# Sunk cost auction modeling.

Consider the following auction concept. I call it a "SUNK COST AUCTION" Each person bids, but you pay all of the money that you bid for every bid you make. So, if you bid \$1 that \$1 is gone, even if you are out-bid by someone bidding \$2. The last person to bid (the highest bid) gets the item. 1. There is an item that you value at \$100 up for auction.

2. You are in the room with 100 people who also want the item and who value it too but you don't know how much they value it. You can safely assume, though, that the level they value it is normally distributed around some well-known value that is not too different from your own evaluation. (Maybe, that well-known value is \$120.) 3. Bidding starts at \$10, the person who bids \$10 WILL (probably) NOT WIN obviously so it's just lost money... but if no one bids except that one person ... well then they do win. So early bidding is low-risk, high reward. 4. The time limit for bidding re-sets after each bid. So, you take \$10 at the last moment... but then there is another 1 min added where you can be outbid.

How would such an auction go down? No one wants to bid until the price is high enough that they have a shot at being "last" ... but, there's always the slim chance that no one will take the risk and you could win a \$100 for only \$10.

Also, as the price rises biding becomes very risky. It would be awful to bid \$100 only to have someone else take it for \$120.

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I think you need to quantify the time, otherwise there is a plenty of room to bid after those $10 (this problem is not well defined). – dtldarek Apr 22 '12 at 13:56 How do you mean? The time limit is 1 min at each bid-level. so each new bid get's a "going going gone" moment. It's not like ebay. I think having a single time limit makes the problem boring. – a little don Apr 22 '12 at 14:29 I mean that the 1 min limit is not a limit at all! There are infinitely many moments you could bid, e.g.$60-1/n$seconds for any$n \in \mathbb{N}$. For such thing to work, you need to define something like bidding close to the time limit is accepted with some probability dependent on the time left. Or you may quantize time, but then you need to define what happens if there is more than one bid in a single step. – dtldarek Apr 22 '12 at 15:01 Once you have bid \$1, if you later want to bid \$3, do you have to pay another \$3 to the auctioneer, or is it enough to pay the difference between your new and old bids? –  Henning Makholm Apr 22 '12 at 15:22
Can you afford an assumption that everyone is rational and bid only to maximize their own expected take-home? (E.g., nobody will outbid you out of spite?) –  Henning Makholm Apr 22 '12 at 15:27

Everyone is obviously very reluctant to bid when they don't think that bid will actually win. I don't think, therefore, that this auction would move like a standard auction: \$10, \$15, \$20, etc. In a regular auction, that \$15 bid is all upside if you value the item at more than \$15, since you either don't pay anything and don't get the item (status quo) or pay \$15 and get the item. In this case, moving in small increments is hugely costly, so I think the rational thing to do is to make a much higher bid right away.
You said to assume that everyone else's valuation of the item is normally distributed about \$120, but first I'm going to assume it's normally distributed around \$80, with a standard deviation of \$5 (you didn't give one, but this process should work for any SD). With 100 people, there's a$50\%$chance that any given person will value the item over \$80, which means that the likelihood of no one outbidding me is $(1/2)^{100}$, i.e., really small. So there's a $1-(1/2)^{100}$ chance of me losing \$80 for nothing, and a$(1/2)^{100}$chance of me paying \$80 for something I give \$100 of value. Obviously not a good bid to make. We can write a function of my bid that compares utility to risk (and painfulness) of being outbid. For a bid of$\$X$, the risk of an $\$X$loss is$1 - (P(z<(X-80)/5))^{100}$, and the likelihood of an$\$100 - \$X$gain is$P(z<(X-80)/5)^{100}$. The expected utility from an \$X bid, therefore, is $$[P(z<(X-80)/5)^{100}] * (100 - X) - [1 - (P(z<(X-80)/5))^{100}] * X$$
After the bidding opens, you should immediately bid the lowest $X$ where this function $=0$. It doesn't look too good -- this is negative most places I've checked, so I think with a 100 person group that has a value that close to yours, you're better off not bidding, since the risk of someone outbidding you is just too high. (I chose \$80 instead of \$120 for the mean of their normal distribution since clearly \\$120 would be even more unlikely).