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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  
\[-(3q-1)=-(3-7q) \rightarrow q=\frac{4}{10}\]
.