Optimum strategy for Dutch auctions?

[andyandy]

dutch auctions - where an item is available, and people make paper bids for that item - and the lowest non-replicated bid wins the prize....

is there anything to say about an optimum strategy? It seems game-theory-ish in nature, but rather complicated in practise...


let's say for simplicity that if there are no non-replicated bids then the prize goes unclaimed....

that each bidder has 1 bid

n=number of bidders

bids are in $1 increments


with the simplest meaningful case, n=2

and for this the optimum strategy is of course to bid $1. Assuming you're both not complete morons, you only have to choose to bid either $1 or $2, but there is no value in bidding $2 - for at best your bid will be replicated. So you both bid $1 and no one wins.

onto n=3

you have a choice of 1,2,3. And yet immediately the situation seems too complicated to predict - do you assume someone will bid $1 and thus bid higher? Or assume that your opponents will assume that someone will bid $1 and thus bid $1? And so on.....

is there any model that can help - or is the outcome completely non-predictable?

[casebro]

I don't know where you are from, but around here "Dutch Auctions" are for when the seller has more than one identical item. And every auction I've ever seen has been 'highest bidder wins', not the lowest like in your OP. In a live Dutch auction, the winner gets to take as many ov the items at his price as he wants. Other bidders can then buy more at that price, then the remaining items get auctioned again, typically for a lower price- if there are any left. I think eBay allows a "second chance offer" at that point.

That said, the only thing to do at auctions is to mentally set your own maximum, and don't go beyond that. If you win, you have fulfiled your goal of getting the item within your price range. If your price range is too low, you can try again when a similar item comes along. Or maybe you are just too cheap to ever get what you want.

[Smike]

According to wikipedia, you mean a Unique Bid Auction (http://en.wikipedia.org/wiki/Unique_bid_auction)

With n=3, I think that it doesn't make a difference whether you choose 1 or 2, as if you have a logical strategy, the other two will also use that strategy and pick the same number.

[andyandy]

ha! Well I call them Dutch auctions :blush:

unique bid auction sounds like what i meant.....:D

[zooterkin]

Parenthetically, neither of the auctions described in the OP and the first response sounds like a Dutch Auction to me, and the wikipedia entry matches what I've thought of as a Dutch Auction:
Dutch auction (http://en.wikipedia.org/wiki/Dutch_auction) is a type of auction where the auctioneer begins with a high asking price which is lowered until some participant is willing to accept the auctioneer's price, or a predetermined reserve price (the seller's minimum acceptable price) is reached. The winning participant pays the last announced price.

[Earthborn]

I don't know where you are from, but around here "Dutch Auctions" are for when the seller has more than one identical item.You are both wrong. A Dutch auction (http://en.wikipedia.org/wiki/Dutch_auction) is an auction where the price begins high and is lowered until someone indicates that he wants it at that price. That Wikipedia page mentions that it called a "Chinese auction" in the Netherlands, but I've never heard anyone call it that and the source for that claim doesn't appear to reliable to me. The Dutch tend to call it "veiling bij afslag", which may be translated as "auction by cut-off".

[Niobe]

The Dutch tend to call it "veiling bij afslag", which may be translated as "auction by cut-off".

This is how we sell flowers, for instance. Bidders sit in the room where a clock counts down from a price per batch and they can hit their buzzer if they want it at the price, stopping the clock. It's a game of wits for the Dutch who want that specific batch but for the lowest possible price :)

[Michael C]

In the case of the lowest unique bid auction, I don't think you can create a mathematically modelled strategy. Your strategy needs to take into account the strategies of the other biddres, and I'd imagine that many people who participate in this sort of auction won't have any strategy more complex than "I'll just pick the first number that comes into my head" or "I always play my lucky number". Probably there are always a few hopefuls who choose the lowest possible bid, thinking that nobody else will dare do something as simple. Your task is to know enough about human psychology to be able to guess how most people will bid.

If you had access to the complete results of many auctions of this kind, and did a lot of statistical analysis, you might be able to find some common bidding patterns and possibly increase your chances of winning by avoiding any of these patterns.

But I don't think it would be worth the effort!

[JoeTheJuggler]

AS far as strategy, wouldn't that depend on how badly you wanted the item? If it's more important to win the auction, you just bid very high. If you're looking for a bargain, but don't mind running a greater chance of not winning, you try a lower bid.

By the way, what if you were trying to sell a Dutch Oven in a Dutch Auction? You'd be doing Double Dutch! I wanted to make a jump rope pun, but I'll just skip it.

[strathmeyer]

The Dutch auction confusion comes from the fact that eBay used the words to describe multi-item auctions for some time (and may still do it.)

The optimum strategy for any action is to bid your buyer value. The different types of auctions just affect the price the winner pays.

Dutch and silent auctions - winner pays their buyer value
"normal auctioneer" auctions - winner pays some small amount more than the person with the second highest buyer value

I don't really understand how the OP is describing an "auction".

[Earthborn]

I don't really understand how the OP is describing an "auction".It isn't an auction, really. It is more a cleverly disguised lottery. It encourages people to "bid" on an expensive item by offering them the option to bid very low, however to bid they must SMS or phone to an expensive number. A lot of people will phone in, raising enough money for the company to offer expensive items for people to bid on. It is a lottery instead of an auction, because every bid costs every bidder money and only one of them can win.

Just like every other lottery, the only mathematically optimum strategy to win it, is not playing.

[blobru]

This is a mighty strange auction. It resembles a silent bid for a contract, where the contractor wants the lowest bid of course, and silent ensures late bidders no unfair advantage. However, by stipulating unique bid the contractor goes against his own interest, so even if we can't assume the bidders are morons (for n=2), we must assume the contractor is (any city that's ever hosted an olympics knows that ain't much of a stretch).
For n=3, if all bidder's budgets = $, i don't think there's an optimum strategy.
For n=3, $=2 -- a bidder's bid B can be 1 or 2.
4 permutations of competing bids: (1,1) (1,2) (2,1) (2,2).
B=1, 1 success (2,2) out of 4, %=25.
B=2, 1 success (1,1) out of 4, %=25.
each of the three bidders will win 25% of the auctions, and 25% no one will win.
But if all bidder's budgets are not the same, what happens?
Call the bidders a, b & c.
Say $a = $b = 2, $c = 3.
Does c now have some advantage? Check out c's new choice:
B=3, 2 successes (1,1) (2,2) out of 4, %=50.
Now c wins 50% of the auctions, a & b 25%, and no one 0%.
So, assuming this simple case holds for more complicated ones (big assumption), where one bidder has a bigger budget and knows it (another big assumption), it appears he should bid next biggest competing budget + 1 to maximize his chances of winning the auction.

[blobru]

assuming this simple case holds for more complicated ones (big assumption)

OoOops... :footinmou this assumption is way off! I'll leave it to others to crunch the numbers.
Interesting puzzle -- maybe it is game theory-ish (if we allow perfect knowledge of competitors' budgets).

[andyandy]

hiya blobru -just to clear up, if there is no unique high bid, then no one will win - so if the bids are $1 $1 (n=2) then no one wins....

conceivably a bid of $4 is also meaningful - if both others bid $3. Indeed, that makes $5 also meanignful....and so on - but limiting the case to just bids of $1 $2 $3 with n=3 seems sufficiently complex for now :)

with n=3 we have the following meaningful combinations;

Cycle 1

A B C
1 1 1
1 1 2
1 1 3
1 2 1
1 2 2
1 2 3
1 3 1
1 3 2
1 3 3


With the above cycle, player A wins 4 times out of 9 with a bid of $1

Cycle 2

A B C
2 1 1
2 1 2
2 1 3
2 2 1
2 2 2
2 2 3
2 3 1
2 3 2
2 3 3


With this cycle player A wins 2 times


Cycle 3

A B C
3 1 1
3 1 2
3 1 3
3 2 1
3 2 2
3 2 3
3 3 1
3 3 2
3 3 3


with this cycle player A wins 2 out of 9 times


from this it seems clear - that player A maximises his chances of success by bidding $1. This seems a bit Prinsoner Dilemma (http://en.wikipedia.org/wiki/Prisoner's_dilemma) -ish - ie by following your own optimum strategy, and by other players following their's [ie also bidding $1] then no one wins ever :)

[blobru]

Hey there Andy. Yeah looks like the Prisoner's Dilemma alright (maybe it's a police auction?)
But even with no optimum strategy there'll still be a pattern to the cycles at least:
I'm guessing it's recursive... (c=cycle)

n=3 $=1 c_1=0
""""" $=2 c_2=1 c_1=1
""""" $=3 c_3=2 c_2=2 c_1=4
*** $=4 c_4=3 c_3=3 c_2=5 c_1=9
*** $=5 c_5=4 c_4=4 c_3=6 c_2=10 c_1=16
etc.

so,
c_x = (c_x one row up) + 1
except
c_1 = ([$^n / n] - $) - (c_$ + c_($-1) + ... + c_3 + c_2)

***haven't "crunched" #'s yet, but adding 1 to $ should just create one extra tie-breaker for every cycle save c_1 [i think]

Man, that's lookin' kinda hairy! oh well.
I don't think increasing n will end the dilemma, but varying $ among the bidders should -- and generate optimum strategies.