## Basic Mixed Strategy Example

A mixed strategy game is one in which there is no dominant strategy for either player. Each player chooses each strategy with some probability.Suppose a two player zero sum game has the following payoff matrix.

A\B | \[B_1\] | \[B_2\] |

\[A_1\] | 2 | -11 |

\[A_2\] | -4 | 3 |

\[A_1, \: A_2\]

with probabilities \[p\]

and \[1-p\]

respectively, then A will expect to receive \[2p-4(1-p)=6p-4\]

for every game he plays. If B plays strategy \[B_2\]

, then A will expect to win \[-p+3(1-p)=3-4p\]

.A will maximise their winnings if he chooses

\[p\]

such that his minimum winnings are as high as possible. This is when \[6p-4=3-4p \rightarrow p=\frac{7}{10}\]

.We can perform the same analysis for player and B.

If A plays strategy 1, and B picks strategies

\[B_1, \: B_2\]

with prbabilities \[q\]

and \[1-q\]

respectively, then B will expect to receive \[-(2q-(1-q))=-(3q-1)\]

for every game he plays. If A plays strategy \[A_2\]

, then B will expect to win \[-(-4q+3(1-q))=-(3-7q)\]

.B will maximise their winnings if he chooses

\[p\]

such that his minimum winnings are as high as possible. This is when \[-(3q-1)=-(3-7q) \rightarrow q=\frac{4}{10}\]

.