# Minimax solution for Zero-Sum Game

I try to understand the way to finding the minimax solution to zero-sum game.

The following example is takes from Wikipedia. Minimax

Wikipedia: The following example of a zero-sum game, where A and B make simultaneous moves, illustrates minimax solutions.Suppose each player has three choices and consider the payoff matrix for A displayed at right. Assume the payoff matrix for B is the same matrix with the signs reversed (i.e. if the choices are A1 and B1 then B pays 3 to A). Then, the minimax choice for A is A2 since the worst possible result is then having to pay 1, while the simple minimax choice for B is B2 since the worst possible result is then no payment.

Question: Why the minimax choice for A is actually A2, the worst possible result is actually A3 (-4), and the worst possible result for B is B3 (+4). In addition the wording "simple minimax choice for B" is very confusing, is it different from the standard minimax choice?

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The worst A takes when taking (across the rows):

• A1: -2
• A2: -1
• A3: -4

The minimum loss is A2.

The worst B takes when taking (along the columns):

• B1: -3
• B2: 0
• B3: -4

It is best for B to pick B2.

Does that help? :)

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It is A2 because the following. A asks oneself "What is the worst thing that can happen to me if I play A1?" The answer to that is -2. "What is the worst thing that can happen to me if I play A2\$" The answer to that is -1. And finally "What is the worst thing that can happen to me if I play A3" The answer to that is -4. Hence, I'm better off in a minimax sense if I play A2. The other thinks the same, her payoffs tho are just the negatives of those above (since one pays them the other receives)

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