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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!

2 Answers 2

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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?

  • 0
    Thanks for your efforts! Do you know the [Monty Hall problem](http://en.wikipedia.org/wiki/Monty_Hall_problem), to which I believe this question is related? Treacherous waters!2012-06-07