See why the winner of a coin-jar auction ends up the biggest loser.
Each bidder's guess scatters around the true value. The winner is whoever guessed the highest. More bidders means the top guess puffs up even further. Winning itself is the news that you overestimated the most. The formula below sizes that puff.
For example, a jar holds 100 🍬 candies, and 10 🧑 classmates each guess the count. One classmate guesses 130 and bids the highest, so they win. Inside are only 100, a loss equal to the 30-candy overcount.
Ten bidders each count the coins, and their signals scatter evenly within ±20 of the truth.
If the winner is someone who bids exactly their own signal, how much do they lose on average?
10 bidders, spread 20 — naive bidding.Show proof
With ten signals spread evenly, the largest one averages 100+20×119≈116.4. Among that many counters, someone almost always overcounts.
The winner is exactly that most optimistic bidder. Paying 116.4 for something worth 100 loses 16.4 on average — not from lack of skill, but because winning itself selects for the biggest overestimate.
More bidders inflate the top signal further (at N=20 the loss is 18.1). The fiercer the bidding, the deeper the naive winner's wound.
In the same jar auction (10 bidders, spread ±20), how should you fix your bid to avoid the loss?
If your signal reads s, what should you bid?
10 bidders, spread 20 — shaded bidding.Show proof
The key is that winning itself is information. If you won, your signal was the largest of ten — and the largest signal is inflated on average.
So subtract the inflation 20×119≈16.4 in advance: if your signal reads s, bid s−16.4. This is arithmetic conditioned on the event of winning.
Then what you pay when you win averages 100, and the expected loss is 0. Shade too much and you lose the auction — but that is a walk-away, not a loss.