Here's the problem:
Let's consider a game for two players in which each player selects a non-negative number, with the maximum value not exceeding 1000. Player 1 chooses even numbers, while Player 2 selects odd numbers. After both players announce their numbers, the player who chose the lower number wins an amount of money equal to the value of their selected number. The other player receives nothing. Your task is to find all the pure Nash equilibria of this game.
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My solution : The Nash Equilibrium of this game is a (0, 1) strategy. Meaning, the first player always chooses 0 and the second one always chooses 1. So they both gain 0 utility. I came to this by thinking that if we start from the top (so strategy (1000, 999)), every player always can descend to increase their utility. Then, when we reach bottom, no player can increase utility by increasing bet. Thus, the strategy is (0,1).
Am I right? And is this the only pure Nash equilibrium?
Best Answer
Hint: If one player chooses a number that exceeds the one chosen by the other by at least 3, that player can improve by increasing their number by 2. It follows that in every pure strategy Nash equilibrium, the two numbers must differ by exactly $1$. The one choosing the larger number must not be able to or not benefit from choosing a smaller number either. There are two such pairs of numbers with this property.