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$A$and $B$play the following game: $A$writes down either number $1$or number $2$, and $B$must guess which one. If the number that $A$has written down is $i$and $B$has guessed correctly, $B$receives $i$units from $A$. If $B$makes a wrong guess, $B$pays $\frac{3}{4}$unit to $A$. If $B$randomizes his decision by guessing 1 with probability $p$and $2$with probability $1-p$, determine his expected gain if (a) $A$has written down number $1$and (b) $A$has written down number $2$.

What value of $p$maximizes the minimum possible value of $B$'s expected gain, and what is this maximin value? (Note that $B$'s expected gain depends not only on $p$, but also on what $A$does.)

Consider now player $A$. Suppose that she also randomizes her decision, writing down number $1$with probability $q$. What is $A$'s expected loss if (c) $B$chooses number 1 and (d) $B$chooses number $2$?

What value of $q$minimizes $A$'s maximum expected loss? Show that the minimum of $A$'s maximum expected loss is equal to the maximum of $B$ 's minimum expected gain. This result, known as the minimax theorem, was first established in generality by the mathematician John von Neumann and is the fundamental result in the mathematical discipline known as the theory of games. The common value is called the value of the game to player $B$.

Step by step solution

01

Calculation

If player A puts number $1$, player $B$will guess the number correctly with the probability $p$and he will miss with the probability $1-p$. Hence, the expected amount of winnings is

$E\left(X_1\right)=1路p-0.75(1-p)=\frac{7}{4}p-\frac{3}{4}$

If player A puts number $2$, player $B$will guess the number correctly with the probability $1-p$and he will miss with the probability $p$. Hence, the expected amount of winnings is

$E\left(X_2\right)=2路(1-p)-0.75p=2-\frac{11}{4}p$

We have to analyse the function $p鈫�\min \left(E\left(X_1\right),E\left(X_2\right)\right)=\min \left(\frac{7}{4}p-\frac{3}{4},2-\frac{11}{4}p\right)$. The minimum is maximized in the point where $\frac{7}{4}p-\frac{3}{4}=2-\frac{11}{4}p$. Solving this we get

$p=\frac{11}{18}鈬�\text{maximied gain}=\frac{23}{72}$

02

Continue Calculation

Now, let's look this problem from the angle of player A. If player $B$chooses number $1$, player $A$will lose one dollar with the probability $q$and will obtain $3/4$dollars with the probability $1-q$. Hence, the expected amount of loss is

$E\left(X_1\right)=1路q-0.75(1-q)=\frac{7}{4}q-\frac{3}{4}$

If player $B$chooses number $2$, player A will lose two dollars with the probability $1-q$and she will obtain $1.5$dollars with the probability $q$. Hence, the expected amount of winnings is

$E\left(X_2\right)=2路(1-q)-\frac{3}{4}q=2-\frac{11}{14}q$

We have to analyse function $q鈫�\max \left(E\left(X_1\right),E\left(X_2\right)\right)=\max \left(\frac{7}{4}p-\frac{3}{4},2-\frac{11}{4}q\right)$. The maximum is minimized in the point where $\frac{7}{4}q-\frac{3}{4}=2-\frac{11}{4}q$. Solving this we get

$q=\frac{11}{18}鈬�\text{minimized loss}=\frac{23}{72}$

Hence, we have proved the claimed.

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