Beauty contest with a known crowd

Question

You are in a two-thirds-of-the-average game with $100$ other players, and you happen to know the room's composition: $50$ of them are "naive" and will pick around $50$ , and the other $50$ are "one-step" thinkers who best-respond to the naive and will pick around $33$ . The winner is closest to two-thirds of the average of everyone's numbers, yours included.

What should you pick?

Accepted Answer

Here you do not need to guess the crowd's reasoning depth, you are told it, so you can compute the best response directly. First find the crowd's total. The 50 naive players contribute about 50 	imes 50 = 2{,}500, and the 50 one-step players contribute about 50 	imes 33 = 1{,}650, for a crowd total of S = 4{,}150 across 100 players. Including your own pick x, there are 101 players, so the average is (4{,}150 + x)/101 and the target is two-thirds of that. You want to sit exactly on the target, so set x = \frac{2}{3}\cdot\frac{4{,}150 + x}{101}. Multiply out: 303x = 2(4{,}150 + x) = 8{,}300 +…

Beauty contest with a known crowd

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