Does a bigger crowd make the winner's curse worse?

Question

Bidders value a common prize whose true worth is $V$ ; each estimate is $V$ plus noise uniform on $[-10, +10]$ . Everyone naively bids their estimate and the top bid wins.

How much must a rational bidder shade in a three-person pit, in a fifty-dealer market, and in the limit of very many bidders? What does the trend tell you?

Accepted Answer

The winner's noise is the maximum of n i.i.d. Uniform[-a, a] draws, with expected value E[\max] = a\,\frac{n-1}{n+1}, \qquad a = 10. This is exactly the amount you should shade below your estimate to break even conditional on winning. Three bidders. E[\max] = 10 \cdot \frac{3-1}{3+1} = 10 \cdot \frac{2}{4} = \5.00. Fifty bidders. E[\max] = 10 \cdot \frac{49}{51} \approx \9.61. The limit. As n 	o \infty, \frac{n-1}{n+1} 	o 1, so E[\max] 	o a = \10.00, the very edge of the noise support. With enough competitors, winning means you almost certainly had close to the maximum possible overestimat…

Does a bigger crowd make the winner's curse worse?

Your answer

More Market Making questions