# value of this possibly Monty Hall-related 2-person zero-sum game?

Player A tosses a fair coin. He knows how it lands; Player B does not. A can now play move 1 or move 2. If he plays move 1, he pays B £1. If he plays move 2, then B can either play move X or move Y. If he plays move X, he plays A £1. If he plays move Y, then he pays A £2 if the coin came up heads and collects £2 from A if it came up tails.

What's the value of this game?

I have been getting tied up, after starting with the idea that A must always play move 2, reaching a contradiction. The answer is supposed to be $\frac{1}{3}$.

Many thanks for any help with this!

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Suppose that $A$ plays strategy $2$ with probability $p_H$ when the coin comes up heads and probability $p_T$ when it comes up tails, and suppose that $B$ plays $Y$ with probability $q$. The expected payoff to $A$ when the coin comes up heads is then

$$-(1-p_H)+p_H(1-q)+2p_Hq=2p_H+p_Hq-1\;,$$

and the expected payoff to $A$ when it comes up tails is

$$-(1-p_T)+p_T(1-q)-2p_Tq=2p_T-3p_Tq-1\;,$$

so the expected payoff to $A$ is

$$p_H+p_T+\frac{q}2(p_H-3p_T)-1\;.$$

For a given $q$ and $p_H$ this is clearly maximized when $p_H=1$, when it is $$f(p_T,q)=\left(1-\frac32q\right)p_T+\frac{q}2\;.$$

Now $$f_{p_T}(p_T,q)=1-\frac32q\quad\text{ and }\quad f_q(p_T,q)=-\frac32p_T+\frac12\;,$$

so the unique critical point of $f$ is at $p_T=\frac13$ and $q=\frac23$. Let $p'=p_T-\frac13$ and $q'=q-\frac23$. Then

\begin{align*} f(p_T,q)&=\left(1-\frac32\left(q'+\frac23\right)\right)\left(p'+\frac13\right)+\frac12\left(q'+\frac23\right)\\ &=p'-\frac32p'q'-p'+\frac13-\frac12q'-\frac13+\frac12q'+\frac13\\ &=\frac13-\frac32p'q'\;. \end{align*}

At the critical point, therefore, the expected payoff to $A$ is $\frac13$. However, this is a saddle point: if $p_T$ and $q$ are either both increased or both decreased, the expected payoff to $A$ decreases, but if they are changed in opposite directions, it increases. Thus, if the opponent’s probabilities are known, either player can punish the other for deviating from those at the critical point, and in that sense the value of the game is $\frac13$ to $A$.

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This is the best I could come up with:

If you build a decision tree you should get something like that:

$1:\quad A \rightarrow_{\text{Move 1}} -1 \quad A \rightarrow_{\text{Move 2}} [2]$

$2:\quad B \rightarrow_{\text{Move X}} +1 \quad B \rightarrow_{\text{Move Y}} [3]$

$3:\quad Coin \rightarrow_{\text{Heads}, {p=0,5}} +2 \quad Coin \rightarrow_{\text{Tails}, {p=0,5}} -2$

All the values are the payoffs of A. Branch $3$'s expected payoff is 0, since the coin is a fair coin:

$3:\quad (0,5\cdot+2) + (0,5\cdot-2) = 0$

B can choose between $+1$ and $0$. Hence, since its goal is to minimize A's payoff, he chooses $0$ (move Y):

$2:\quad B \rightarrow_{\text{Move X}} +1 \quad B \rightarrow_{\text{Move Y}} 0$

A must maximize its payoff so between $-1$ and $0$ he will always choose $0$ (move 2) :

$1:\quad A \rightarrow_{\text{Move 1}} -1 \quad A \rightarrow_{\text{Move 2}} 0$

Does it makes sense to you?

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So you get the value of the game as 0? But can A possibly improve on that by playing move 1 some of the time when he knows the coin has come up tails? This is given that B moves without knowing whether it's heads or tails. This is sounding very like Monty Hall. Maybe I should run a Monte Carlo simulation. – Harry Macpherson Jun 7 '12 at 10:47
What would be A's best strategy? Maybe to play Move 2 and let B decide only when he is sure to get a positive payoff (i.e., when the coin is tail)? In that case B will always play X. And so you get (0.5 * -1) + (0.5 * +1) = 0. Maybe you can achieve an higher value with a "random strategy" but I can't see a "rational strategy" that achieves more than 0. Let me know if you have other ideas. – Pietro M. Jun 7 '12 at 11:27
You can also consider this: A chooses to let B make its move when B doesn't actually have a choice (i.e. the expected payoff of [3] is +1). This happens when A plays "2" 3/4 of times he gets head and 1/4 of times he gets tails: the expected payoff of [3] in this case is +1. So it is indifferent what choice B makes. Ultimately A gets a payoff of 0,5*-1 + 0,5*+1 = 0, because 1/4*1/2 (chance of Tails) + 3/4*1/2 (chance of Heads)= 0,5. – Pietro M. Jun 7 '12 at 11:45
Corrected: Let A's strategy be to play move 2 whenever it's heads, and to play move 2 $\frac{1}{6}$ of the time when it's tails. Let B's optimal strategy be to play move X with probability $p$ and move Y with probability $1-p$. When it's heads, A gains $2p+(1−p)=p+1$. When it's tails, A gains $−2∗(\frac{1}{6})∗p+\frac{1}{6}*(1−p)−\frac{5}{6}=−\frac{2}{3}−\frac{p}{2}$. The coin is fair, so A's expected gain is $\frac{1}{2}(p+1)+\frac{1}{2}(−\frac{2}{3}−\frac{p}{2})=\frac{p}{4}+ \frac{1}{6}$, which B minimises by setting p equal to 0, which gives A an expected gain of $\frac{1}{6} > 0$. – Harry Macpherson Jun 7 '12 at 14:43
I'm sorry, I realize I don't have enough knowledge to help you. Hope you'll find someone smarter than me. I tried. – Pietro M. Jun 7 '12 at 15:11