### The Good, the Bad and the Weird: Duels, Truels and Utility Functions

In the excellent (and highly recommended) book "Fifty Challenging Problems in Probability with Solution", Frederick Mosteller poses "The Three-Cornered Duel":
A, B and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unit. What should A's strategy be?
This is problem 20 in Mosteller's book, and it also appears (with an almost identical solution) in Larsen & Marx "An Introduction to Probability and Its Applications".

Mosteller's solution:
A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly shoots the more dangerous C first, and A gets one shot at B with probability 0.3 of succeeding. If he misses this time, the less said the better. On the other hand, suppose A hits B.  Then C and A shoot alternately until one hits. A's chance of winning is $$(.5)(.3) + (.5)^2(.7)(.3) + (.5)^3(.7)^2(.3) + \ldots$$ . Each term corresponds to a sequence of misses by both C and A ending with a final hit by A. Summing the geometric series we get ... $$3/13 < 3/10$$. Thus hitting B and finishing off with C has less probability of winning for A than just missing the first shot. So A fires his first shot into the ground and then tries to hit B with his next shot. C is out of luck.
Is this right? What if B were to follow A's example and fire into the ground? What if all three were to keep firing into the ground? That this type of an outcome isn't unreasonable for certain sets of shot accuracy probabilities can be illustrated by considering the case where A's accuracy is 0.98, B's accuracy is 1.0 and C's accuracy is 0.99. Mosteller's argument is equally applicable in this case, but if B shoots C after A deliberately misses he'll be shot by A with probability 0.98. Is that reasonable?

Under the assumption that deliberately missing is allowed, there are $$2^3-1=7$$ possible outcomes - each participant can be shot or not shot, and there must be at least one participant not shot. The lack of clarity for what the ideal strategies are for A, B and C in the general case arises from the utility of 2- or 3-way ties to each of the participants being undefined.

In the next article I'll analyze 2-way duels where deliberate missing is allowed by using such fully-defined utility functions. These results will be used in a third article on 3-way duels (truels); in particular, I'll re-examine Mosteller's solution.

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows $\frac{\partial W}{\partial P}(P=1) = … ### Solving a Math Puzzle using Physics The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds. The first two parts require students to interpret the meaning of the components of the formula $$T(x) = 5 \sqrt{36+x^2} + 4(20-x)$$, and the final "challenge" component involves finding the minimum of $$T(x)$$ over $$0 \leq x \leq 20$$. Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities. One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},$ where $$\theta_1, \theta_2$$ are the angles to the normal and $$v_1, v_2$$ are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …
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