Consider a guessing game with N people and the person closes

Consider a guessing game with N people, and the person closest to 20 plus one-half of the average is awarded the prize, which is split in the event of a tie. The range of guesses is from 0 to 100. Show that 40 is Nash equilibrium. (Hint: Suppose that N-1 people choose 40 and one person deviates to x < 40. Calculate the target as a function of the deviant’s decision, x. The deviant will lose if the target is greater than the midpoint between x and 40, i.e., if the target is greater than (40 + x)/2. (Check to see if this is the case.)

Solution

To show that everyone choosing 40 is Nash equilibrium, we will have to consider a situation where everyone chooses 40 and then show that any deviant would lose by choosing any number other than 40.

The target number is as follows:

Target              =          20        +          [(x₁+ x₂ + … + x_N)/2N]

where x_i­ is the number chosen by the ith person.

Suppose everyone chooses 40. Then,

            Target = 40N/N = 40

Since everyone’s number is equally close to the target, the prize is equally split between everyone.

Now suppose the deviant chooses a number x other than 40. The target as a function of deviant’s decision is as follows.

            Target              =          20        +          [x + 40(N – 1)]/2N

Case 1: x < 40

In this case, the target will lie between 40 and x. The deviant will win if the target is closer to x and lose if the target is closer to 40. In other words: since x is less than 40, the deviant will win if the target is less than the average of x and 40 and lose if the target is greater than the average of x and 40. For everyone choosing 40 to be Nash equilibrium, the deviant must lose. That is, we have to show:

            Target > (x+40)/2

Substitute the value of target in the above expression and then simplify it in the form of x.

            20        +          [x + 40(N – 1)]/2N     >          (x+40)/2

            20 + (x/2N) + 20 – (40/2N)                 >          (x/2) + 20

            (x/2N) – (x/2)                                     >          (20/N) – 20

            (x – Nx)/2N                                         >          (20 – 20N)/N

            x(1 – N)                                               >          40 – 40N

            x(1 – N)                                               >          40(1 – N)

Since (1 – N) is negative, canceling it out on both the sides will require the sign to be changed. Therefore,

            x          <          40

which is indeed the case. Therefore, the deviant will surely lose.

Case 2: x > 40

In this case also, the target will lie between 40 and x. Again the deviant will win if the target is closer to x and lose if the target is closer to 40. But since x is greater than 40, the deviant will win if the target is greater than the average of x and 40 and lose if the target is less than the average of x and 40. For everyone choosing 40 to be Nash equilibrium, the deviant must lose. That is, we have to show:

            Target < (x+40)/2

Substitute the value of target in the above expression and then simplify it in the form of x.

            20        +          [x + 40(N – 1)]/2N     <          (x+40)/2

            20 + (x/2N) + 20 – (40/2N)                 <          (x/2) + 20

            (x/2N) – (x/2)                                     <          (20/N) – 20

            (x – Nx)/2N                                         <          (20 – 20N)/N

            x(1 – N)                                               <          40 – 40N

            x(1 – N)                                               <          40(1 – N)

Since (1 – N) is negative, canceling it out on both the sides will require the sign to be changed. Therefore,

            x          >          40

which is indeed the case. Therefore, the deviant will surely lose.

Conclusion: When everyone chooses 40, no one has any incentive deviate, implying that everyone choosing 40 is Nash equilibrium.