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Current events
Remember that game where two people bid sucessively for a dollar bill? The highest bidder takes the dollar home. You also pay your highest bid whether or not you win the dollar. The Nash equilibrium is an infinite bid from both players, or alternatively the equilibrium is undefined.
Posted by Tyler Cowen on August 1, 2006 at 04:26 AM in Current Affairs | Permalink
Comments
There is no current event that this is a good model for.
Posted by: joe o at Aug 1, 2006 1:51:07 AM
Joe: I assume the current event is meant to be the conflict between Israel and Hezbollah. Each side commits more and more resources in an attempt to win, because neither wants to be seen to have committed so many resources in vain.
The analogy is hardly perfect. It is not entirely clear that the two sides are contributing more resources than they originally planned. And there is a quite likely scenario in which each side "wins" in its own way: Israel wins the military battle, but Hezbollah wins increased power and influence in the Arab world.
Posted by: David Wright at Aug 1, 2006 3:28:42 AM
Hezbollah has said they expected Israel to respond with the usual tit for tat, which seem likely, as the cost for them has been very high. Israel military has has said they had not expect the fighting would be so difficult. Maybe one or both are not being truthful, and some clever plan is being executed which we will see only in hindsight.
Posted by: joan at Aug 1, 2006 5:26:46 AM
Isn't this broken if the first bidder bids exactly one dollar?
Posted by: Mo at Aug 1, 2006 7:45:16 AM
I think you are right Mo.
They say you can't cheat an honest man.
Posted by: Neil Craig at Aug 1, 2006 8:02:31 AM
I've heard this done as a radio show game with 2 callers and $100. The DJ goes up by $5 every few seconds, the first caller to call out their name, gets the last mentioned amount.
Every time I listen to it I think of game theory. ;-)
Posted by: Chris Meisenzahl at Aug 1, 2006 8:24:24 AM
If it is an open auction, then faced with an opening bid of $1, couldn't the auctioneer bid $1.01?
Posted by: Tom at Aug 1, 2006 8:28:39 AM
I think what Tyler means is the dollar auction with non-zero initial stakes. Suppose that for exogneous reasons, each player begins with an initial bid that is greater than zero... (consistent with the current event that is apparently being modeled.)
Posted by: jeff at Aug 1, 2006 8:55:53 AM
I confess to failing to understand the point of the game. Wouldn't a rational actor recognize the infinite-bidding possibility and refuse to participate in the first place, or if forced to participate refuse to bid more than one penny and take that loss?
Posted by: Paul Gowder at Aug 1, 2006 9:18:37 AM
Mike Munger uses that logic to explain the K Street phenomenon.
Posted by: Jessica Pickett at Aug 1, 2006 9:31:33 AM
Even with a non-zero bid, might a rational actor make a bid of 1.00 (or maybe 99 cents) more than the other bidder's initial bid and be done with the game? The other player has no possibility of being better off than his or her initial position.
Or, if this is a repeated game with many players, make a habit of bidding $5 if another player bids anything. Then you might win a lot. Say, Dr. Cowen, who was your dissertation adviser again? (Also, who was the last secretary of state? I seem to recall him saying something about the matter ...)
Posted by: ryan at Aug 1, 2006 9:58:23 AM
I question the assumption of this game. It is stating that one will always bid higher to try and decrease their loss, however, there is a cost associated with bidding higher and that is the cost of being outbid again. For example:
Every time one makes a bid there is a chance that it will be the winning bid and a chance that it will be a losing bid. As such, every bid has an expected value that can be calculated. If we assume every bid has a 50% chance of being a winning bid and a 50% chance of not being the winning bid, then at 50 cents we have $0 expected return (50% chance of making 50 cents and a 50% chance of having to pay 50 cents).
All bids above $.50 will a negative expected value and thus no rational actor should make those bids. Perhaps 50/50 isn’t the correct percentages to use, however, at any percentage there should be a point where the expected return goes from positive to negative.
Anyone see where my logical is incorrect?
Posted by: Rob at Aug 1, 2006 9:59:59 AM
joe says "There is no current event that this is a good model for."
Probably true, but surely an experiment could be run that would permit this Nash equilibruim to be tested.
Posted by: jim at Aug 1, 2006 10:25:35 AM
A model that some think better describes such current events is the "red queen hypothesis", which is used to model coevolution in predators and prey, ideas, and even state reactions to insurgencies. See Van Valen, L. 1973. A new evolutionary law. Evol. Theory 1: 1-30.
Posted by: Andy at Aug 1, 2006 10:35:02 AM
Rob: I think it's a matter not of logic and assumptions but of what people actually do in experiments.
http://erraticwisdom.com/2006/05/12/interesting-psychology-experiments
Posted by: Damien at Aug 1, 2006 11:53:45 AM
Also, has anyone ever tried that experiment permitting people to bid fractions of a penny? One would imagine that the bids in that case would converge on 0.99999...
Posted by: Paul Gowder at Aug 1, 2006 12:00:12 PM
Mo,
Why would you choose to bid $1?
Posted by: Maria at Aug 1, 2006 1:54:59 PM
I have thought about this game a little more than is healthy.
No one should ever really play this game.
Betting a buck, at the margin, appears the same as not betting; the expected return is o. Any rational opponent will not bid if Mo's bid is first. Mo however, having joined the game, now carries the risk of having an opponent who is irrational and bids a little more, say $1.01. Of course the opponent is worse off as they now have an expected return of 0.01 instead of 0, but then, they are irrational. If this is the case, Mo just lost his buck except if Mo keeps betting and loses some greater amount, x or x-1.
Given the route the game takes, why would one even place the dollar bet? It is only by not playing that one's expected return is strictly 0.
Posted by: disaggregated at Aug 1, 2006 2:18:31 PM
I think there's a little confusion about equilibrium here.
If there is no equilibrium, avoiding the game altogether is
not an equilibrium, either. After all, if nobody bids at
all, then you should a penny and win.
It's a paradox. Playing is a mistake. If nobody plays,
then not playing is a mistake, too.
Posted by: Keith at Aug 1, 2006 3:08:15 PM
"Given the route the game takes, why would one even place the dollar bet? It is only by not playing that one's expected return is strictly 0."
You would want to do that if you wanted to come out even and also didn't want your opponent to profit (which he would do if you refused to bid at all).
Posted by: Slocum at Aug 1, 2006 3:09:53 PM
This is about the Tullock Auction--where all bidders pay the amount of their bid, not just the winner. This podcast at econtalk.org
http://www.econtalk.org/archives/2006/06/giving_away_mon.html
provides an argument about why spending to get grants ('free money") is based on teh same principle.
Posted by: Simon at Aug 1, 2006 3:33:59 PM
Is this what we see in reality? Not the experimental setting -- actual geopolitics (or, in Simon's example, lobbying)? Do states consistently commit far more resources than they could possibly gain? Do lobbyists actually spend more than they get? Even in the original Tullock model, where the higher bids only guarantee a higher chance of winning the rent rather than the entire prize, a game of two rent-seekers should conclude with each player consuming 1/4 of the rent. Is that happening?
Posted by: ryan at Aug 1, 2006 4:39:16 PM
Obviously, there are many equilibria in practice, because we have different
focal points. Some of us work on K street, some of choose not to. Some of
us vote, some of us don't.
Posted by: will mcbride at Aug 1, 2006 4:51:40 PM
Keith:
But that can't be right. If we assume complete information and rationality on behalf of the bidders, any given bidder A will know the extent of the universe of other potential bidders. So A's utility calculuation while considering her first bid will be as follows. If she fails to bid, her expected gain is zero. If she does bid, then she must assume that everyone else has exactly the same utility calculation, and conclude that her expected gain is - all her money.
Can someone who knows more game theory than I chime in here? (Is anyone actually reading this blog yet?) I don't think I understand how nonparticipation is an equilibrium. The claim that there is no equilibrium seems to rely, sub rosa, on each actor's not having access to the utility calculations undertaken by the others.
Posted by: Johannes Climacus at Aug 1, 2006 4:54:18 PM
(ignore the "(Is anyone actually reading this blog yet?)" above -- I copied and pasted that paragraph in from my own new blog)
Posted by: Johannes Climacus at Aug 1, 2006 4:55:57 PM
Mo is close. The opening bidder should bid $0.99 (or, more generally, the largest possible bid that is strictly less than the amount that the players can win). Then the rational opponent will not bid, since any higher bid cannot result in a profit and might result in a loss. So the opening bidder walks away with a penny profit, and the other player breaks even. And I think that Ryan is right that you can end the game at any point by raising $0.99.
Of course, this strategy doesn't work if biddable values can be arbitrarily close to the amount at stake and raises can be arbitrarily small (so that there is no largest possible bid less than the amount at stake). It also wouldn't work if you abandon the assumption that other people are rational, but then again, neither would the rest of economics*.
*joke!
Posted by: Blar at Aug 1, 2006 5:06:45 PM
Blar... I think you've solved it on the opening bid. Bids less than that amount will suffer from the same irrationality that the one penny opening bid will -- it requires the false assumption that the other players will give up before you do.
Posted by: Johannes Climacus at Aug 1, 2006 5:27:05 PM
And I think that Ryan is right that you can end the game at any point by raising $0.99.
After a little more thinking, I'm pretty sure this is wrong. Once the bidding is in progress, it is never rational to raise your opponent by $0.99, since that would require bidding at least $1.00 more than your previous bid, which can never bring you a marginal profit and at best (if your opponent has bid a penny more than you and is certain to be rational) can be expected to have no impact on your net winnings. The only time when the 99-cent raise makes sense is when both players have the same bid, which only occurs at the beginning of the game, when both bids are 0. Once bidding has started, the better way to stop the bidding is just to stop bidding.
Posted by: Blar at Aug 1, 2006 5:40:10 PM
Actually, I believe two major results from economic theory apply here. By Nash, there MUST be an
equilibrium, although it is in mixed strategies. This is true even if the game has started and two
players have both bid. Secondly, the Revenue Equivalence Theorem tells us that this outcome will be
the same as if the auction were conducted using a more standard format (e.g. first-price auction).
I worked this out with some fellow economists and what you get is that each bidder must randomize their
probability of either raising their bid or dropping out in such a way that the other bidder is also
indifferent to staying in (and risking continued escalation) or dropping out (and taking a guaranteed
loss).
Of course, this isn't what we see empirically, either in classroom experiments or in current events.
Posted by: Peter at Aug 1, 2006 6:07:35 PM
Peter: interesting stuff. Where does the equilibrium land in your calculation? And what about more than 2 bidders?
Posted by: Johannes Climacus at Aug 1, 2006 6:20:51 PM
The prediction is that, on average (i.e. in expectation), the auctioneer will receive the same revenue as in a 2nd-price sealed-bid auction, which is $1. But because the player are engaged in mixed strategies, sometimes in practice we should see lower bids, and sometimes we should see the bidding spiral very high. We didn't try to work it out for more than two bidders, but the revenue equivalence theorem should still hold in that case, also.
Posted by: Peter at Aug 1, 2006 6:29:27 PM
Peter wrote: "By Nash, there MUST be an equilibrium"
Nash's theorem does not appply to games with an infinite strategy space. For example, the "Name the highest number" game does not have a Nash equilibrium. In this game, two players simultaneously name a real number (not including infinity). The player that names the higher number recieves one dollar, and the player that names the lower number receives nothing. If they tie they split the dollar. This game hash no Nash equilibrium, even in mixed strategies, because it is always a profitable deviation for the player who sometimes loses or ties to bid high enough that he always wins.
Of course, it is still possible that this game does have a Nash equilibrium. I would be interested to see the equilibrium you derived.
Posted by: MITdude at Aug 1, 2006 6:39:17 PM
I stand corrected on the point of existence. I think we assumed 2 players making alternating offers. Bid increments are fixed (e.g. just enough to exceed the other bidder).
Posted by: Peter at Aug 1, 2006 6:52:31 PM
I thought about this more. The game has a pure strategy equilibrium under the following assumptions.
(A1) At least one player has finite wealth.
(A2) One player has higher wealth than the other.
(A3) Bids can only occur in discrete increments (i.e., no fractions of a penny).
Proposition: Under (A1), (A2), and (A3), there is a pure strategy equilibrium in which the player with higher wealth wins by bidding one penny.
Proof: Denote the less wealthy player by L, and the wealthier player by H. Define the following variables.
W_L: the wealth of player L
W_H: the wealth of player H
B_L: the current bid of player L
B_H: the current bid of player H
If B_H > W_L, the game ends because player L can no longer outbid player H. Therefore, once B_H > W_L - $0.99, player H is guaranteed to win the dollar because he strictly prefers to win by bidding W_L + $0.01 rather than lose with his current bid. Knowing this, player L will never outbid player H once B_H > W_L - $0.99. Now imagine B_H > W_L - $1.98 and B_H > B_L. Player L will never outbid player H in this situation because such a bid cannot lead to higher payoffs for player L than losing with his current bid. First note that if player L places a bid that is greater than B_H but less than B_H + $0.99, player H is guaranteed to win because he strictly prefers to win by bidding MAX (W_L - $0.98, B_L + $0.01) rather than lose with his current bid. Alternatively, player L could bid more than B_H + $0.99, but this lowers his payoff even if he wins. Finally, player L could bid exactly B_H + $0.99, but I assume player H will then outbid him by one penny in this case (since he would be indifferent between this and losing with his current bid). By repeatedly iterating this argument, we can show that player L will never outbid player H, so player H can win by bidding one penny.
QED
According to this model, Hezbollah should immediately surrender after Israel throws one rock into Lebanon, so this obviously is not an adequate model of the current situation there.
Posted by: MITdude at Aug 1, 2006 8:00:18 PM
MITdude's model and the basic "dollar auction" model both assume that the size of the payoff is the same for both sides and is commonly known. Probably neither is true. Israel doesn't know how badly Hezbollah has to be beaten before they will abandon their cause, and Hezbollah doesn't know how many civilian casualties Israel is willing to cause in pursuit of its goal. By its current actions Israel is demonstrating that this number is a lot higher than some people thought.
Posted by: Peter at Aug 1, 2006 8:16:19 PM
Good point. The proof given above breaks down if there is even a small difference in the payoffs depending on who wins.
Posted by: MITdude at Aug 1, 2006 8:43:04 PM
MITdude: you rock. Your first iteration (and presumably subsequent ones too) is off by a penny though. Assume W_L = $100. Then by your first iteration of the reasoning, H should prefer to squash L when B_H reaches $99.01 rather than lose that amount. However, he doesn't prefer that. The squashing occurs at B_H = 100.01, where H's net loss is the same as it would be losing the $99.01 bid.
The original squishpoint thus should be B_H > W_L - $0.98.
God, that was embarassingly nitpickey. I'm glad I'm posting this under a fake name. :-)
Actually, on re-reading, I think there's a slightly larger flaw in the reasoning, one that actually counts.
Finally, player L could bid exactly B_H + $0.99, but I assume player H will then outbid him by one penny in this case (since he would be indifferent between this and losing with his current bid).
Why would H outbid him by one penny if H is indifferent between that and the loss? What's H's motivation in the face of indifference? (Perhaps to punish L for the bid? If so, that makes the model a bit more complicated.)
Posted by: Johannes Climacus at Aug 1, 2006 8:43:49 PM
Thanks for checking the proof. I did it quickly and could have made errors. However, I believe the analysis is correct.
> should be B_H > W_L - $0.98
By the notation B_H > W_L - $0.99, I meant to denote strict equality, meaning B_H must be at least W_L - $0.98, as you point out.
> Why would H outbid him by one penny if H is
> indifferent between that and the loss?
To demonstrate existence of a Nash equilibrium, you can assume any optimal action when a player is indifferent. Having two optimal actions might stop this from being a *unique* Nash equilibrium though. The point you raise is an important one because the proof depends on player H's payoff being at least as high as player L's in the event that he wins. If player L has a slightly higher payoff, the proof is no long valid.
Posted by: MITdude at Aug 1, 2006 11:38:50 PM
That's what I get for being nitpickey late at night. You're right, of course. I read > as >/=.
WRT your second point, I suppose this is where someone brings up the diminishing marginal utility of wealth or some such.
Posted by: Johannes Climacus at Aug 2, 2006 1:04:15 AM
This model seems silly to me and unlikely to yield the results it predicts if tested empirically. People attend auctions voluntarily and it would not take long to figure out that we should stop bidding. Also, few people would attend an auction with the requirement that all losing bidders would end up paying the highest amount.
Let's assume that this is really meant to refer to the Israel conflict. I don't think it applies either because Hizbollah is not a rational actor in the sense that we think of Israel. Many of us in the West are too secular to understand the intense power of true religious belief to motivate someone to do things that do not fit into economic models.
Check out this interview with a Hizbullah fighter.
http://www.guardian.co.uk/israel/Story/0,,1832930,00.html
"Hizbullah prides itself on its secretiveness and discipline. "We don't take anyone who knocks at our door and says 'I want to join'. We raise our fighters. We take them when they are young kids and raise them to become Hizbullah fighters. Every fighter we have believes that the ultimate form of being is martyrdom." The three men nod their assent."
...
"Every one of those fighters is a true believer, he has been not only trained to use guns and weapons but [indoctrinated] in the Shia faith and the Husseini beliefs," Ali says.
If Ali were at the dollar auction, he would just shoot you and grab the dollar.
Posted by: Contributor X at Aug 2, 2006 5:01:45 AM
The proof above is valid, but it is fragile to trembling hands and relies heavily on both players having shared and common knowledge of each other's wealth. If you are H and W_L is a large sum of money for you to lose, then would you really sit there as the bidding escalated to play L out?
Posted by: dsquared at Aug 2, 2006 6:35:32 AM
Here's a thought. (As you can see from my previous wrong comments, I'm just floating things out there and seeing if they stick.)
Where the parties have equal (or unknown) wealth, doesn't the first person win if he adopts a brinksmanship sort of deterrent strategy? If A bids 1¢ to start and credibly makes it known that he has adopted a strategy of punishing anyone else who outbids him by bidding them back up heedless of the destruction to his own position, can't he run away with the prize? A could adopt such a strategy if he could find some way to communicate an irrevocable committment to the other parties. Then B couldn’t respond in kind because A would be un-deterrable (being irrevocably committed).
Posted by: Johannes Climacus at Aug 2, 2006 10:46:17 AM
dsquared wrote:
> If you are H and W_L is a large sum
> of money for you to lose, then would
> you really sit there as the bidding
> escalated to play L out?
No, I would not. As with any game where the equilibrium depends on a long string of backward induction (e.g., the centipede game), it is not reasonable to assume that an opponent that has deviated from the equilibrium n times will then stick to the equilibrium behavior on play n+1, so subgame perfect nash equilibrium is not an appropriate solution concept for this game.
Climacus wrote:
> credibly makes it known that he has
> adopted a strategy of punishing anyone
> else who outbids him
Such a threat is not credible if all players are rational, but one way of operationalizing this idea is with a reputation model in which some players are irrational. For example, assume a small fraction of players are "crazy" and will always outbid their opponent until their wealth runs out. A rational player might initially want to pool with these crazy players by outbidding his opponent for a certain number of rounds. Therefore, we could have a situation where two rational players continue outbidding each other, with each trying to convince the other he is crazy enough to lose all his wealth in the game.
Posted by: MITdude at Aug 2, 2006 12:10:03 PM
MITdude:
I don't think a game with such a strategy precludes all players being rational. If A can make the threat credible, it's an optimum strategy, so A doesn't need to be able to establish his own craziness or irrationality to do it. He simply needs a good committment technique to make the threat credible. (Brinksmanship can be rational.)
Such techniques are well-known in the negotiation world. For example, bidder A could make a legally binding contract with the auctioneer to increase his bids without limit.
Posted by: Johannes Climacus at Aug 2, 2006 12:45:26 PM
Yes, that would work too. The "doomsday device" in Dr. Strangelove is another good example!
Posted by: MITdude at Aug 2, 2006 1:55:47 PM
Ok, in this comment thread we've collectively broken this game for:
- Players have unequal wealth
- Players have ability to make binding committments
- Players have ability to refuse to bid at all
- Players have ability to bid increasingly small amounts without bound
- Players have the ability to open by bidding exactly one biddable unit below the prize value
Am I missing anything? Anyone else want to beat it up a bit?
Posted by: Johannes Climacus at Aug 2, 2006 9:01:52 PM
This game is not so irrational as it seems. It might be explained like this:
I agree to commit 1 cent in order to win a dollar. If I am outbidden by 1 cent, the game starts again - I can commit 2 more cents in order to win a dollar - perfectly rational (the first cent is a sunk cost and it doesn't matter for the next bid).
In reality, we can see this in the lottery - you give some money every week because you expect to win the next time and you may continue doing it infinitely. However, not all people do it infinitely - some of them stop buying lottary tickets irrespective of the money they already gave up and irrespective of the possibility to win (others become addicted...).
For sure, you increase your chances for winning the lottery if you invest all your money in it. But most people just don't do it - because they have other needs also and these needs compete for the same money they have. So, irrespective of "anecdotal evidences" for any games played in classrooms, I'm not pessimistic on the results (surely, people are not giving up all money for lottery tickets).
Posted by: Georgi at Aug 3, 2006 10:46:13 AM
Thinking on the fly here, so I might be wrong, but I think you can broaden that theorem by combining (A1) and (A2) into "both players have finite wealth" -
then we just cant't say which player will win (first mover I'd guess), but someone will with the same outcome (bid penny, other drops out) since (A3) still gurantees that someone tops out first and both players can calculate that out.
Requiring finite wealth means that we make the strategy space finite so Nash applies again.
Also I don't think that differences in payoffs would be that much of a problem - somehow they'd map into differences in the wealth levels. For example if player A values each penny twice as much as player B in utility terms that basically means he really has twice the "money" that he really does. If utility is concave in wealth that shouldn't be much of a problem either - just change units from dollars to von-Neumann utils. Similar things hold for uncertainty about opponent's wealth under usual assumptions about risk aversion, priors, etc.
Oh, and isn't this why Dixit gets two hour standing ovations at the end of his course?
Posted by: radek at Aug 3, 2006 5:58:55 PM
I am surprised nobody mentioned yet a very old version of the same game, called blood feud.
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Posted by: levan at Sep 11, 2006 4:23:07 AM
