If we assume that we live in a world where everyone is perfectly rational and that there is a single "optimal bid", and if we assume that everyone was be capable of calculating what that optimal bid would be.
If this was the case, everyone would make the optimal bid at the auction. But then the rational person would know that he could simply bid a tiny bit more than the "optimal bid" and be guaranteed to win. But this would mean that this new bid was better than the "optimal bid" which leads to a contradiction.
Therefore there can't be a single "optimal bid" that could be rationally calculated.
If we assume that we live in the real world, we could hold this auction repeatedly and find out what actual people bid on average - and then use the result to calculate an optimal strategy.
Originally posted by wolfgang59Interesting question! I think the dynamics would definitely be affected by whether or not the top bidder gets any satisfaction from making the penultimate bidder pay. For instance, in a two-bidder situation, a bidder with no compunction to punish his opponent would never bid above £9.99 (or £10), as there would be a net loss for any bids above £10. However, if that same bidder took great glee in punishing his opponent, then bids above £10 would be expected since there would be some value associated with spending money to gain the satisfaction of making the other bidder pay. Multiple bidder scenarios would complicate things even further.
I am auctioning a £10 note. Highest bidder gets it (for their bid).
The twist is the penultimate bidder (ie the guy that gave up) also pays me but gets nothing.
What am I bid?
I believe the original question posed an auction where each person would be allowed only one bid. You probably wouldn't even know who else was bidding.
More interesting are cases when you have a a more traditional auction like this. I've heard of cases where they've actually done this, and the winner ends up paying like two or three times or more what the original bill was worth. Imagine that you've just bid 9.50 and someone else bids 10.00. If you bid 10.50 and win, you end up losing 0.50, but if you don't bid, you will lose 9.50 (now that the top bid equals the value of the bill, no third party is going to be making any bids to get you off the hook). After you've bid 10.50, the other guy faces the same situation and bids higher, and it goes on and on until someone finally gives up.
Originally posted by MelanerpesThis is the scenario I intended.
I believe the original question posed an auction where each person would be allowed only one bid. You probably wouldn't even know who else was bidding.
More interesting are cases when you have a a more traditional auction like this. I've heard of cases where they've actually done this, and the winner ends up paying like two or three times or more wha ...[text shortened]... the same situation and bids higher, and it goes on and on until someone finally gives up.
The optimal strategy (if no collusion is allowed) is NOT TO BID.
Even if you start with a bid of one penny as soon as someone bids 2p you are potentially looking at a loss ... so you bid 3p ... your rival is facing a 2p loss so he bids 4p .............................
Its a nice game for charity events 😀
Originally posted by Melanerpes0th possibility
I believe the original question posed an auction where each person would be allowed only one bid. You probably wouldn't even know who else was bidding.
More interesting are cases when you have a a more traditional auction like this. I've heard of cases where they've actually done this, and the winner ends up paying like two or three times or more wha ...[text shortened]... the same situation and bids higher, and it goes on and on until someone finally gives up.
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No-one bids. What crazy person would action money?
1st possibility
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First person bids $10
Second person will not bid at all (no loss, else he will lose whatever amount he goes above the $10).
2nd possibility
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The first person will not bid more than $10 (guaranteed loss, whatever happens next)
3rd possibility
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The first person bids below $10 but $N or more.
I'm assuming it an open action, everyone can see what others have bid.
Second person should bid (if anything) $(10+N) or more, else the first person overbid (minimising his loss). This results in at least $N loss, so the second person will not bid at all.
Originally posted by wolfgang59Suppose, in a world of completely rational people, the rational person concludes that the optimal strategy is "not to bid" -- then no one will bid.
This is the scenario I intended.
The optimal strategy (if no collusion is allowed) is NOT TO BID.
Even if you start with a bid of one penny as soon as someone bids 2p you are potentially looking at a loss ... so you bid 3p ... your rival is facing a 2p loss so he bids 4p .............................
Its a nice game for charity events 😀
But then a rational person could say, since no one else is bidding, I will bid 1.00 and make an easy profit of 9.00.
But that would mean that "not bidding" isn't the optimal strategy.
In the real world, I agree - a rational person would probably stay far away from a game like this.
If the auction was for charity, it gets especially interesting. The loser won't be all that upset because the money's going to a good cause (and they can take a tax deduction on it). In fact, it might turn into a battle to see which person is willing to donate the most money.
Lets assume everyone is rational and profit maximizing. Lets also say that when one is indifferent between bidding and not bidding one chooses to refrain from bidding to minimize administration and risk he might die in the process.
First lets consider the case a couple of you suggested: no one bids. Now that's easy, I bid $0.01, the rest of you rational people refrain from bidding and I net $9.99.
Obviously everyone not bidding can't be optimal.
Now let's consider the case where someone overbids me by bidding Y1, where $0.01<Y1<$10.00, since his EV of bidding is never higher than $10. Let Xn denote my nth bid and Yn denote his nth bid and let Xi=0 when one stops bidding, assuming there are only 2 bidders, which is without loss of generality, but I am too lazy to proof it.
Now, what is the EV of his bet? It is the infinite series Sum(n=1->oo)P(Yn>Xn+1|X1,Y1,...,Xn,Yn)(10-Yn)-P(Xn+1>Yn|X1,Y1,...,Xn-1,Yn)(Yn)
I have a hard time evaluating this equation, so I will first evaluate the simpler case in which Y2=0 (note Yi=0 implies Yk=0 for all k>=i). Then his EV becomes P(Y1>X2|X1,Y1)(10-Y1)-P(X2>Y1|X1,Y1)(Y1)=Z(10-Y1)-(1-Z)Y1=10Z-Zy1-Y1+Zy1=10Z-Y1, where Z=P(Y1>X2|X1,Y1) or in words the probability that he wins the auction with his bid. Now EV=0 => Z=Y1/10, so when he bids $5 I have to overbid him less than 50% of the time to make it a profitable bid for him. Now, my way of thinking is very weird since it is a game with complete information in which there are no mixed strategies and hence no probabilities different from 1 and 0 and I am using Yn and Xn as some sort of stochast and a normal variable at the same time, but I am just thinking out loud here. I have the sneaking suspicion that the optimal strategy is being the first to bid $0.01, but I don't know the proper way to respond to an overbid to, say $9. One could argue that a bid of $9.01 is clearly profit maximizing, but the probability of winning with such a bid is very small, considering he also has the exact same strategy available to him as me. The thing is, if a rational person would overbid, then both him and me are overbidding till infinity, but if a rational person wouldn't overbid, then none of us would be bidding at all after someone else has bid. The value of the game of the first situation is very negative, and the value of the latest 0 for all but the first one to bid something. So I think the latter should be optimal, but I don't know a solid way to proof it.
The equilibrium here is one with mixed strategies. You need more information (e.g. number of players) to calculate it exactly, but the idea is that there is a set of bets from which you randomize from. An example of a mixed strategy would be to betting 0 with probability p or x with probability 1-p. The equilibrium with two players will most likely be of this (example) form.
Originally posted by PalynkaThe original problem was a traditional auction where every 'player' can continue to bid. In this scenario I believe there is no better strategy than not competing. (Given that all players adopt the optimal strategy).
The equilibrium here is one with mixed strategies. You need more information (e.g. number of players) to calculate it exactly, but the idea is that there is a set of bets from which you randomize from. An example of a mixed strategy would be to betting 0 with probability p or x with probability 1-p. The equilibrium with two players will most likely be of this (example) form.
Palynka's idea would seem to be appropriate for the 'single-bid' scenario.
As Palynka says the strategy must depend on number of layers.
This strategy would also depend on goal - is it to maximise your own profit (in which case the range of bets would not exceed £10) or is it to beat your opponent (in which case the range would extend to infinity with shrinking probability).
So for starters lets consider a two player game where the goal is to beat your opponent. Anyone up for the math?
Originally posted by wolfgang59Even if the goal is just to maximize one's own profit, the bids can still get a lot higher than 10 -- the guy who's behind would probably prefer to bid say 11 (for a net loss of 1), than to stay with his losing bid of 9.
The original problem was a traditional auction where every 'player' can continue to bid. In this scenario I believe there is no better strategy than not competing. (Given that all players adopt the optimal strategy).
Palynka's idea would seem to be appropriate for the 'single-bid' scenario.
As Palynka says the strategy must depend on number of laye ...[text shortened]... consider a two player game where the goal is to beat your opponent. Anyone up for the math?
The optimal strategy might involve always making the smallest possible bid that exceeds your opponent -- and then using a probability table to determine whether you would make a bid or take your losses. If the current bid is low (0.10), your probability of bidding may be something like 99.9% -- and as the bidding went up, your probability would drop at some sort of rate until it reached some point where it would be extremely close to 0% (but never actually reaching 0% unless the bid was inifinity)
With this strategy, your opponent and your opponent would know the probability that the other guy will make another bid - but would never know for certain what the decision would be.
The problem with this auction is that
1. it is either not a constant sum game, or a zero sum game with multiple players who are not all the same.
2. You have to consider the infinite sum of infinite variables who are depending on eachother. Note that theres an order in the bidding and thus the optimal strategy need not be symmetric.
To make the problem more static, consider the following game:
1. The game is between 2 players, player 1 and player 2.
2. The players alternate in bidding, where the highest bidder gets $10 (which is undividable and untransferable to avoid collusion), but both pay their highest bid. Betting increments with $1
3. Players 1 starts of with bidding $1
Originally posted by DarrieIn reply to previous two posts I think we have established that the best course of action with a traditional auction (ie repeated bids) was NOT TO BID.
The problem with this auction is that
1. it is either not a constant sum game, or a zero sum game with multiple players who are not all the same.
2. You have to consider the infinite sum of infinite variables who are depending on eachother. Note that theres an order in the bidding and thus the optimal strategy need not be symmetric.
To make the proble ...[text shortened]... t both pay their highest bid. Betting increments with $1
3. Players 1 starts of with bidding $1
I think it interesting to investigate the optimal strategy if one is allowed only a single bid (a tender if you wish) as previously discussed by Palynka.