Game Theory: Penalty Shot Game Given a game matrix for the penalty shot game:
(1/2,-1/2)           (-1,1)
(-1,1)              (1/3,-1/3)
What is the minimax strategy and expected value for this game?
I calculated the minimax strategy to be -4 for both the shooter and goalie. However, I'm not sure if I did the calculation right. Could anyone shed some light on this. Thanks
 A: Suppose the row player chooses top with probability $p$ and bottom with probability $1-p$.
If the column player chooes left, the row player's expected payoff is $$\frac{1}{2}p-(1-p)\cdot 1.$$ If the column player chooses right, the row player's expected payoff is $$-p\cdot1+\frac{1}{3}(1-p).$$
The minimax strategy is the $p$ that causes these two payoffs to be equal. $$\frac{1}{2}p-(1-p)=-p+\frac{1}{3}(1-p),$$ so $p=8/17$ is the minimax strategy.
The value is found by substituting this value of $p$ into the payoffs from above so the value is $-5/17$.
You can read a summary of minimax solutions here: http://www.mit.edu/~jcrandal/16.499/GameTheoryBasics.pdf.
Edit: fixed mistake in the algebra spotted by memo.
A: 
Blockquote What is the minimax strategy and expected value for this game?

In zero-sum games minimax strategy and Nash equilibrium strategies are equivalent and the value of the game is just the expected payoff resulting from minimax strategies. 
The Nash equilibrium of the game is $(\frac{8}{17},\frac{9}{17})(\frac{8}{17},\frac{9}{17})$, i.e. the minimax strategy of player 1 and of player 2 are the same and is $(\frac{8}{17},\frac{9}{17})$.
The value of the game is thus $\frac{-5}{17}$
If you want to calculate the optimal strategies in the usual way you can solve the minmax problem.
$\min\max\{\frac{3}{2}q-1,\frac{1}{3}-\frac{4}{3}q\}=\bar{v}$
where $\frac{3}{2}q-1$ is the expected payoff to row player if s/he plays the 1st action, $\frac{1}{3}-\frac{4}{3}q$ is the expected payoff to row player if s/he plays the 2nd action and  $q$ is the probability that column player plays the first action. If you solve it, you get $q^*=\frac{8}{17}$ and the upper value $\bar{v}$  equals the lower value (maxmin) because the game is finite so the value is $v=\frac{-5}{17}$  
