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## 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
If B plays strategy 1, and A picks strategies
$A_1, \: A_2$
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
$-(3q-1)=-(3-7q) \rightarrow q=\frac{4}{10}$
.