Question

3.The “Guess-the Mean Game” is played as follows: Each of the N players writes on a piece of paper an integer between 1 and 10 without showing it to any other players. The players then reveal the numbers they have written down. The winner is that person whose number comes the closest to equaling half the average of the number submitted by the other players. For example, if there are four players and they write down the numbers 1, 3, 5, and 6 respectively, then |1 - (1/2)(3 + 5 + 6)/3| =1.33,         |3 - (1/2)(1 + 5 +6)/3| = 1, |5 - (1/2)(1 + 3 + 6)/3| = 3.33, and |6 - (1/2)(1 + 3 + 5)/3| = 4.5. The winner in this case is the second player, who wrote down the number “3.” Every player strictly prefers winning to tying, and strictly prefers tying to losing. What is the unique Nash equilibrium of this game?

Answer 1

The Nash equilibrium of this game can be found out by eliminating weakly dominated strategies.

Lets see for a three player game. If other two players choose a number more than 5, then the sum would vary from 10 to 20 and the average would vary from 5 to 10. Hence, a person who has to predict a number closest to half the average should write a number from 2.5 to 5. This means that guessing a number above 5 is weakly dominated by any other guess. An individual would never guess a number more than 5.

Similarly, one can check that numbers above 5/2= 2.5 are weakly dominated by any other guess believing that no body chooses a number above 5. If everyone believes that nobody would choose 5 and should choose a number below 5, the sum of other two players would vary from 0 to 10, the average would vary from 0 to 5, half of average would vary from 0 to 2.5. Hence, nobody chooses a number above 2.5.

Moving in the same fashion, all numbers above 0 will be weakly dominated strategies and hence believing that others stick to their strategies, the best response is to choose 0.

One can verify this n>3, the nash equilibrium will be to choose 0