Answer to Question #106117 in Operations Research for Riya

Question

Player A and B play a game in which each has three coins, a 5p, 10p and a 20p.
Each selects a coin without the knowledge of the other’s choice. If the sum of the
coins is an odd amount, then A wins B’s coin. But, if the sum is even, then B wins
A’s coin. Find the best strategy for each player and the values of the game.

Answer 1

Create a table

$\def\arraystretch{1.5} \begin{array}{c:c:c:c} A/B & B_1 & B_2 & B_3 \\ \hline A_1 & -5 & 10 & 20 \\ \hdashline A_2 & 5 & -10 & -10\\ \hdashline A_3 & 5 & -20 & -20\\ \end{array}$

$p_1$ - probability of choosing $A_1$

$p_2$ - probability of choosing $A_2$

$q_1$ - probability of choosing $B_1$

$q_2$ - probability of choosing $B_2$

$-5p_1+5p_2=10p_1-10p_2$ or

$p_1=p_2$

$p_1+p_2=1$

Therefore, $p_1=0.5$ & $p_2=0.5$

The same we do with $q_1$ & $q_2$

$-5q_1+10q_2=5q_1-10q_2$ or

$q_1=2q_2$

$q_1+q2=1$

Therefore, $q_2=1/3$ & $q1=2/3$

First value of rows is the choice of player A. The first value of columns is the choice of player B. All others elements are values of winner(and the money he won).

Now we see that the best strategy for A is to choose 5p, thus he wins in 2 of 3 cases ( win - 0.6(6), win10 - 0,3(3), win20 - 0,3(3)) 

For player B the best strategy is to choose 10p, as when he chooses 10 or 20 he wins twice. But at the same time, if he loses it is better to lose 10p instead of 20p (win - 0.6(6), win10 - 0,3(3), win20 - 0,3(3))

Value of the game = $[a_{11}a_{22}-a_{21}a_{12}]/[(a_{11}+a_{22}-(a_{12}+a_{21}))]=$

$=[(-5)*(-10)-(5)*(10)]/[{-5+(-10)}-{10+5}]=0$