Perfect Probabilistic Encryption still requires key length about as long as message Let $(E,D)$ be a probabilistic encryption scheme with $n$-length keys (given a key $k$, we denote the corresponding encryption function by $E_k$) and $n+10$-length messages. Then, show that there exist two messages $x_0, x_1 \in \{0,1\}^{n+10}$ and a function $A$ such that 

$$\mathrm{Pr}_{b \in \{0,1\}, k \in \{0,1\}^n}[A(E_k(x_b)) = b ] \geq \frac{9}{10}.$$ 

(This is problem 9.4 from Arora/Barak Computational Complexity.)

My gut intuition says that the same idea from the proof in the deterministic case should carry over. WLOG let $x_0 = 0^{n+10}$, and denote by $S$ the support of $E_{U_n}(0^{n+10})$. We will take $A$ to output $0$ if the input is in $S$. Then, assuming the condition stated in the problem fails to hold for all $x \in \{0,1\}^{n+10}$, we conclude that  $\mathrm{Pr}[E_{U_n}(x) \in S] \geq 2/10$ for all $x$. This implies that there exists some key so that $E_k$ maps at least $2/10$ of the $x$ into $S$ (the analogue of this statement in the deterministic case suffices to derive a contradiction), but now I don't really see how to continue. Is my choice of $A$ here correct, or should I be using a different approach?
It seems you have to use some different approach. There is a hint in Arora/Barak's book. From that hint, I got the following solution.

Again we set $x_0=0^{n+10}$. The algorithm $A$ works as follows. Let $y=E_k(x_b)$ be its input. Let $Y_{x_1}$ denote the set of ciphtertexts $y$ such that there exists some $k$ satisfying $D_k(y)=x_1$ ($D$ is the decryption algorithm). The algorithm $A$ outputs $1$ if and only if $y\in Y_{x_1}$ (so $A$ runs in polynomial time if P = NP). Then we have

$$\mathrm{Pr}_{b \in \{0,1\}, k \in \{0,1\}^n}[A(E_k(x_b)) = b ]=\frac{1}{2}\left(1-\mathrm{Pr}_{k\in\{0,1\}^n}[E_k(x_0)\in Y_{x_1}]\right)+\frac{1}{2}.$$

If this probability is less than $9/10$ for all $x_1\in \{0,1\}^{n+10}$, then we have $\mathrm{Pr}_{k\in\{0,1\}^n}[E_k(x_0)\in Y_{x_1}]&gt;1/5$. This means $\mathrm{Pr}_{k\in\{0,1\}^n,k'\in\{0,1\}^n}[D_{k'}(E_k(x_0))=x_1]&gt;2^{-n}/5$. Note this inequality holds for all $x_1\in \{0,1\}^{n+10}$, which means

\begin{align}
1&amp;=\mathrm{Pr}_{k\in\{0,1\}^n,k'\in\{0,1\}^n}[D_{k'}(E_k(x_0))\in\{0,1\}^{n+10}]\\
&amp;=\sum_{x_1\in\{0,1\}^{n+10}}\mathrm{Pr}_{k\in\{0,1\}^n,k'\in\{0,1\}^n}[D_{k'}(E_k(x_0))=x_1]\\
&amp;&gt;\frac{2^{n+10}\cdot2^{-n}}{5},
\end{align}

a contradiction!