Recall the game “matching pennies“. Player I has to chose between ‘0’ or ‘1’, player II has to chose between ‘0’ and ‘1’.No player knows what is the choice of the other player before making his choice. Player II pays to player I one dollar if the two number match and gets one dollar from player I if they do not match.

The optimal way to play the game for each of the player is via a mixed strategy. Player I has to choose ‘0’ with probability 1/2 and ‘1’ with probability 1/2. And so is player II.

Now, suppose that the game is the same except that if the outcomes match and both player chose ‘1’ player II pays 5 dollars to player I and not one dollar. (All other payoffs remain at one dollar.)

Test your intuition: What is the probability player I should play ‘1’. (More than 1/2? less than 1/2? more than 1/3? less than 1/3? more than 2/3? less than 2/3?)

According to Abraham Neyman, failing to have the right intuition in this case caused Barcelona to fail in the second game against Inter Milano, in the semi finals of the European champion leagues last year. Do you see a connection? I will come back to that in a later post.

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ufo makerIn Line4

Player II pays to player II one dollar~

↑

Ⅰ？

(GK: Corrected, thanks)ufo makerI think at first site,under such exception, optimal strategy for Ⅰ is to choose 1 everytime,because it assure Ⅰ not lose any money.

But if some condition exist, (for example Ⅰ have to yield above ~dollar

in ~games ) then, maybe optimal strategy have to be changed.

If no such factor of “hurry”exist, then it is optimal for Ⅰ to choose 1 all time.

Colin ReidThe usual strategy in these games, as I understand it, is a mixed one such that you are indifferent to what your opponent does on average. For player I to be indifferent between player II picking ‘0’ and ‘1’, he has to pick ‘1’ himself with low probability (1/4 or less?) because the payoff for matching ‘1’s is so much higher. Likewise player II will pick ‘1’ with low probability.

Albert JiangNice puzzle! In a completely-mixed Nash equilibrium of this game, player I’s strategy has to be such that player II is indifferent between picking 0 or 1. Solving the corresponding linear equation, player I should choose ‘1’ with probability 1/4. Similarly player II should choose ‘1’ with probability 1/4. This is somewhat counter-intuitive since if player I believes II is playing an evenly mixed strategy (1/2,1/2), then her best strategy would be to play ‘1’ with a high probability (1). It would be interesting to run experiments on this game with human players…

Johan RichterThere is a variant of this question that is also a good test of intuition. Suppose that we change the original game so that an outside person will pay

player I an extra dollar if both coins are tails. What is the equilibrium strategy now?