Question

3. On the first day of class we played the beauty contest in which n players submitted a number in the interval [0, 10 and the player closest to won (with ties broke by randomization). Here š denotes the average strategy played 3 (a) What strategies survive iterative deletion of strictly dominated strategies? (b) What are the Nash equiliburia of the game?

Answer 1

a) The only strategy which will survive repeated iterations is 0

In the first iteration, all values more than 3.33 will be dominated and hence be eliminated. In the second iteration, any value between 1.11 and 3.33 will be similarly eliminated and so on. Finally only 0 will remain.

Explanation

For the players, the highest value which could be chosen is 10.

Even if all players chose the highest value, the average of it would be 10. (Remember, this is the maximum average. Even if one player chose less than 10, the average will drop)

So, the winning value can have a maximum of 10/3 (or 3.33). Any value more than 3.33 is dominated by 3.33 (In other words, under no circumstance is a player better off by choosing a value more than 3.33 instead of 3.33 itself)

Hence, all players should eliminate numbers above 3.33 from their strategy list.

Now, since all players are rational and eliminates all numbers greater than 3.33 from their strategy list, the maximum value of the winning number would be 3.33/3 ~ 1.11

That is, in the second iteration, we eliminate all values greater than 1.11 from our strategy list.

And going ahead with more and more iterations, we can eliminate all numbers other than 0 from the strategy list. Hence, only 0 will remain in the end.

b) Nash equilibrium is when all players go with number 0

Definition of NE: State from where no player would want to deviate (i.e change strategy) in a unilateral manner.

In the event of all players choosing 0, the average is 0 and the winning number itself is 0. And each player has 1/n chance of winning. (Since tie is brocken with randomisation).

From this state, if anyone wishes to deviate and choose 'x' instead of 0 unilaterally, the average would be (x/n)

And the winning number would be, (x/3n). This would be nearer to 0 (which is chosen by all other players) than 'x'. Hence the player who chose will lose. Therefore, no rational player will deviate from the original choice of 0.

Therefore, 0 is the Nash equilibrium