*}[X]$ namely is the average value won when applying the trivial `always switch' or `never switch' strategies.

\section{Cover's switching strategy}\label{sec:EnvelopeSwitching}
If we  adapt our probability of switching to the observation $A=a$, we can obtain on average higher results than $\frac{3}{2}\E_{\P}[X]$. The technique of switching smartly is first devised by Cover and in 2003 personally communicated to McDonnell and Abbott \cite{McDonnell09}, who published this technique with numerous examples \cite{McDonnell09,Abbott10,McDonnell11}. The techniques used were already lightly touched upon by Christensen and Utts in 1992 \cite{Christensen92} and by Ross in 1994 \cite{Christensen94} as response on the article of Christensen and Utts. Since McDonell and Abbott fleshed out the switching strategy and describes it in more detail, we will use their articles as foundation.

In short, Cover's switching strategy devises a function $P\colon(0,\infty)\to[0,1]$ which picks an observed $a\in(0,\infty)$ and assigns a probability $P(a)\in[0,1]$ to it. The player then switches his envelope with probability $P(a)$. Note that, in contrary to what initially may be assumed, the function $P$ does not need to integrate to $1$; it is not a probability density function. It implies after observing an $a\in(0,\infty)$ a probability of whether or not to switch the envelopes.

\begin{definition}[Switching strategy]
A function $P\colon(0,\infty)\to[0,1]$ is called a \emph{switching strategy}.
\end{definition}

Assume that the prior $f$ on $X$ is \emph{continuous} with finite expectation value and let $\P\in\Pmod$ be its accompanying distribution on $\Omega$. The function $P$ can be picked in such a way that, while not knowing the distribution $\P\in\Pmod$, we can always on average get a higher result than $\frac{3}{2}\E_{\P}[X]$.

Consider for a short while a more general problem where envelope $A$ receives $2X$ with probability $1-p$ and receives $X$ with probability $p\in[0,1]$. The original problem can then be retrieved by putting $p=\frac{1}{2}$. After observing value $A=a$, the player randomly chooses envelope~$B$ with probability $P(a)$. Let $R$ be the value returned to the player at the end of the game, then the probability distribution of $R|X=x$ is given by
\begin{align}
\P[R=x|X=x]&=\P[A=x|X=x](1-P(x))+\P[A=2x|X=x]P(2x)\\
&=p(1-P(x))+(1-p)P(2x),\\
\P[R=2x|X=x]&=\P[A=2x|X=x](1-P(2x))+\P[A=x|X=x]P(x)\\
&=(1-p)(1-P(2x))+pP(x).
\end{align}
Therefore the conditional expectation of the return given $X=x$ is
\begin{align}
\E_{\P}[R|X=x]&=x\P[R=x|X=x]+2x\P[R=2x|X=x]\\
&=x(2-p)+pxP(x)-x(1-p)P(2x).
\end{align}
Following \cite{McDonnell09}, we get
\begin{align}
\E_{\P}[R]=\E_{\P}[\E_{\P}[R|X]]&=\int_0^\infty f(x)\E_{\P}[R|X=x]dx\\
&=(2-p)\E_{\P}[X]+\int_0^\infty xf(x)(pP(x)-(1-p)P(2x))dx.
\end{align}
When the player never or always switches, thus $P\equiv0$ or $P\equiv1$, his expected return is $(2-p)\E[X]$. Define
\begin{equation}
G=\int_0^\infty xf(x)(pP(x)-(1-p)P(2x))dx
\end{equation}
as the players gain of his switching strategy. One can immediately see that for all $x$ the requirement $pP(x)\geq(1-p)P(2x)$ is a sufficient condition for a non-negative gain. The gain by a coordinate transformation can be rewritten as
\begin{align}
G&=\int_0^\infty xf(x)(pP(x)-(1-p)P(2x))dx\\
&=\int_0^\infty xP(x)\left(pf(x)-\frac{1-p}{4}f\left(\frac{x}{2}\right)\right)dx,
\end{align}
where both representations will become useful in many different ways. This will be summarized in the following definition.
\begin{definition}
Let $f$ be a continuous prior on $X$ with finite expectation value. Let $P$ be a switching strategy. The \emph{gain} of a player is defined by
\begin{align}
G&=\int_0^\infty xf(x)(pP(x)-(1-p)P(2x))dx\label{eq:EnvelopeGainP}\\
&=\int_0^\infty xP(x)\left(pf(x)-\frac{1-p}{4}f\left(\frac{x}{2}\right)\right)dx.\label{eq:EnvelopeGainF}
\end{align}
\end{definition}
Equation~\ref{eq:EnvelopeGainF} immediately gives us the optimal switching strategy.
\begin{theorem}[McDonnell, Abbott \cite{McDonnell09}]\label{thm:switchopt}
The optimal switching strategy $P^*(x)$ is
\begin{equation}
P^*(x)=\1_{(0,\infty)}\left(pf(x)-\frac{1-p}{4}f\left(\frac{x}{2}\right)\right).\end{equation}
\end{theorem}
\begin{proof}
We can split the integrand of \eqref{eq:EnvelopeGainF} in two functions: $x\mapsto P(x)$ and $x\mapsto x\left(pf(x)-\frac{1-p}{4}f\left(\frac{x}{2}\right)\right)$. Function $P$ can be chosen at will, but the image must lie in $[0,1]$. The integration in \eqref{eq:EnvelopeGainF} is therefore maximized when $P(x)=1$ is chosen if $x\left(pf(x)-\frac{1-p}{4}f\left(\frac{x}{2}\right)\right)>0$ is positive and $P(x)=0$ otherwise. Since $x\mapsto x$ is positive for all $x\in(0,\infty)$, it is sufficient to require that the function $x\mapsto pf(x)-\frac{1-p}{4}f\left(\frac{y}{2}\right)$ must be positive when $P(x)=1$. Switching strategy $P^*$ implements this strategy.
\end{proof}

From now on we will put $p=\frac{1}{2}$ as in our model the envelope is  distributed to the player uniformly. In this case Equation~\ref{eq:EnvelopeGainP} leads to $P(x)\geq P(2x)$ for all $x\in(0,\infty)$ being sufficient for non-negative gain. We will now look at two example strategies, namely Cover switching and threshold switching. More examples, comparisons between the examples and simulations of the examples can be found in \cite{McDonnell09,Abbott10,McDonnell11}.

\begin{example}[Cover switching]
This example is taken from \cite{McDonnell09,Abbott10}. Let $a\in(0,\infty]$ be arbitrary and put $P(x)=e^{-ax}$. This $P$ is called \emph{Cover switching}. The inequality $P(x)\geq P(2x)$ is automatically satisfied as the requirement $e^{-ax}>e^{-2ax}$ holds for all positive $x$.

Suppose that $X\sim\mathrm{Exp}(c)$ is exponentially distributed. Let $R_C$ be the random variable of the return when Cover switching is used. Switching using Cover's strategy is maximized for $a=\left(\frac{1}{3}\sqrt[3]{2}+\frac{1}{6}\sqrt[3]{4}-\frac{1}{3}\right)c$, with a return of $\E[R_C]=\frac{3}{2c}\left(2-2\sqrt[3]{2}+\sqrt[3]{4}\right)\approx\frac{1.6013}{c}$. The return of this prior for never or always switching strategy is $\E[R]=\frac{3}{2}\E[X]=\frac{1.5}{c}$, thus Cover switching does improve on never or always switching.
\end{example}

\begin{example}[Threshold switching]
This example is taken from \cite{McDonnell09}. Let $b\in(0,\infty)$ be arbitrary and define $P(a)=\1_{(0,b]}(a)$. This strategy is called \emph{threshold switching}. The player tries to switch all the low values he gets, but always holds on high values. Of course, if the prior on $X$ is unknown, no advice on which $b$ to choose can be given. However, as $P(a)\geq P(2a)$ always holds, this strategy has non-negative gain for all possible values $b$ independent of the prior $f$.

Let $X\sim\text{Exp}(c)$ be exponentially distributed. Looking at Equation~\ref{eq:EnvelopeGainF}, we want that $P(x)=1$ when $f(x)\geq\frac{1}{4}f\left(\frac{x}{2}\right)$ holds, thus when $ce^{-cx}\geq\frac{1}{4}ce^{-\frac{c}{2}x}$ holds. This is satisfied when $x<\frac{4\ln(2)}{c}$ holds, thus take $b=\frac{4\ln(2)}{c}$. Let $R_T$ be the value returned to the player when using threshold switching. The expected return will be $\E[R_T]=\frac{51+4\ln(2)}{32c}\approx\frac{1.6804}{c}$, which is higher than Cover switching's $\E[R_C]=\frac{1.6013}{c}$ and the $\E[R]=\frac{1.5}{c}$ of never or always switching. Threshold switching is in fact the optimal switching function here, as $f(x)=\frac{1}{4}f\left(\frac{x}{2}\right)$ is obtained for a unique $x=\frac{4\ln(2)}{c}$ and $P^*$ from Theorem~\ref{thm:switchopt} then becomes threshold switching.
\end{example}
\subsection{Optimizing Cover's switching strategy}
Having two strategies, namely Cover and threshold switching, we would like to know how to optimize them while knowing as little of the distribution on $X$ as possible. McDonnell et al.~\cite{McDonnell11} considered various cases and here we will take a look at the ones with the least information on the prior of $X$. Assume for the rest of this subsection that the prior $f$ on $X$ is absolutely continuous and differentiable on $(0,\infty)$.

\subsubsection{Cover switching optimized}
This example is taken from \cite{McDonnell11}. Let $a\in[0,\infty]$ and consider Cover's switching strategy, $P(x)=e^{-ax}$. The gain as a function of $a$ can be written as
\begin{equation}
G(a)=\int_0^\infty pxf(x)e^{-ax}-(1-p)xf(x)e^{-2ax}dx.
\end{equation}
The optimal value $a^*$ solves the functional equation
\begin{equation}
\int_0^\infty e^{-a^*t}x^2f(x)dx=\frac{2(1-p)}{p}\int_0^\infty e^{-2a^*t}x^2f(x)dx,
\end{equation}
where $\int_0^\infty e^{-at}x^2f(x)dx$ is the Laplace transform of $x\mapsto x^2f(x)$ with parameter $a$. This leads to the following lemma:
\begin{lemma}[McDonnell et al.~(2011)]
Let $p\in[0,1]$ be the probability the lowest envelope is given to the player. Let $P$ be Cover's switching strategy with parameter $a$. The following cases must be distinguished for the optimal value $a^*$ of $G(a^*)$:
\begin{itemize}
    \item[{$p\in\left[0,\frac{1}{5}\right)$:}]
    Take $p\in\left[0,\frac{1}{5}\right)$, then the optimal value is $a^*=\infty$. This makes never switching is the optimal strategy. The optimal gain is $G(\infty)=0$.
    \item[{$p\in\left[\frac{1}{5},\frac{2}{3}\right)$:}]
    Take $p\in\left[\frac{1}{5},\frac{2}{3}\right)$. If the function $x\mapsto pf(x)-\frac{1-p}{4}f\left(\frac{x}{2}\right)$ has a unique sign change, then $a\mapsto G(a)$ has a unique stationary point $a^*\in[0,\infty)$. This $a^*$ maximizes $a\mapsto G(a)$.
    \item[{$p\in\left[\frac{2}{3},1\right]$:}]
    Take $p\in\left[\frac{2}{3},1\right]$, then the optimal value is $a^*=0$. This makes always switching is the best strategy. The optimal gain is $G(0)=(2p-1)\E[X]$.
\end{itemize}
\end{lemma}
\begin{proof}
The proof can be found in \cite{McDonnell11}.
\end{proof}
Do note that there is an asymmetry in the previous lemma. When the probability of receiving the envelope with the lowest value is below $\frac{1}{5}$, never switching is the best strategy. However, when the probability is above $\frac{2}{3}$, always switching is the best strategy. One should expect the turning points to lie symmetric around $\frac{1}{2}$ such that they are $\frac{1}{3}$ and $\frac{2}{3}$ or $\frac{1}{5}$ and $\frac{4}{5}$. This phenomenon is widely studied in \cite{McDonnell11}.

\subsubsection{Threshold switching optimized}
The first part of this example is taken from \cite{McDonnell11}. Let $b\in[0,\infty]$ and consider threshold switching, $P(x)=\1_{[0,b]}(x)$. The gain as a function of $b$ can be written as
\begin{equation}
G(b)=\int_0^\infty pxf(x)\1_{[0,b]}(x)-(1-p)xf(x)\1_{\left[0,\frac{b}{2}\right]}(x)dx.
\end{equation}
Taking the derivative yields $G'(b)=b\left(pf(b)-\frac{1-p}{4}f(b)\right)$, resulting to a possible set of optimal values. Let $B=\{b\in[0,\infty)|G'(b)=0\}$ and take $b^*\in B$, then differentiating $G$ twice yields
\begin{equation}
G''(b^*)=b^*\left(pf'(b^*)-\frac{1-p}{8}f'\left(\frac{b^*}{2}\right)\right),\
\end{equation}which must be negative for $b^*$ to be a local maximum. This results to the condition
\begin{equation}
\frac{8p}{1-p}f'(b^*)< f'\left(\frac{b^*}{2}\right).
\end{equation}
Note that \cite{McDonnell11} does not have a strict inequality, but the second derivative test is inconclusive when $G''(b^*)=0$. Thus it is still possible for $b^*$ to be a saddle or a minimum in case of $G''(b^*)=0$, which is why we write a strict inequality here. This all results into the following lemma:
\begin{lemma}
Let $p\in[0,1]$ be such that the probability of receiving the lower envelope is $p$. Let $P$ be the threshold switching strategy. The set
\begin{equation}
B=\left\{b\in[0,\infty]\middle|pf(b)=\frac{1-p}{4}f\left(\frac{b}{2}\right)\land\frac{8p}{1-p}f''(b)<f'\left(\frac{b}{2}\right)\right\}\cup\{0,\infty\}
\end{equation}
is the set of all $b$ that locally maximize $b\mapsto G(b)$.
\end{lemma}
\begin{proof}
The proof can be found in \cite{McDonnell11}.
\end{proof}

This example however requires full knowledge on $f$. Can we optimize $P$ with as little knowledge as possible? The answer is yes and is given by \cite{Egozcue15}. Rephrase the gain as
\begin{align}
G(b)&=\int_0^\infty pxf(x)\1_{[0,b]}(x)-(1-p)xf(x)\1_{\left[0,\frac{b}{2}\right]}(x)dx\\
&=\frac{1}{2}\E[XP(X)-XP(2X)]=\frac{1}{2}\E\left[X\1_{\left(\frac{b}{2},b\right]}(X)\right],
\end{align}
then we can derive a lower bound of $G$ when knowing only $\E[X]$ and $\V[X]$.
\begin{theorem}[Egozcue and García (2015)]\label{thm:EnvelopeSwitchBound}
Let $X$ be a continuous random variable with $\E[X]=\mu$ and $\V[X]=\sigma^2$. Let $m=\mu^2+\sigma^2$. Let $P$ be the threshold strategy with parameter $b\in(0,\infty)$.
\begin{enumerate}
    \item The gain has lower bound

       \begin{equation}
    G(b)\geq\frac{3+2\sqrt{2}}{b}\left(\frac{3}{2}b\mu-\frac{1}{2}b^2-m\right).
    \end{equation}
    \item If $\mu^2\geq 8\sigma^2$ holds, then $b^*=\sqrt{2m}$ is the optimal threshold strategy with gain
        \begin{equation}
    G(b^*)\geq(3+2\sqrt{2})\left(\frac{3}{2}\mu-\sqrt{2m}\right)\geq0.
    \end{equation}
    The gain is strictly positive when $\mu^2>8\sigma^2$.
\end{enumerate}
\end{theorem}
\begin{proof}
The proof can be found in \cite{Egozcue15}.
\end{proof}
Thus if $\E[X]^2\geq8\V[X]$ holds, we can find an optimal $b^*$ which always results in non-negative gain. The next question is whether there is a $b\in(0,\infty)$ which guarantees us a strictly positive gain for all priors on $X$. This is not possible, as the following theorem will point out.
\begin{theorem}[Egozcue and García (2015)]
Let $b\in(0,\infty)$. A random variable $X$ with $\E[X]^2<8\V[X]$ such that $G(b)=0$ always exists.
\end{theorem}
\begin{proof}
The proof can be found in \cite{Egozcue15}.
\end{proof}

\subsection{Discrete priors and the switching strategy}
We only considered switching strategies for continuous priors thus far. If $X$ is distributed discretely by distribution $\P\in\Pmod$, then switching strategies are still possible. As in the continuous case, we have
\begin{equation}
\E[R|X=x]=x(2-p)+pxP(x)-x(1-p)P(2x).
\end{equation}
The expected return is therefore
\begin{align}
\E[R]&=\sum_x f(x)\E[R|X=x]\\
&=(2-p)\E[X]+\sum_x xf(x)\big(pP(x)-(1-p)P(2x)\big).
\end{align}
A change in variables like the continuous case now becomes a bit awkward. Let $S$ be support of $X$, then
\begin{equation}
\sum_{x\in \supp_{\P}(X)}xf(x)P(2x)=\sum_{x\in2\supp_{\P}(X)}\frac{x}{2}f\left(\frac{x}{2}\right)P(x)
\end{equation}
holds. Thus we get
\begin{align}
G&=\sum_{x\in \supp_{\P}(X)}xf(x)(pP(x)-(1-p)P(2x))\\
&=p\sum_{x\in \supp_{\P}(X)}xP(x)f(x)-\frac{1-p}{2}\sum_{x\in2\supp_{\P}(X)}xf\left(\frac{x}{2}\right)P(x),
\end{align}
and since in the discrete case $\supp_{\P}(X)=2\supp_{\P}(X)$ does not necessarily hold the latter formula cannot be easily written in a single sum.

Put $p=\frac{1}{2}$, then it becomes clear that $P(x)\geq P(2x)$ is sufficient for non-negative gain. Thus any decreasing switching strategy will again be better than always keeping the envelope or always switching. Since this is also the case for continuous priors, we can recap both cases in the following theorem.

\begin{theorem}\label{thm:switchall}
Let $f$ be a continuous or discrete prior on $X$ with finite mean and suppose the envelope is handed to the player with uniform distribution. Let $P$ be a decreasing switching strategy that is strictly decreasing on at least one subset $U\subset(0,\infty)$ that has positive probability measure. Then $P$ will lead to a positive gain and is a better strategy than either always keeping the envelope or always switching the envelope.
\end{theorem}
\begin{proof}
This theorem has already been proven in parts in the previous sections, so we'll only give a summary here. The gain, the difference between strategy $P$ and always keeping or switching envelopes, is defined as
\begin{equation}
G=\frac{1}{2}\int_0^\infty xf(x)(P(x)-P(2x))dx
\end{equation}
when $f$ is continuous and as
\begin{equation}
G=\frac{1}{2}\sum_xxf(x)(P(x)-P(2x))
\end{equation}
when $f$ is discrete. As $P$ is decreasing, $P(x)\geq P(2x)$ will always hold, making $G$ non-negative. If $P$ is also strictly decreasing on a subset $U\subset(0,\infty)$ with positive probability measure, $G$ becomes is strictly positive.
\end{proof}

\section{The Ali Baba problem}\label{sec:EnvelopeAli}
We thus far only studied the two envelope problem where there is only one player. Consider now the following Ali Baba problem mentioned by Nalebuff in 1989 \cite{Nalebuff89}. In the Ali Baba problem, envelope $A$ is filled first and then given to Ali. Then a hidden fair coin is tossed to decide whether envelope $B$ must contain value $2A$ or $\frac{1}{2}A$, where after filling the envelope it is given to Baba. Ali and Baba are allowed to privately look at the contents of their envelope. If they both agree that switching is the better choice, they are able to do so.

Take a look at the following reasoning taken from the introduction of \cite{Nalebuff89}. First consider Ali's choice after observing value $x$. Baba has value $\frac{1}{2}x$ with probability $\frac{1}{2}$ and value $2x$ with probability $\frac{1}{2}$. He will reason that on average Baba has value $\frac{5}{4}x$, thus Ali always proposes a switch. Consider Baba's viewpoint next. Say he has a value $y$, then $A$ has values $2y$ or $\frac{1}{2}y$. Thus Baba expects a value of $\frac{5}{4}y$ in Ali's envelope, making him also wanting to switch. They both want to switch, while Ali obtained a fixed value and Baba's value is linked to Ali's. This is considered paradoxical.

The problem in this paradox again arises from wrongly applying probability theory like in Section~\ref{sec:EnvelopeNaive}. Call $x$ the amount of money in envelope $A$. The probability space is
\begin{equation}
\Omega=\left(\X=\left\{(x,2x),\left(x,\frac{1}{2}x\right)\right\},2^\X,\P\right)
\end{equation}
where $\P$ is the probability measure with $\P\left[\left(x,\frac{1}{2}x\right)\right]=\P[(x,2x)]=\frac{1}{2}$. Since Ali always gets an amount $x$, the expected value of Baba's envelope calculated from Ali's viewpoint remains the same:
\begin{equation}
\E[B]=\sum_{b\in\left\{\frac{x}{2},2x\right\}}b\P[B=b|A=x]=\frac{x}{2}\cdot\frac{1}{2}+2x\cdot\frac{1}{2}=\frac{5}{4}x.
\end{equation}
Therefore Ali will always want to switch.

Here it does not matter how value $x$ is picked from $(0,\infty)$ in contrast to the original two envelope problem discussed. Since Ali knows his value $x$ is fixed, he correctly deduces that Baba has either $2x$ or $\frac{1}{2}x$. An extension like in Section~\ref{sec:EnvelopeFormal} is therefore not necessary.

Consider now Baba's viewpoint. He either receives an amount $2x$ or $\frac{1}{2}x$. The corresponding expected values are
\begin{align}
\E[A|B=2x]&=\sum_{a\in\{x,4x\}}a\P[A=a|B=2x]=1\cdot x+0,\\
\E\left[A\middle|B=\frac{1}{2}x\right]&=\sum_{a\in\left\{\frac{x}{4},x\right\}}a\P[A=a|B=\frac{1}{2}x]=0+1\cdot x.
\end{align}
Thus the expected value of Ali's envelope is
\begin{align}
\E[A]=\E[\E[A|B]]&=\sum_{b\in\left\{\frac{x}{2},2x\right\}}\P[B=b]\E[A|B=b]\\
&=\frac{1}{2}x+\frac{1}{2}x=x.
\end{align}
It is unclear for Baba whether he should switch, as $A$ is expected to have on average value $x$. Baba does lose $x$ if he switches when having value $2x$ in contrast to gaining $\frac{1}{2}x$ when having value $\frac{1}{2}x$, thus he may never switch as he finds the risk of losing money to be too high.

Consider lastly the values gained by Ali and Baba compared to their original envelope. Let $p\in[0,1]$ be the probability that Ali and Baba agree to switch. We now have the probability space
\begin{equation}
\Omega'=\left(\Y=\left\{(x,-x),\left(-\frac{1}{2}x,\frac{1}{2}x\right),(0,0)\right\},2^\Y,\P\right)
\end{equation}
with $\P[\{(x,-x)\}]=\P\left[\left\{-\frac{1}{2}x,\frac{1}{2}x\right\}\right]=\frac{1}{2}p$ and $\P[\{(0,0)\}]=1-p$. When there is no switch, $A$ and $B$ gain nothing. The probability of no switch is $1-p$, therefore $\{(0,0)\}$ has mass $1-p$. The probability of switching is $p$ and $B$ has $2x$ with probability $\frac{1}{2}$, therefore $\{(x,-x)\}$ has mass $\frac{1}{2}p$. Let $G$ be the vector of gains. The expected gains are
\begin{align}
\E[G]=\begin{pmatrix}x\\-x\end{pmatrix}\P[\{(x,-x)\}]+\begin{pmatrix}-\frac{1}{2}x\\\frac{1}{2}x\end{pmatrix}\P\left[\left\{\left(-\frac{1}{2}x,\frac{1}{2}x\right)\right\}\right]=\begin{pmatrix}\frac{1}{4}px\\-\frac{1}{4}px\end{pmatrix}.
\end{align}
The expected gains sum to $0$, thus no money is magically put into the system. Furthermore, $A$ will always profit in the long run for any switching strategy with $p>0$ as he initially obtains a fixed value $x$ and the expected value obtained by $B$ is $\E[B]=\frac{5}{4}\E[A]$. Thus Baba must always refuse the switching request from Ali in order to lose no money, giving a trivial non-paradoxical game.

\subsection{Different versions of the Ali Baba problem}
\begin{table}
\centering
\begin{tabular}{cc|cc}
Envelope known&Contents known&Ali's choice&Baba's choice\\\hline
Yes&Yes&Trade&Prior\\
Yes&No&Trade&Keep\\
No&Yes&Prior&Prior\\
No&No&Indifferent&Indifferent
\end{tabular}

\caption{Table taken from \cite{Nickerson06} giving advice to Ali and Baba whether they should switch their envelopes. `Envelope known' means that both players know Ali received the first envelope. `Contents known' means both players are allowed to look at their envelope's contents. When `Prior' is stated, Ali or Baba should decide on the prior they impose on the value of their envelope.}
\label{tbl:alibaba}
\end{table}

This is just one version of the Ali Baba problem; at least four different variations can be imposed upon the problem. Those problems are discussed by Nickerson and Falk in 2006 \cite{Nickerson06} and we will give a quick overview. For some variations the optimal strategy is independent of a prior on Ali's value, for others a prior needs to be taken into account. Two out of the four variations even completely define the optimal strategy of Ali and Baba.

Nickerson and Falk considered the following three variants on the Ali Baba problem in \cite{Nickerson06}:
\begin{enumerate}
\item is the distribution on Ali's value unknown or from the uniform distribution between $0$ and $100$,
\item do the players know which envelope they received and
\item are the players able to look at the contents?
\end{enumerate}
The first variant is not treated here as we assume that the players do not know the distribution on Ali's value. Even if they do know, \cite{Nickerson06} only considers the uniform prior between $0$ and $100$ which I find too restrictive and not realistic. Here we always assume that the prior on the first envelope is unknown, possibly unbounded, reducing the number of different problems to four.

Table~\ref{tbl:alibaba} gives all optimal strategies. Unfortunately, two out of four cases depend on the prior Ali and Baba impose on their envelope. If both players know the contents of their envelope, the prior on the observed value is needed to give an advice to switching.\\
Note that Ali must always trade if he knows he received the first envelope. In this case the expected value of Baba's envelope is $\E[B]=\frac{5}{4}\E[A]$. In both cases however Baba must either decide on his prior or always keep the envelope, thus blindly switching for all values is only fortunate when HIS prior has infinite mean. This debunks the paradox in most cases and when the mean is infinite, the paradox can be classified as St.~Petersburg-like.\\
If both players have no information, the symmetry in the game dictates that there is no information to base a player's strategy on. No paradoxical results can be obtained here.

All variants can be investigated further to obtain optimal strategies. in particular, Cover's switching strategy can most likely be applied here. I believe that similar results as in the initial two envelope problem will rise here as well, which will be left open for further research.

\section{Concluding remarks}\label{sec:EnvelopeConcl}
The two envelope problem is another example of why performing conditional probability incorrectly can lead to contradicting and paradoxical results. Here paradoxical results arise by wrongly applying conditional probability. Most calculate the expected value of envelope $B$ as
\begin{equation}
\E[B]=\frac{A}{2}\P\left[B=\frac{A}{2}\right]+2A\P[B=2A],
\end{equation}
however random variable $A$ is used here twice with a different context. As observed in Section~\ref{sec:EnvelopeNaive}, the actual calculation should be
\begin{equation}
\E[B]=\E[A|A<B]+\frac{1}{4}\E[A|A>B],
\end{equation}
where $\E[A|A<B]$ and $\E[A|A>B]$ are impossible to calculate without having more information. When the lowest value of both envelopes $X=\min\{A,B\}$ is considered, we observe that
\begin{equation}
\E[A]=\frac{3}{2}\E[X]=\E[B]
\end{equation}
holds, which is not considered as paradoxical.

We now only need to know the distribution of $X$. All probability distributions on $(0,\infty)$ are however possible, therefore we first resorted to safe probability in order to get some insights on a `true' distribution. In Section~\ref{sec:EnvelopeSafe} however we saw that in contrast to the problems in Chapter~\ref{chap:DiscPara} safe probability will not be helpful. Proposition~\ref{prop:EnvelopeIndSafe} even states that a safe distribution does not exist when using the set of distributions $\Pmod$. The set $\Pmod$ must be pruned to only allow distributions that have finite support on the possible observations in order for a safe distribution to exist. This safe distribution becomes almost trivial and non-informative as it states that $B=2A$ always happens with probability $\frac{1}{2}$.

When we state that $B=2A$ happens with probability $\frac{1}{2}$ independent of $A$'s value and we base our decisions on this assumption, we will on average get a return of $\frac{3}{2}\E_{\P^*}[X]$ where $\P^*$ is the true distribution. If we drop this assumption and assume the random variable $X$ of the minimal value being continuous, we can improve on this return. Cover switching states that we create a switching strategy $P\colon(0,\infty)\to[0,1]$ which assigns a probability of switching to each observed value. The gain $G$ relative to the trivial strategies can then be defined by
\begin{equation}
G=\frac{1}{2}\int_0^\infty xf(x)(P(x)-P(2x))dx
\end{equation}
with $f$ the probability density function on $X$. For the never or always switching strategies we have zero gain. Theorem~\ref{thm:switchall} states that we only need to make $P$ decreasing in order to perform as well as the `never switching' strategy. This is quite intuitive, as you are less likely to swap the envelopes when you initially obtain a high value. When a low value is observed, you will lose less if you swap it for an even lower envelope.\\
Initially it was unknown how much gain a switching strategy yields, as nothing is known of the distribution of $X$. The gain $G$ can therefore become arbitrarily close to zero. You can for example apply threshold switching with $31$ as threshold, while $X$ only takes values higher than $94$. Recently in 2015, Egozcue and Fuentes García \cite{Egozcue15} proved that in certain conditions only depending on the expectation value and variance of the distribution on $X$ an optimal threshold value for threshold switching can be obtained, resulting in a lower bound for the gain which is strictly positive as described in Theorem~\ref{thm:EnvelopeSwitchBound}. In this case one is able to actually improve on the trivial `never switching' strategy by at least a certain amount.

The most important observations for this thesis are in Sections~\ref{sec:EnvelopeNaive} and \ref{sec:EnvelopeFormal}. When explicitly writing down the probability spaces used for studying this problem, one first observes that the calculation performed in Equation~\ref{eq:EnvelopeIntroWrong} is wrong. Furthermore, one observes that the naive probability space of Section~\ref{sec:EnvelopeNaive} with only $\Omega=\{(x,2x),(2x,x)\}$ as sample space is too restrictive. Only then you are able to extend this probability space to the one in Section~\ref{sec:EnvelopeFormal} including all possible distributions on the lowest value $X$. This enables you also to observe that $\P[B=2A|A=a]=\frac{1}{2}$ for all $a\in(0,\infty)$ is simply not possible; an assumption that does follow from safe probability and that everyone starting to study this problem makes. This underlines the message that when performing conditional probability, the probability space with the accompanying $\sigma$-algebra should always be taken into account to avoid paradoxical results.

\chapter{The conclusions}
After looking at each paradox, we can confirm the existence of a common thread among them. For every paradox a paradoxical result is obtained by wrongly applying conditional probability and in most if not all cases the mistakes in application would not be made when correctly considering the sub-$\sigma$-algebra one wants to condition on. This thesis has argued that when performing conditional probability, the full probability space must be taken into account and the sub-$\sigma$-algebra must never be ignored.

The first paradox considered in Chapter~\ref{chap:BorelKolmogorov} was the Borel-Kolmogorov paradox, where conditioning on the $\sigma$-algebra of latitudes gave a different conditional probability distribution on a great circle than conditioning on the $\sigma$-algebra of meridians. As Kolmogorov \cite{Kolmogorov33} already stated in 1933, the difference in conditional probability distributions is due to great circles on a sphere with uniform probability measure having zero mass and conditioning on events with zero probability is not meaningful. Gyenis et al.~\cite{Gyenis17} proved that the conditional probability space of latitudes is fundamentally different from the conditional probability space of meridians, which explains why the two conditional probability distributions are different. Therefore, the Borel-Kolmogorov paradox is merely an example of why the sub-$\sigma$-algebra must always be given along with the conditional probability distribution, especially when concerning events with zero probability.

The next paradoxes considered in Chapter~\ref{chap:DiscPara} were the so-called discrete conditional paradoxes. One example is the famous Monty Hall problem. Many argue that the probability of door $a$ having the car behind given door $c$ is opened is~$\frac{1}{2}$, which can mathematically be obtained by applying Bayes' rule. However, if the car is behind door $a$, the game master has a choice to open either door $b$ or $c$ and this choice must be taken into account. Therefore, the sub-$\sigma$-algebra conditioned on must not only consist of the event $\{a,b\}$ that door $c$ is open, but the event $\{a,c\}$ that door $b$ is open as well. After considering this we find the probability of door $a$ having the car behind it being dilated to an interval from~$0$ to~$\frac{1}{2}$, which is the correct answer of the Monty Hall problem.\\
The question of the Monty Hall problem was whether a meaningful single probability distribution rather than an interval can be given here. After resorting to safe probability, which was defined in Chapter~\ref{chap:SafeProp}, we found out that distributions behave like a `true' distribution when they put a probability of $\frac{1}{3}$ on door $a$ having the car independent on the actually opened door. It is therefore safe to state that the probability of door $a$ having the car is $\frac{1}{3}$ or equivalently that the game master bases his choice of door $b$ or door $c$ when the car is behind door $a$ on the flip of a fair coin.\\
Moreover, we proved Theorem~\ref{thm:DiscMainThm}, which enables us to not only find a safe probability distribution for the Monty Hall problem, but a safe distribution for all similar problems as well. As long as the problem has a countable number of outcomes, a finite number of possible observations and a large enough set of possible probability distributions with fixed distribution on the outcomes, Theorem~\ref{thm:DiscMainThm} can be applied to find for each outcome a safe distribution denoting the probability it occurs given any observation. If an outcome has probability~$p$ of occurring, then safe probability states that it is safe to state that the probability of that outcome remains $p$ no matter the observation obtained. This may not be the true probability and in most cases is not, but it behaves in an average sense as if it were true.

The last paradox considered in Chapter~\ref{chap:TwoEnvelope} was the two envelope problem. The initial result that the other envelope holds a value $\frac{5}{4}$ times the value of your envelope is a misapplication of conditional probability. First a sufficient probability space was created where the event that envelope $A$ holds value $a$ is measurable for all $a\in(0,\infty)$, from which the conclusion is drawn that the expectation value of envelope $A$ is equal to the expectation value of envelope $B$. Furthermore, we quickly observed that the initial assumption most make, namely that the other envelope holds twice or half your observed value with equal probability, is not true as this implies that value $a$ is picked from an infinite set with uniform probability, which is impossible. The two envelope problem is also an example of where a vast set of probability distributions are possible, however this set is too large for safe probability to construct a single safe distribution that is informative. It is however possible to construct strategies which systematically improves the average value won against the trivial strategies as McDonnell and Abbott \cite{McDonnell09,Abbott10,McDonnell11} pointed out. Most initially discuss that this improvement can be arbitrarily close to zero debating the worth of employing such strategies. However, for threshold switching under certain yet quite easy to meet requirements a strictly positive lower bound on the improvement and an optimal threshold value can be computed, making it worth to employ this strategy.

None of these problems would be considered paradoxical if one correctly applied conditional probability and kept the sub-$\sigma$-algebra into account. Therefore, when applying conditional probability one is advised to carefully address the whole probability space to not make any unnecessary mistakes. Once conditional probability is applied correctly, each problem becomes much more interesting in its own way. From dilated probabilities to optimal switching strategies and fundamentally different probability spaces, once the paradoxical statements are resolved a vast field of interesting research is presented for further exploration.

\bibliographystyle{alpha}
\bibliography{../Referenties/Referenties}

\appendix
\chapter{Corrections to [GHSR17]}
\section{Corrections to appendix 1}\label{app:CorLong}
\pagenumbering{Alph}
In appendix 1, equation 94 of the proof of the conditional expectation on latitudes in \cite{Gyenis17}, they claim that the volume of the unit sphere is $2\pi$, whereas they should have claimed that the surface area of the unit sphere is $4\pi$. Luckily this constant is not actually used in the computations, thus their mistake has no impact on the rest of their article.

\section{Corrections to appendix 2}\label{app:CorMer}
I suggest the following corrections on appendix 2, the proof of conditional expectation on meridians:
\begin{enumerate}
\item In the verification \cite{Gyenis17} claims that $\int_0^\pi\sin\theta d\theta=1$ holds between equations (107) and (108) and between equations (115) and (116), while that integral actually has value $2$.
\item The conditional distribution of \cite{Gyenis17} is verified on a half meridian arc on page 2614, while this distribution must be verified on a full circle in order to be compared with the latitudes. Verification on a full meridian, e.g.~computation of $q(C)$ in formula (85), quickly reveals that the conditional distribution of \cite{Gyenis17} integrates to $2$.
\item The random variable $X$ is integrated on the domain $[0,2\pi)$, while $X$ is only defined on $[0,2\pi)\times[0,\pi]\in\mathfrak{M}$. The set $A\times[0,2\pi)$ is not an element of $\mathfrak{M}$.
\item Between equations (105) and (106) of \cite{Gyenis17}, they implicitly claim that the identity $X(\phi,\theta)=X(\phi,\theta+\pi)$ holds for all $\F$-measurable $X$, where now notation of \cite{Gyenis17} is used. This is most certainly false without any more restrictions on $X$.
\item Before equation (85) a measure $q_\mathcal{M}$ is defined on a whole meridian. Since the integral is taken from $\psi=0$ to $\psi=2\pi$, the integral of $q_\mathcal{M}$ over $S$ becomes $1$. However, as pointed out earlier, $\psi>\pi$ is not in our domain. Thus the integral must be split up in two arcs with $q_{\mathcal{M}}$ taking value $\frac{1}{2}$ on each arc.
\item The normalization constant is used in equation (103) of \cite{Gyenis17} is $2\pi$, where it must be $4\pi$. This does however not impact further calculations, like the same mistake made from equation (109) to (110).
\end{enumerate}

\chapter{Proofs}
\section{Proof of Proposition~\ref{prop:BorelConFalse}}\label{app:ProofBorelConFalse}
\begin{propositionBorelConFalse}
Conjecture~\ref{con:BorelConjecture} is false.
\end{propositionBorelConFalse}
\begin{proof}

Suppose Conjecture~\ref{con:BorelConjecture} is true. In sections \ref{sec:BorelLong} and \ref{sec:BorelMer} we have seen that the parametrization using latitudes gave the uniform distribution and the parametrization using meridians gave a cosine. However, in Section~\ref{sec:BorelLong} the set $[0,2\pi)\times\left\{\frac{\pi}{2}\right\}$ is considered, the red horizontal line in Figure~\ref{fig:BorelVis}. Section~\ref{sec:BorelMer} treats the set $\{0,\pi\}\times[0,\pi]$, the blue vertical lines in Figure~\ref{fig:BorelVis}. Those sets differer, however we can find a set which has two different probabilities using a rotation.

Call $f$ the transformation from polar to Euclidean coordinates from Equation~\ref{eq:BorelPolar}. Let $O$ be a rotation such that
\begin{equation}
(f^{-1}\circ O\circ f)\left([0,\pi]\times\left\{\frac{\pi}{2}\right\}\right)=\{\pi\}\times[0,\pi],
\end{equation}
thus where the largest half latitude is rotated to a half meridian. Recall that $\mathfrak{C}$ is the $\sigma$-algebra of latitudes. We now rotate those latitudes to get the following $\sigma$-algebra of rotated latitudes:
\begin{equation}
\mathfrak{C}'=\left\{F'\in\B\mid F'=\left(f^{-1}\circ O^{-1}\circ f\right)(F),F\in\mathfrak{C}\right\}.
\end{equation}
Now $F=\{\pi\}\times[0,\pi]$ is an element of both $\mathfrak{C}'$ and $\mathfrak{M}$. We will prove that different methods of converging to $F$ yield to different $\P[E|F]$ for a measurable $E$. We will take the set $E=\{\pi\}\times\left[\frac{1}{4}\pi,\frac{3}{4}\pi\right]$ as our example.

First take a look at $\mathfrak{C}'$. Let $\epsilon\in\left(0,\frac{\pi}{2}\right)$ be arbitrary and consider the sets $C_\epsilon=[0,\pi]\times\left[\frac{\pi}{2}-\epsilon,\frac{\pi}{2}+\epsilon\right]$ and $R_\epsilon=\left[\frac{1}{4}\pi,\frac{3}{4}\pi\right]\times\left[\frac{\pi}{2}-\epsilon,\frac{\pi}{2}+\epsilon\right]$. Let $\{R'_{n^{-1}}\}_{n\in\N}$ be with $R'_{n^{-1}}=(f^{-1}\circ O^{-1}\circ f)(R_{n^{-1}})$ and let the sequence $\{C'_{n^{-1}}\}_{n\in\N}$ be with $C'_{n^{-1}}=(f^{-1}\circ O^{-1}\circ f)(C_{n^{-1}})$. Since $R_{n^{-1}},C_{n^{-1}}\in\mathfrak{C}$ holds for all $n\in\N$, we get $\{R'_{n^{-1}}\}_{n\in\N},\{C'_{n^{-1}}\}_{n\in\N}\subset\mathfrak{C'}$. Furthermore, $f^{-1}\circ O^{-1}\circ f$ has determinant~$1$ since $O$ is a rotation and $f$ is almost surely a bijection on $F$, resulting to
\begin{align}
\P[R'_\epsilon|C'_\epsilon]&=\frac{\P[R'_\epsilon]}{\P[C'_\epsilon]}=\frac{\iint_{R'_\epsilon}\sin\psi d\psi d\phi}{\iint_{C'_\epsilon} \sin\psi d\psi d\phi}=\frac{\iint_{R_\epsilon}g(\sin\psi)\cdot1 d\psi d\phi}{\iint_{C_\epsilon}g(\sin\psi)\cdot1d\psi d\phi}\\
&=\frac{\int_{\frac{1}{4}\pi}^{\frac{3}{4}\pi}\int_{\frac{\pi}{2}-\epsilon}^{\frac{\pi}{2}+\epsilon}\sin (g(\psi))\cdot1 d\psi d\phi}{\int_{0}^{\pi}\int_{\frac{\pi}{2}-\epsilon}^{\frac{\pi}{2}+\epsilon}\sin (g(\psi))\cdot1d\psi d\phi}=\frac{\frac{3}{4}\pi-\frac{1}{4}\pi}{\pi}=\frac{1}{2}
\end{align}
where $g$ is the result of the transformation $f^{-1}\circ O^{-1}\circ f$ on the integrand. The result is $\P[R'_0|C'_0]=\frac{1}{2}$, as $\P[R'_\epsilon|C'_\epsilon]$ is constant in $\epsilon$. The sequence $\{R'_{n^{-1}}\}_{n\in\N}$ converges to $E$ and the sequence $\{C'_{n^{-1}}\}_{n\in\N}$ converges to $F$, thus we have $\P[E|F]=\frac{1}{2}$ as well.

For $\mathfrak{M}$ we will use Proposition~\ref{prop:BorelMerBayes}. Let $x=\pi$ and $\epsilon\in(0,\pi)$, then we have $M_\epsilon\in\mathfrak{M}$ and $R_\epsilon=[\pi-\epsilon,\pi+\epsilon]\times\left[\frac{1}{4}\pi,\frac{3}{4}\pi\right]\subset M_\epsilon$. Furthermore, the sequence $\{M_{n^{-1}}\}_{n\in\N}$ converges to $F$ and $\{R_{n^{-1}}\}_{n\in\N}$ converges to $E$, thus Proposition~\ref{prop:BorelMerBayes} states that
\begin{equation}
\P[E|F]=\frac{1}{2}\left(\cos\left(\frac{1}{4}\pi\right)-\cos\left(\frac{3}{4}\pi\right)\right)=\frac{\sqrt{2}}{2}.
\end{equation}

We now have both $\P[E|F]=\frac{1}{2}$ and $\P[E|F]=\frac{1}{2}\sqrt{2}$, which are clearly not equal. Conjecture~\ref{con:BorelConjecture} is therefore false.
\end{proof}

\section{Proof of Proposition~\ref{prop:SafeDice}}\label{app:ProofSafeDice}
\begin{propositionSafeDice}
Let $\X$, $\Y$, $\Omega$, $U$, $V$ and $\Pmod$ be as before. Let
\begin{equation}
\tilde{\mathcal{P}}=\left\{\P\middle|
\begin{array}{l}
\P[U=1\mid V=y_1]=\P[U=2\mid V=y_1],\\
\P[U=3\mid V=y_1]=\frac{1}{2}-5\P[U=1\mid V=y_1],\\
\P[U=5\mid V=y_2]=\P[U=6\mid V=y_2],\\
\P[U=3\mid V=y_2]=3\P[U=5\mid V=y_2]+\frac{1}{2},\\
\P[U=1\mid V=y_1],\P[U=5\mid V=y_1]\in\left[0,\frac{1}{10}\right]
\end{array}
\right\}
\end{equation}
be a set of probability distributions on $\Omega$. Let $\Psafe$ be a probability distribution on $\Omega$ with $\Psafe[U=1|V=y_1]=\Psafe[U=2|V=y_1]$, $\Psafe[U=5|V=y_2]=\Psafe[U=6|V=y_2]$ and $\Psafe[U\in y|V=y]=1$ for all $y\in\Y$. The following are equivalent:
\begin{enumerate}
    \item $\Psafe$ is a member of $\tilde{\mathcal{P}}$.
    \item $\Psafe$ is safe for $\langle U\rangle|[V]$.
    \item $\Psafe$ is safe for $\langle U\rangle|\langle V\rangle$.
\end{enumerate}

Furthermore, distribution $\Psafe$ is not safe for $U|[V]$.
\end{propositionSafeDice}
\begin{proof}
Here we will abbreviate $\Psafe[U=u|V=v]$ to $\Psafe[u|v]$ for all $(u,v)\in\Omega$. Let $\P\in\Pmod$ be arbitrary. We will start with the implication from 1 to 2. Firstly we have
\begin{equation}
\E_{\P}[U]=\sum_{u=1}^6 u\P[U=u]=\frac{7}{2}.
\end{equation}
Take an arbitrary $\Psafe\in\tilde{\mathcal{P}}$ and consider $y_1$ first. Writing $\Psafe[u|y_1]$ in terms of $\Psafe[1|y_1]$ for all $u\in\{2,3,4\}$ gives
\begin{align}
\E_{\Psafe}[U|V=y_1]&=\Psafe[1|y_1]+2\Psafe[1|y_1]+3\left(\frac{1}{2}-5\Psafe[1|y_1]\right)+4\left(3\Psafe[1|y_1]+\frac{1}{2}\right)\\
&=\frac{7}{2}+15\Psafe[1|y_1]-15\Psafe[1|y_1]=\frac{7}{2}.
\end{align}
The same calculation holds for $V=y_2$ as well. Therefore we have the equality $\E_{\P}[U]=\E_{\Psafe}[U|V=v]$ for all $\P\in\Pmod$ and $v\in\supp_{\Psafe}(V)$, letting $\Psafe$ fulfil the requirement of safety for $\langle U\rangle|[V]$.

The implication from 2 to 3 follows from Proposition~\ref{prop:SafeImply}.

Lastly, we will prove the implication from 3 to 1. Let $\Psafe$ be safe for $\langle U\rangle|\langle V\rangle$. Take $p,q\in[0,1]$ with $\P[V=y_2|U=3]=p$ and $\P[V=y_2|U=4]=q$. Definition~\ref{def:SafeProp} states that we need $\E_{\P}[\E_{\Psafe}[U|V]]=\E_{\P}[U]$. The right-hand side is already computed as $\E_{\P}[U]=\frac{7}{2}$. For $\E_{\P}[\E_{\Psafe}[U|V]]$, we first focus on $V=y_1$. Conditioning on $U$ yields
\begin{equation}
\P[V=y_1]=\sum_{u=1}^6\P[V=y_1|U=u]\P[U=u]=\frac{1}{6}\sum_{u=1}^6\P[V=y_1|U=u].
\end{equation}
The game master cannot lie, thus $\P[V=y_1|U=1]=\P[V=y_1|U=2]=1$ and $\P[V=y_1|U=5]=\P[V=y_1|U=6]=0$ immediately hold. We further have $\P[V=y_2|U=3]=p$ and $\P[V=y_2|U=4]=q$, thus the value of $\P[V=y_1]$ is
\begin{equation}
\P[V=y_1]=\frac{1}{6}\sum_{u=1}^6\P[V=y_1|U=u]=\frac{2}{3}-\frac{p+q}{6}.
\end{equation}
Since $\Y$ only has two elements, $\P[V=y_2]=\frac{1}{3}+\frac{p+q}{6}$ follows. We can now write out $\E_{\P}[\E_{\Psafe}[U|V]]$ to
\begin{align}
\E_{\P}[\E_{\Psafe}[U|V]]&=\E_{\Psafe}[U|V=y_1]\P[V=y_1]+\E_{\Psafe}[U|V=y_2]\P[V=y_2]\\
&=\left(\Psafe[1|y_1]+2\Psafe[2|y_1]+3\Psafe[3|y_1]+4\Psafe[4|y_1]\right)\left(\frac{2}{3}-\frac{p+q}{6}\right)\nonumber\\
&\phantom{=}+\left(3\Psafe[3|y_2]+4\Psafe[4|y_2]+5\Psafe[5|y_2]+6\Psafe[6|y_2]\right)\left(\frac{1}{3}+\frac{p+q}{6}\right).
\end{align}
This collapes to
\begin{align}
\E_{\P}[\E_{\Psafe}[U|V]]&=\frac{p+q}{6}\left(\sum_{u=3}^{6}u\Psafe[u|y_2]-\sum_{u=1}^4u\Psafe[u|y_1]\right)+\sum_{u=1}^4\frac{2u}{3}\Psafe[u|y_1]\nonumber\\
&\phantom{=}+\sum_{u=3}^6\frac{u}{3}\Psafe[u|y_2].
\end{align}
As we have $\E_{\P}[U]=\frac{7}{2}$, we need to get rid of the dependence on $p+q$. This results into the condition
\begin{equation}
\sum_{u=3}^{6}u\Psafe[u|y_2]=\sum_{u=1}^4u\Psafe[u|y_1].
\end{equation}
When this condition is satisfied, we need
\begin{equation}
\sum_{u=1}^4\frac{2u}{3}\Psafe[u|y_1]+\sum_{u=3}^6\frac{u}{3}\Psafe[u|y_2]=\frac{7}{2}
\end{equation}
for $\Psafe$ to be safe for $\langle U\rangle|\langle V\rangle$. Impose lastly the conditions $\Psafe[5|y_2]=\Psafe[6|y_2]$, $\Psafe[1|y_1]=\Psafe[2|y_1]$ and $\Psafe[U\in y|V=y]=1$ to ensure the game master never lies and there is no preference between $1$ and $2$ and between $5$ and $6$ to make $\Psafe$ more realistic. This all results into the following system of equations:
\begin{equation}\label{eq:SafeDiceSystem}
\begin{pmatrix}
\frac{2}{3}&\frac{4}{3}&2&\frac{8}{3}&1&\frac{4}{3}&\frac{5}{3}&2\\
-1&-2&-3&-4&3&4&5&6\\
1&-1&0&0&0&0&0&0\\
0&0&0&0&0&0&1&-1\\
1&1&1&1&0&0&0&0\\
0&0&0&0&1&1&1&1
\end{pmatrix}\begin{pmatrix}
\Psafe[1|y_1]\\\Psafe[2|y_1]\\\Psafe[3|y_1]\\\Psafe[4|y_1]\\
\Psafe[3|y_2]\\\Psafe[4|y_2]\\\Psafe[5|y_2]\\\Psafe[6|y_2]
\end{pmatrix}=\begin{pmatrix}
\frac{7}{2}\\0\\0\\0\\1\\1
\end{pmatrix}.
\end{equation}
The set $\tilde{\mathcal{P}}$ is the set of all probability distributions that are solution to this system. As $\Psafe\in\tilde{\mathcal{P}}$ needs to be a valid probability distribution, we need to have $\Psafe[u|v]\in[0,1]$ for all $(u,v)\in\Omega$. Solving \eqref{eq:SafeDiceSystem} gives
\begin{align}
\Psafe[3|y_1]&=3\Psafe[1|y_1]+\frac{1}{2},\\
\Psafe[4|y_1]&=5\Psafe[1|y_1]-\frac{1}{2},
\end{align}
thus $\Psafe[3|y_1]\in[0,1]$ holds when $\Psafe[1|y_1]\in\left[0,\frac{1}{6}\right]$ and $\Psafe[4|y_1]\in[0,1]$ holds when $\Psafe[1|y_1]\in\left[0,\frac{1}{10}\right]$. Without loss of generality $\Psafe[5|y_2]\in\left[0,\frac{1}{10}\right]$ must hold as well. This proves the last requirement $\Psafe[1|y_2],\Psafe[5|y_2]\in\left[0,\frac{1}{10}\right]$ from $\tilde{\mathcal{P}}$ and completes our proof that if a distribution $\Psafe$ is safe for $\langle U\rangle|\langle V\rangle$ with the requirements in the proposition, it is a member of $\tilde{\mathcal{P}}$.

Lastly we turn to safety for $U|[V]$. Pick probability measure $\Psafe$ on $\Omega$ arbitrarily with $\Psafe[1|y_1]=\Psafe[2|y_1]$, $\Psafe[5|y_2]=\Psafe[6|y_2]$ and $\Psafe[U\in y|V=y]=1$ for all $y\in\Y$. Proposition~\ref{prop:SafeProperties} implies $\Psafe[u|y_1]=\frac{1}{6}$ for all $u\in\X$, which is a clear contradiction to $\Psafe[U\in y|V=y]=1$. Thus there is no probability measure $\Psafe$ with the requirements in the proposition that is safe for $U|[V]$.
\end{proof}

\section{Proof of Theorem~\ref{thm:DiscMainThm}}\label{app:ProofDiscMainThm}
\begin{theoremmain*}
Let $\X$ be countable and $\Y$ be finite. Let $U$ be an $\X$-valued random variable and $V$ be a $\Y$-valued random variable. Let $\{p_u\}_{u\in\X}\subset[0,1]$ with $\sum_{u\in\X}p_u=1$. Let
\begin{equation}
\Pmod\subseteq\{\P\mid\forall u\in\X:\P[U=u]=p_u\}
\end{equation}
be our set of probability distributions on $\X\times\Y$ such that $|\Y|$ distributions $\P_1,\ldots,\P_{|\Y|}\in\Pmod$ exist imposing $|\Y|$ linearly independent vectors $(\P_i[V=v])_{v\in\Y}$ with $i\in\{1,\ldots,|\Y|\}$. Let $u\in\X$ be arbitrary and let $\Psafe$ be a distribution on $\X\times\Y$ with full support on $V$, then the following are equivalent:
\begin{enumerate}
    \item For all $v\in\Y$ we have $\Psafe[U=u|V=v]=p_{u}$.
    \item $\Psafe$ is safe for $\GeneralInd|[V]$.
    \item $\Psafe$ is safe for $\langle\GeneralInd\rangle|\langle V\rangle$.
\end{enumerate}
\end{theoremmain*}
\begin{proof}
The implication from 1 to 2 is satisfied by Proposition~\ref{prop:SafeProperties}. Let $v\in\Y$ and $\P\in\Pmod$ be arbitrary, then we have
\begin{equation}
\Psafe[U=u|V=v]=p_{u}=\P[U=u].
\end{equation}
Proposition~\ref{prop:SafeProperties} states that $\Psafe$ is safe for $\GeneralInd|[V]$.

The implication from 2 to 3 is by Proposition~\ref{prop:SafeImply}.

Consider lastly the implication from 3 to 1. Safety for $\langle\GeneralInd\rangle|\langle V\rangle$ implies
\begin{equation}
\E_{\P}[\GeneralInd]=\E_{\P}[\E_{\Psafe}[\GeneralInd|V]]
\end{equation}
for all $\P\in\Pmod$. We will construct a sufficient $\Psafe$ from this requirement and prove that $\Psafe[U=u|V=v]=p_{u}$ for all $v\in\Y$ is necessary.\\
Let $\P\in\Pmod$ be arbitrary. Take a first look at $\E_{\P}[\GeneralInd]$, then
\begin{equation}
\E_{\P}[\GeneralInd]=\P[U=u]=p_{u}.
\end{equation}
Let $\Psafe$ be an arbitrary distribution on $\X\times\Y$ with full support on $V$ that is safe for $\langle\GeneralInd\rangle|\langle V\rangle$ and let $v\in\Y$ be arbitrary, then we can write out
\begin{equation}
\E_{\Psafe}[\GeneralInd|V=v]=\Psafe[U=u|V=v].
\end{equation}
The value $\Psafe[U=u|V=v]$ always exists since $\Psafe$ has full support on $V$. The expectation $\E_{\P}[\E_{\Psafe}[\GeneralInd|V]]$ can now be expanded to
\begin{align}
\E_{\P}[\E_{\Psafe}[\GeneralInd|V]]&=\sum_{v\in\supp_{\P}(V)}\E_{\Psafe}[\GeneralInd|V=v]\P[V=v]\\
&=\sum_{v\in\supp_{\P}(V)}\Psafe[U=u|V=v]\P[V=v].
\end{align}
We now need to abbreviate our notation. From now on we write $\Psafe[u|v]$ instead of $\Psafe[U=u|V=v]$ for all $(u,v)\in\X\times\Y$. For safety for $\langle\GeneralInd\rangle|\langle V\rangle$ we require
\begin{equation}
p_{u}=\E_{\P}[\E_{\Psafe}[\GeneralInd|V]]=\sum_{v\in\supp_{\P}(V)}\Psafe[u|v]\P[V=v].\label{eq:DiscSafeCondition}
\end{equation}
We will also abbreviate the index of the sum to $v\in\Y$ instead of $v\in\supp_{\P}(V)$ as $\P[V=v]=0$ will be zero for every $v\in\Y\setminus\supp_{\P}(V)$ and this will have no impact on the summation.\\
Note that equation~\ref{eq:DiscSafeCondition} is an inner product of the vectors $(\Psafe[u|v])_{v\in\Y}$ and $(\P[V=v])_{v\in\Y}$ and the value of the latter vector is known, thus the equation $\sum_{v\in\Y}\Psafe[u|v]\P[V=v]=p_{u}$ is linear. Since $\Pmod$ has $|\Y|$ distributions $\P_1,\ldots,\P_{|\Y|}$ that have linearly independent vectors $(\P_i[V=v])_{v\in\Y}$ with $i\in\{1,\ldots,|\Y|\}$, we can create a system of $|\Y|$ linearly independent equations
\begin{equation}
\sum_{v\in\Y}\Psafe[u|v]\P_i[V=v]=p_{u},\qquad i\in\{1,\ldots,|\Y|\}.
\end{equation}
Since we have a system of $|\Y|$ linearly independent equations with $|\Y|$ unknowns $\Psafe[u|v_1],\ldots,\Psafe[u|v_{|\Y|}]$, there is a unique solution. The coefficients $\P[V=v]$ with $v\in\Y$ sum up to $1$, thus $\Psafe[u|v]=p_{u}$ is the unique solution to this system.\\
Let now $\P\in\Pmod$ again be arbitrary, then the requirement in Equation~\ref{eq:DiscSafeCondition} still holds as
\begin{equation}
\E_{\P}[\E_{\Psafe}[\GeneralInd|V]]=\sum_{v\in\supp_{\P}(V)}\Psafe[u|v]\P[V=v]=p_{u}\sum_{v\in\supp_{\P}(V)}\P[V=v]=p_{u}.
\end{equation}
Therefore $\Psafe[u|v]=p_{u}$ for all $v\in\Y$ is required for $\Psafe$ when $\Psafe$ is safe for $\langle\GeneralInd\rangle|\langle V\rangle$, thus 3 implies 1.
\end{proof}

\end{document}