# Winner's curse Nash equilibrium

I read a problem in a book describing the 'Winner's curse'. If you were to bid for a 100 USD bank note (auction, 1 opponent), how much would you bid? The special rules are that your first bid is final, all bids have to be made in the same moment and that the bidder who bid the lower amount (lost the auction) has to pay what he bid. The winner doesn't have to pay, but receives the 100 USD plus the bid from the looser. For example, you bid 110 USD for the 100 USD bank note and your opponent bid only 20 USD. Thus he has to pay you what he bid (20 USD). My question: what would an equilibrium strategy look like (if there is one)?

• I think you mean Winner's curse, not "course." The equilibrium strategy will depend on the players' resources (how much can they borrow, what is their wealth, etc.). With no constraints on resources or information about their wealth distributions, you'd have something like this: your opponent bets $x$, then you bet some $y > x$, then if you bet $y > x$, the other player would wager some $z > y$. That would spiral forever. The constraints matter here. Jun 27, 2021 at 7:46

In a case where there are no constraints on how much players can bet, their best should go to infinity. You want to bet more than the second player, and he wants to do the same.

Now let's analyze the situation where there is a constraint on how much each player can bet, and it is public. The player who can bet less should bet 0. At the same time, the other player can bet any amount higher than the limit for the first player. This way, the player with more resources assures his win, while the other participant reduces his losses.

• Sounds convincing. What, if both players start with the same potential. Shouldn't it be a zero-sum-game somehow, given equal knowledge and possibilities? Jun 28, 2021 at 12:00
• Probably both should always bet the (identical) maximum. Jun 28, 2021 at 12:00
• If there are constraints and both players have the same maximum bet, the strategy would depend on rule defining what happens if they place the same bet. Here it is not defined, but let's suppose that they have to pay what they bid and then get the half of the reward (v/2). In this case all depend on their maximum possible bid (B). If v/2>=B then they should bid the maximum. If v/2<B then it seems that pure solutions would be that one player bids the maximum and other 0. They are symmetric so mixed equilibrium strategy would be for each of the players to bet B or 0 with 50% probability. Jun 28, 2021 at 15:47