EXAMPLE 5 Another Matching Game

In this game, Players I and II can again show either heads or tails . When two Hs appear, Player II pays $5 to Player I. When two Ts appear, Player II pays $1 to Player I. When one and one are displayed, then Player II collects $3 from Player I. The payoffs (that Player I receives from Player II) for this game are given in Table 15.6.

A worst-case analysis, like that which solved our initial location/schedule game, is of little help here. Player I may lose $3 whether he plays or , making his maximin . Player II can keep down her losses to $1 by always playing (and thus avoiding the loss of $5 when two Hs appear), so Player Il’s minimax is 1. However, if Player II chooses and Player I knows this, then Player I will also play and collect $1 from Player II. Can Player II do better than lose $1 in each play of the game?

Consider the situation where Player I uses a mixed strategy , which involves playing with probability and playing with probability , where . Against Player Il’s pure strategy , Player I’s expected value is

Against Player Il’s pure strategy , Player I’s expected value is

These two linear equations in the variable are depicted in Figure 15.3. Note that the four points where these two lines intersect the two vertical lines, and , are the four payoffs appearing in the payoff matrix.

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The point at which the lines given by and intersect can be found by setting , yielding

so . To the left of , , and to the right, ; at , . If Player I chooses , he can ensure an average payoff of

regardless of what Player II does.

In other words, Player I’s optimal mixed strategy is to pick and with probabilities and , respectively, which gives Player I an expected value of and prevents him from having a bigger average loss. As can be seen from Figure 15.3, is the highest expected value that Player I can guarantee against both strategies and of Player II. Although yields Player I a higher expected value for , and yields him a higher expected value for , Player I’s choice of protects him against an expected loss greater than , which neither of his pure strategies does (each may produce a maximum loss of −3). Put another way, the intersection of and at is the minimum of the function given by to the left and to the right (shown by the dashed line in Figure 15.3). If Player II had more than two strategies, this approach to finding a minimum that puts a floor on Player I’s expected loss can be extended.

A similar calculation for Player II results in the same optimal mixed strategy and expected value . But because the payoffs for Player II are losses, means that she gains on the average. It is a coincidence that Player I and Il’s optimal mixed strategies are identical; in the penalty shootout that we will return to in Example 6, this is not the case.

Therefore, this game is unfair, even though the sum of the amounts ($6) that Player I might have to pay Player II when he loses is the same as the sum that Player II might have to pay Player I when she loses. Interestingly, Player II, who will win an average of cents each time the game is played, is favored, even though she may have to pay more to Player I when she loses (a maximum of $5) than Player I will ever have to pay her (a maximum of $3).