Question #106117
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.
1
Expert's answer
2020-03-31T10:44:23-0400

Create a table

A/BB1B2B3A151020A251010A352020\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}

p1p_1 - probability of choosing A1A_1

p2p_2 - probability of choosing A2A_2

q1q_1 - probability of choosing B1B_1

q2q_2 - probability of choosing B2B_2

5p1+5p2=10p110p2-5p_1+5p_2=10p_1-10p_2 or

p1=p2p_1=p_2

p1+p2=1p_1+p_2=1

Therefore, p1=0.5p_1=0.5 & p2=0.5p_2=0.5

The same we do with q1q_1 & q2q_2

5q1+10q2=5q110q2-5q_1+10q_2=5q_1-10q_2 or

q1=2q2q_1=2q_2

q1+q2=1q_1+q2=1

Therefore, q2=1/3q_2=1/3 & q1=2/3q1=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 = [a11a22a21a12]/[(a11+a22(a12+a21))]=[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=[(-5)*(-10)-(5)*(10)]/[{-5+(-10)}-{10+5}]=0



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