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The dollar auction is a type of auction in which players bid money for a dollar.

  • Whoever bids the most pays what they bid and gets the dollar.
  • Whoever bids the second most pays what they bid and get nothing.

Ideas that don't work

If the players follow a naive strategy, everyone will try to bid 1 cent more than the current bid until it hits 99 cents, after which the top two players will keep overbidding each other, ad infinitum.

This however would obviously not be a Nash equilibrium, since the top two players would switch to not bidding at all.

However, everyone not bidding at all is also not a Nash equilibrium, since then any individual player would benefit from deciding to bid, since then they get a dollar for a penny. My question is, what are the (sub-game perfect) Nash equilibrium then?

Simplifications

Here are some simplifications to make it a finite extensive game, and to avoid trivial solutions:

  1. So that one of the players does not immediately bid 99 cents, the players can only bid 1 cent at a time.

  2. There will only be two players. They will alternate turns. On each turn, they can either bid one more cent than the other player (or one cent on the first turn), or not bid.

  3. The first player to not bid pays their highest bid (or 0 if they did not bid) and gets nothing. The other player pays their highest bid and gets a dollar (100 cents). (Note: Yes, this means if the first player passes on the first turn, the second player gets a free dollar).

  4. As it stands, this game is not well defined since it could go on forever. To prevent this (and also make the number of choices finite), there will be maximum bid. After the bid reaches a million dollar (100,000,000 cents), a fair coin is flipped. If it heads, the next player must not bid. If tails, the next player may bid, but if they do, then the other player must not bid. (The coin ensures symmetry.)

  5. The players' payoffs are just their gains minus their loses (i.e. their utility is a linear function of their wealth).

Since this is now a finite extensive game, we know a sub-game perfect Nash equilibrium will exist.

Problem

What are these equilibriums?

Benjamin Wang
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Christopher King
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1 Answers1

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I can propose the following backward game analysis looking for a conditional optimal strategy at each turn.

Let $D=100$ cents and $M=10^8$ cents.

If tails then the last player always bids, because then their gain is $D-1$ cents bigger.

Then if the current bid is $M-1$ cents and the next player bids then their expected gain is $D/2-M$ cents and if they does not bid then their gain is $2-M$ cents, so the next player bids.

Then if the current bid is $M-2$ cents and the next player bids then their expected gain is $D/2-M$ cents and if they does not bid then their gain is $3-M$ cents, so the next player bids.

Continuing, we see that if $k<D/2-1$ and the current bid is $M-k$ cents then the next player bids.

But if the current bid is $M+1-D/2$ cents then the expected gain for the next player is always $D/2-M$, so they are free to bet or not to bet.

Suppose that the current bid is $M-D/2$ cents. If the next player bids and the other player bids then the next player expected gain is $D/2-M$ cents, if the next player bids and the other player does not bid then the next player expected gain is $D-M-1$ cents, if the next player does not bid then their gain is $D/2-M+1$ cents. So the guaranteed next player expected gain is $D/2-M+1$ cents and it is achieved when they does not bid. So the next player does not bid.

Then if the current bid is $M-D/2-1$ cents then the next player always bid, because then their gain is $D-1$ cents bigger.

Then if the current bid is $M-D/2-2$ cents and the next player bids then their gain is $D/2-M+1$ cents and if they does not bid then their gain is $D/2-M+3$ cents, so the next player does not bid.

Then if the current bid is $M-D/2-3$ cents then the next player always bid, because then their gain is $D-1$ cents bigger.

And so forth.

Since the initial bit $0$ is an even number, we conclude that the first player does not bid and their gain is $0$ cents, whereas the second player bids and their gain is $D-1$ cents.

To justify the found strategy, I note that although the first player gives up the dollar to the second, it is much better for them than an expectation to lose a half of a million in a chase for the dollar. I hope this conclusion also is practical.

Alex Ravsky
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