Is there a rational way to bid in a dollar auction?

Inspired by a recent thread on bid fee auction sites.

Martin Shubik invented the dollar auction. It’s a simple concept: two people are bidding for a dollar which is given to the highest bidder. It’s presumed that both people want to maximize the amount of money they have and they cannot collude together. The catch is that when the auction is completed, both people have to pay the highest amount they bid. But only the one who made the highest bid overall gets the dollar.

When it’s played, a dollar auction is a good demonstration of how sunken costs work. Once a person has started bidding they have an incentive to keep bidding.

Let’s say their current bid is thirty-three cents and the opposing bidder comes back with a thirty-four cent bid. If the first bidder drops out now, he’ll have to pay thirty-three cents and get nothing. But if the first bidder bids thirty-five cents and wins, then he’ll have a gain of sixty-five cents rather than a loss of thirty-three cents. So he makes the bid.

But his opponent faces the same situation. He has to make a choice of dropping out and losing thirty-four cents or making a thirty-six cent bid in the hope of getting a sixty-four cent profit.

The logic still applies even after the bids go over a dollar. The person who drops out will still do worse than the person who makes the final bid. Assuming they bid up one cent at a time, the person who drops out will always come out with ninety-eight cents less than the person who makes the final bid.

People playing this game tend to get caught up in it. So let’s take human emotions out of it and assume it’s a competition between two computers. They have no emotional involvement in the game. They’re just being programmed to play whatever rational strategy will achieve the maximum amount.

But what is that rational strategy?

Should both computers be programmed with the knowledge that this is a sucker’s game and they should rationally refuse to play? The dollar auction starts and neither computer makes a bid.

Or does it depend on who makes the first bid? Let’s say computer 1 is chosen at random to make the first bid. Can it make a bid and “know” that computer 2 will not respond with a higher bid?

Suppose computer 1 makes an initial bid of ninety-nine cents. Computer 2 analyzes the possibilities and calculates that if it bids nothing it will make zero profit. If it bids a dollar and computer 1 quits, it will also make a zero profit. If it bids a dollar and computer 1 responds with another bid, then regardless of which computer ends up winning they will both end up with a negative amount.

So it appears that a computer will not submit a bid if the other computer makes an initial bid of ninety-nine cents. Under that circumstance, there can be no strategy that will give it more money than doing no bidding will.

And that means that a computer can be programmed to make a ninety-nine cent bid if it has the initial bid. The other computer will be programmed to not bid against it so the first computer will make a profit of one cent.

But now comes the tricky part. Should you program a computer that has the initial bid to make a bid of one cent?

The question comes down to what the response would be from the other computer. What is the rational thing to program a computer to do if it has the second bid and the opposing computer made an initial bid of one cent? Should a computer be programmed to never bid unless it gets the initial bid? If so, then the best program would be for a computer to bid one cent if it gets the initial bid.

But now consider this. Assume as a hypothetical, that the two computers have arrived at a situation where each has submitted bids. Both computer would analyze the facts I described above for human players and realize that further bidding will not improve their profit but just sink them further into the hole. So a computer should be programmed that if it is in a situation where both computers have made a bid, it should respond by not making a bid.

Now go back a step. If both computers are programmed to stop bidding once each computer has made a bid, what is the rational way to program computer 2 when it has the second bid and computer 1 made an initial bid of one cent? In this situation computer 2 should bid two cents. Because then computer 1 will be presented with the situation that both computers have made bids so it should stop bidding. Computer 2 therefore makes a profit of ninety-eight cents.

So it appears that a computer should not make an initial bid of one cent.

This OP is getting overly long so I’ll stop now. Hopefully someone else will wish to pick up the analysis from this point.

Short answer is that it varies wildly based on the ruled of the game (including communication between both players) and individual utility functions that measure risk. And with typical utility functions the correct decision is usually to not play.

Okay, now I’m not on my phone so I can give a better answer. Note that by “players” I mean computer 1 and computer 2. I like that you restricted it to computers because humans are irrational. For example humans are more likely to bet many times in increments of one cent than respond to a bet of a large increase because they might be scared off by a substantial bet.

Anyways, if both players have linear (risk-neutral) utility functions, and we assume they make decisions based on expected values, and both players know this, then neither player will play. Ho hum.

The only time when it is logical to play is if you think that you are more willing to take risks than your opponent. In certain scenarios then you would be able to force him out with a sufficiently high bid such that it is profitable for you but not profitable for him to attempt to contest your bid.

But what about the point I made that if a computer “knows” the other computer is programmed not to play, then it becomes rational for it to make a bid of one cent.

This brings up another point I was going to raise. What’s the best strategy if you become part of an auction already in progress?

Let’s say that two people started the auction and both have already submitted bids. Then they both die while the auction is still in progress. You and another rational person are their designated heirs. You inherit not only their opportunity to bid, you also inherit the liability for the bids they’ve already made.

So basically treat it as the scenario where you have no choice about whether or not you wish to play the game. You’re just trying to finish it with the maximum gain and/or minimum loss.

Let’s say the last bid made was by the original opposing player and his bid was X. What should you bid?

If you bid X+1 cent, you’re just going to be falling in to the original pattern of incremental increases. My idea is that you should bid X+99 cents.

Your opponent will then have the option of quitting and losing amount X (his last bid). Or he can bid X+100 cents, which will give him the same amount X if you then quit and he wins. (And if you don’t quit, both of you are eventually going to end up losing more than X regardless of who wins.) So confronted with your bid of X+99, he has no possibility of doing better than X so he might as well quit at this point.

A video of the auction in action.
I suppose you can scare the others off if you jump to the end and bid high right away…

That would be my ninety-nine cents opening bid strategy. At that point, everyone else should realize there’s no possibility of making a profit in this auction.

But if I bid $1 you’re out 99cents and I have an advantage in the next auction.

Except if your opponents wanted to spite you and see you take a loss. I bid $1! etc…

You need to start the bidding at $200 and take a $199 dollar loss. No one wants to lose $199 just to see you lose $200…:wink:

Didn’t the original exercise have more than two potential bidders–a classroom–and a specified opening bid? “Who will bid five cents for this fine new dollar bill?”

I don’t like this statement, and to me represents the irrational aspect of the game. For example: If you know I always go all in when I’m bluffing, you’ll always call. But if I know I’ll go all in with pocket aces. But if you know that I know you know what I know you’ll go cross eyed.

In other words, you’re trying to apply a “known” to an indeterminate system without any basis. Isn’t this where game theory kicks in? The other computer has a probability for each action, with various factors influencing it, self preservation being one of them. This auction is a bit stupid and rather reckless, so self preservation will keep most people out.

I think in the end no matter what gets programmed each computer would say that there is an uncertainty to each situation, such that the best outcome is to not get involved.

Or I should say: No one wants to lose $200 (pay $201) just to see you lose the same amount ($200). :o

Since I currently have risk assessment on my mind, is it possible that the rational way to bid is to assess the various risk potential for each bidder?

If the other computer has $1million putting 99cents at risk isn’t an issue. If the other computer only has $1, it would go past 10cents.

As we learned, $1 means more to some than to others.

The counter to that is the risk element.

If you don’t bid you will be guaranteed a zero profit. If you do bid, the best result you can achieve is a zero profit but you also have the risk that there might be further bids in which case you will achieve less than a zero profit. So there’s no rational basis to make the bid - it adds the possibility of loss without adding any possibility of gain.

Make it simple. Assume both computers are running the same program. So each knows exactly how the other will act in any given situation and there’s no element of bluff.

What does the other computer know about that? The devil’s in the details here. The problem was originally constructed as a static game with complete and perfect information, and in that formulation, you can use backward induction to show that rational players won’t play. If you formulate the problem differently, you’ll get different results.

There are variations. I chose the simplest one.

The programmer can start with a simple program:

  1. Do not make a bid

And this program will always achieve a profit of zero.

But then he can modify the program:

  1. If you have the initial bid, submit a bid of one cent.
  2. If you do not have the initial bid, do not submit a bid.

This program will achieve a higher result than the first program.

My question is what program will achieve the highest results.

I’d go with the .99 strategy, or basically attempt to bid .01 more than the highest bid under 1.00 at the very last second and not bid at all if there is a bid over .99 in the last second of the bid. If you are playing with an entire classroom, my guess is that if you simply hold onto your $1.00 then you will end up coming out ahead in either the short, medium or long run…IOW, to paraphrase War Games, sometimes it’s better not to play at all. :wink: Or, if you are programming a computer to play, have it play conservatively initially, waiting until the last minute to bid, or simply watching to see how the bidding is going. My guess is that initially, people will get caught up in the game, but when they are ‘winning’ and getting either a couple of cents for their efforts or actually losing money in either a ‘winning’ or ‘losing’ bid, they will stop playing, or set some type of limits on what they are prepared to lose, so you could, theoretically, play to that. Unless they are also computer generated AI’s, in which case, I’d probably stick with ‘don’t play at all’ as the only winning strategy.


The bid 1 cent only program will achieve the highest result because you are essentially playing against the auctioneer because both programs are the same, according to:

So you’ve essentially made the two programs collude, even though the OP stated no collusion. Because if you run this program against itself many times, you’ll get the highest profit.

I’m thinking that a good strategy might be:
1: If I’m the opening bid, bid 99 cents
2: If I’m not the opening bid, and the current bid is 49 cents or less, bid 50 cents.
3: If the current bid is 50 cents or more, make no bid (or no further bids).

Any number of actors all using this strategy will be guaranteed a positive (non-zero) expectation value (making it superior to a “don’t ever play” strategy). On the other hand, if you know everyone else is using this strategy (and you know they won’t change strategies), then you’ll maximize your profit by starting with a 50 cent bid, not a 99 cent one. True optimum play, like in Paper, Rock, Scissors, would have to involve an element of randomness.