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The game is described as follows :

In a simplified version of the game Morra each of the players shows one finger or two fingers, and simultaneously guesses how many fingerss the other player will show. If both players guess correctly, or both players guess incorrectly , there is no payoff. If just one player, guesses correctly, that player wins a payoff equal to the total number of fingers shown by both players

I know that this game has to be zero sum but I want to understand well why it is the case, after all one player gains something which teoretically is not lost by the other in the sense that he had it before the game started. In this case, there is a situation in which the player who 'lost' the point could have win them but it is just the case because the game it's simmetric, it not always happens.

I thought about the following abstraction to explain what is happening, say the two players had before starting the game an amount of points so that to cover all the possible negative pay-off for them, and that who wins receives the number of points equal to the sum of the fingers from the other playes, now it seems to me that the game is zero-sum. Otherwise it seems to me that the pay-offs would be possible to be represented by

$$(\text{sum of fingers in the case of first wins something }, 0 ) \text{ or } $$ $$(0, \text{sum of fingers in the case of second wins something} )$$

which would make the game not zero-sum.

Should it be obvious? Am I totally wrong with my reasoning? Why?

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    $\begingroup$ The description of the payoffs seems incomplete: What is the payoff to the player who guesses incorrectly when the other player is correct? $\endgroup$ – Herr K. Jan 28 at 1:17
  • $\begingroup$ It is not described at all ! The exercise is written exactly as I reported it $\endgroup$ – Tortar Jan 28 at 9:41
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    $\begingroup$ If the payoffs are not fully described, you cannot determine whether the game is zero-sum or not. Since you "know" it has to be zero-sum, it must be that the payoffs in the case in which player $1$ guesses $X$ correctly and $2$ is incorrect must be $(X,-X)$ and vice-versa. $\endgroup$ – Bayesian Jan 28 at 9:44
  • $\begingroup$ This is Exercise 7 on p. 22 of Philip Straffin's book "Game Theory and Strategy". It appears in Chapter 3, "Zero-Sum Games". So though the payoffs are not fully specified, they are obviously meant to be zero-sum: The loser pays the winner. $\endgroup$ – VARulle Jan 29 at 10:20
  • $\begingroup$ yes, you're right it was obvious but I wanted to be sure I didn't miss something $\endgroup$ – Tortar Jan 29 at 13:17
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A zero-sum game is one in which the players' payoffs sum to zero in every possible outcome of the game.

The given payoff description is incomplete. But if we assume that in the outcomes where only one player guesses correctly, the other player (who guesses wrong) would have to lose an amount equal to the total number of fingers shown (denoted $X$, say, to use @Bayesian's notation), then the game is indeed zero-sum:

  • In outcomes where both guess correctly or incorrectly, they each get $0$, which sum to zero.
  • In outcomes where only one guesses correctly, the correct guesser gets $X$ and the incorrect guesser gets $-X$, which also sum to zero.
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