0
$\begingroup$

Suppose there is an auction website. The rules are that the winner pays second-highest bid, and if a bid is made with less than 3 minutes left on the timer, the timer is reset to 3 minutes.

A seller is auctioning a large collection of items as many individual listings, but they all have the same closing time. (Until and unless the 3-minute-bid rule is triggered.) There are multiples of some items. Suppose you want to buy one of the multiples of a given item, but only one. Is there a game-theoretically sound way to guarantee that either you win exactly one item and pay no more than your max bid, OR that all items sell to other parties for more than your max bid?

In a single-item auction, you can meet the criteria simply by placing your max bid. (I've read assertions that you should place your bid right before the end, to avoid emotional bidders who will exceed their max bid just to outbid you, but with the 3-minute-bid rule, that may not be applicable, and maybe isn't a cleanly game-theoretic consideration, anyway.) However, in a duplicate-item scenario, that can fail in several ways:

  1. If you put your max bid on multiple listings, you may end up accidentally winning multiple listings and exceeding your budget.
  2. If you don't, then you might instead arbitrarily pick one item to go after. Let's say for sake of example that all the items have one bidder on them, including you, and so all the items are tied in price. There are a few last minute bids (say, by bidders not carefully watching for the cheapest copy, or perhaps there's some other combination of events that could make the resulting actions optimal for each involved agent), pushing your listing's timeout past when the other listings close. You can't switch to a cheaper listing yet, because you might still win this one. Then, say, a latecomer outbids you. One way or another, you may find yourself outbid, and the other listings (which sold for cheaper than your max) are already closed.
  3. There may be some tactic where you jump between listings, incrementally raising your bid so that you're only ever highest-bid on one listing at a time, and reducing (eliminating?) the chance you'll get stuck waiting for a single listing while another sells for cheaper. Perhaps this allows you to keep listings open long enough to guarantee an acceptable outcome? In any case, it is not clear to me that there exists such a strategy that guarantees you don't end up in one of the undesirable results.

(This question is not purely academic, haha, but I'm interested in the game-theoretic solution.)

The best I've found on google is this: http://answers.google.com/answers/threadview/id/503078.html But it seems like it's addressing sequentially-sold listings, and so isn't quite applicable.

To reiterate: Is there a game-theoretically sound way to guarantee that either you win exactly one listing (of multiple duplicates) and pay no more than your max bid, OR that all listings sell to other parties for more than your max bid?

$\endgroup$

1 Answer 1

1
$\begingroup$

It can't be done.

  1. You certainly can't guarantee that you'll win even a single listing, as there's always the possibility that the sale price is higher than your max bid on every listing.

  2. You also can't ever bid on more than one listing at a time, or else you run the risk that you're topped by exactly one bid in both. If both of those first-price bids are higher than your max bid, there's nothing you can do, and you'll win both listings.

  3. In order to prevent any listing from being sold for less than your max bid, you must bid on any listing that is about to be sold for less than your max bid. This can happen for multiple listings simultaneously, which would force you to submit bids on multiple listings. This violates #2, as you can't bid on multiple listings and not risk winning more than one of them.

  4. Since this is a second-price auction, submitting your max bid can only make somebody else pay less than your max bid. There's absolutely nothing you can do to raise the final price above your max bid amount, unless you bid more than your max bid (which is a contradiction).

You can't guarantee that you win even a single listing. You also can't guarantee that you'll win any listing that does sell for less than your maximum bid, unless you risk the possibility of winning multiple listings. Finally, there's nothing at all you can do to force others to pay more than your maximum bid, since your max bid can only force them to pay less than that in a second-price auction.

So, there is no way to guarantee that you win any listing for less than your max bid, and also no way to guarantee that any listing sells for more than your max bid. It's entirely possible that every single listing sells for less than your max bid, and that you win none of them.

$\endgroup$
2
  • $\begingroup$ Hmm...you're right. I'm not sure I follow all of your points, but nevertheless it's not too hard to construct a two-listing scenario where pretty clearly there is no guaranteed solution. I'll mark this answer correct. Failing a guaranteed solution, then - anybody have a method that should be nearly optimal most of the time? This is not really a well-formed question, and more a request for thoughts as to how I should deal with such auctions - the thought of trying to keep track of multiple auctions is kindof overwhelming, haha. $\endgroup$
    – Erhannis
    Mar 2, 2021 at 21:34
  • $\begingroup$ Your point 2 is not quite right as the highest bidder gets the item but pays the second highest bit. I don't think that changes the result that you might win both if you bid on a second one while the first is still yours to win. $\endgroup$ Mar 4, 2021 at 16:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .