Talk:El Farol Bar problem

Actual Place?
This is my first time contributing to wikipedia so pardon me if i fail to follow any guidelines. Should the page provide a reference at the bottom to the URL for the actual Bar El Farol? As it is a real place. Their URL is http://elfarolsf.com Also should the page refer to this page http://en.wikipedia.org/wiki/Santa_Fe_Institute. I may be mistaken but i believe The El Farol Bar problems relationships with The Santa Fe Institute, Santa Fe and the actual bar are all relevant in a historical context. Saint141 (talk) 21:47, 1 August 2011 (UTC)

Prisoner's Dilemma?
Thanks for adding this entry, I think its a really cool problem and I'm glad its here. I'm also happy to have folks working on the game theory section, we are an elite few :) I removed two references comparing this problem to the prisoner's dilemma.  I'm not sure I see the similarity between the two games.  In the el farol problem, the nash equilibrium is the most socially efficient among all strategy choices, right?  That is society as a whole will not do (strictly) better if a non-nash equilibrium set of strategies is chosen.  The interesting thing in the prisoner's dilemma is that self-interest leads to socially inefficient strategy choices, which doesn't seem to be a problem here.  You might say that the mixed strategy equilibrium is inefficient, but this is true for many games (like for instance battle of the sexes) which I think are more similar to the el farol problem.  I may be misunderstanding the relationship, but it seems more misleading than helpful. Please add it back in if I'm wrong, but it would be good to say more about the relationship. best, --Kzollman 06:23, Jun 26, 2005 (UTC)

This article needs a quote from Yogi Berra.

Emphasis on repeated plays
It seems that there should be more emphasis that this game is a repeated game. That is, participants do not simply play once on one Thursday afternoon; it is critical to the character of the game that it is infinitely repeated and that the attendence of previous Thursdays is available to the same participants.

Also, it seems that the El Farol bar problem is not a specific case of the "more general" Minority Game; there are at least three fundamental differences. In the minority game, only an odd number of people can play (whereas any number of people may be participate in El Farol). Also, there are three outcomes possible in the El Farol bar (positive utility from attending the bar, negative utility from attending the bar, and zero utility from staying home) whereas there are only two outcomes possible in the minority game (positive utility from being in the minority and negative utility from being in the majority). Finally, the minority is the not the group that gains utility in the El Farol bar; if 59% attend the bar, then the majority will gain utility. Basically, I believe the character of the El Farol bar problem is substantively different than that of the Minority Game, even if both games demonstrate the same impossible-to-devise-a-superior-strategy characteristic. --Bjp716 02:08, Nov 7, 2006 (UTC)


 * Thank you for your suggestion. It sounds like you know more about this game than I do.  I'd like to invite you to make any changes you see fit to the article, I suspect they will be most welcome.  If other editors have concerns with your changes, they will discuss them here.  Thanks again and welcome to wikipedia.  --best, kevin [kzollman][talk] 02:22, 8 November 2006 (UTC)

Mixed Strategy Solution
I agree that the important aspect of the game is iteration and there are no solutions based on analysis of past attendance. However there is a solution based on a 'mixed strategy' although it is a decision theoretic problem not a game - if each player decides to attend 50% of days say by flipping a coin then for sufficiently large populations the bar will virtually never be crowded. If each player has access to a random number generator then they can evolve an optimal strategy using the following algorithm:

1. Choose a threshold 0 > t < 1 at random

2. Generate a random number n

3. If n > t

go to the bar

If bar crowded increase t     Endif Else stay at home If bar not crowded decrease t     Endif Endif 4. Go to 2

You could use bayes' formula to adjust t but any incremental change will work.

This is easy to model on a spreadsheet. It is also possible to introduce elements of heterogeneity amongst players such as risk aversion and crowd tolerance.

The Bar problem is a useful teaching tool because it emphasizes the importance of analysis a problem to determining the right data to support the decision and the danger of collecting data just because it is available and easy to do so.-- 62.31.119.101 (talk) 17:41, 22 January 2008 (UTC)

Equilibrium

 * In the case of the El Farol Bar problem, however, no mixed strategy exists that all players may use in equilibrium.

Why is going with 60% probability not an equilibrium? Even Brian's article seems to say it is. --Tgr (talk) 19:03, 22 January 2008 (UTC)

If each member of the population attended 60% of the time then the average attendance would be 0.6P (where P is the size of the population) which means that the bar would be crowded 50% of the time. Assuming the negative utility of a crowded bar is equal to the positive utility of an uncrowded bar then going 60% of the time would yield zero utility. Everybody would be better off if everybody went somewhat less than 60% of the time. --NigelPhil (talk) 11:20, 25 January 2008 (UTC)NigelPhil


 * I would have thought that an optimal shared strategy is not the question - the tragedy of the commons is well known. I think that Tgr's point is that a random 60% of the time approach is a stable equilibrium; any deviation from 60% by the population at large will encourage contrarian behaviour by everyone, while a 60% strategy does not. --Rumping (talk) 21:15, 25 October 2008 (UTC)

I mostly agree with Rumping and Tgr. In fact [Duncan Whitehead's September 2008 paper] shows on page 12 that there is a unique symmetric mixed strategy Nash equilibrium (in addition to multiple asymmetric equilibria) for the single stage bar problem. The exact probability with which each player should go in the symmetric equilibrium strategy should be close to 60% but might be slightly different because of:

-possible differences between the marginal utility (over staying home) of going to an uncrowded bar and the marginal disutility of going to a crowded bar

-the fact that Whitehead's formulation (and Brian Arthur's?) of the problem treats "exactly 60%" as part of the crowded scenario

-for any finite number of people, the binomial distribution will not be quite symmetric about the mode (unlike the normal)

I believe (but have not verified) that assuming staying home has utility exactly halfway between going to a crowded and an uncrowded bar, 60% is the equilibrium probability in the limit as the number of players goes to infinity. I will amend the quoted (false) sentence in the article and link to Whitehead's paper. (Should I give the exact formula, which is somewhat messy? I will leave it up to a more experienced editor to decide.) Matt Atwood (talk) 00:01, 11 February 2009 (UTC)

Minority Game
I think that the Minority Game section would be better in a separate article, especially given debate here that the El Farol Bar problem is not really an example of MG. (I agree it is not a true example, although there is some overlap of properties, which would be useful to highlight. Regretably, I am not expert enough on the subject to produce a full article.)

Here too, the effect of the expected number of further iterations should be indicated. (Not just the number. Or more accurately, I think the expected probability distribution of further iterations would be needed, since the optimium for the average number is probably not exactly the average of the optima for each number. A fixed number is merely a special case, though (ignoring tractability), a method to compute given a fixed number obviously allows computation given a probability distribution. This also incorporates definite non-iteration as a special case.)

The Minority Game section states: "Allowing for mixed strategies in the single-stage minority game produces a unique symmetric Nash equilibrium, which is for each player to choose each action with 50% probability, as well as multiple equilibria that are not symmetric." The all-50% NE is obvious, but I cannot believe (at least without previous plays to analyse, without further evidence, maybe in a reference) that there are other NEs. Even if there are some, there must be a symmetrical set, due to symmetry of the problem, so I would say, "... as well as a symmetrical set of asymmetric equilibria." (Maybe "other" would be better (simpler) than "asymmetric" with any wording, since all-50% is obviously the only possibility that is symmetrical w.r.t all players - unless all other NEs have no symmetry at all.)

The issue of how to analyse previous plays is a complex issue in its own right, and I cannot find anything definitive in the literature - indeed it is hardly mentioned! But some reference to the issue ought to be found and given (in any Games Theory article that mentions NEs), otherwise NEs do not make sense! The whole concept of NEs assumes that any deviation will tend to be exploited, but does not (Nash himself included!) say how that should be done. If I am sure that all other players are using NEs (as conventional GT says rational player should) then (even though I am rational!) I do not need to bother to compute them myself - I can even play a fixed strategy! (E.g., consider an equal-payoff binary choice game when one (NE) player always tosses a coin and the other (lazy) player always calls heads.) The problem is that, by NE definition, trying to exploit deviation itself leaves the exploiter open to exploitation! I "know" that no rational player will try to exploit me, so apparently can be rational and lazy! (Actually, in general, even a lazy player may need to eliminate some defintely bad plays, but apparently, if you believe conventional GT, need not, after all, play a mixed strategy. That is, conventional GT is paradoxical, so cannot be true as is stands!)

John Newbury (talk) 11:04, 2 July 2009 (UTC)