QuoteOriginally posted by: rmexicoIt's been a long time, but thanks to Beero, I remember some game theory now.10 isn't an equilibrium. If I knew everyone else would choose 10, I could choose 8 (or 12.5) and win. Alternatively, I could be a nice guy and choose 10, so that everyone wins, but there's no guarantee that I would. Everyone else would see this coming, so nobody would choose 10.1 is an equilibrium. If everyone else chooses 1, and I choose something different from 1, I don't win. None of the other numbers in the interval [1, 90] has that property.Agreed ... 1 is the more stable equilibrium (minimizing distance to the 4/5 and 5/4 points) ... the choices I wrote below are not the more stable equilibrium.

so is 1 really the answer? i'm still not convinced. this might be picking bones from an egg, but seems to me you guys made some assumptions on how a genius would reason, i.e., his utility function.but if all you geniuses would pick 1, then i definitely learned a lesson.

The answer may be that, by stating that everyone is a genius, everyone will think in the same way. The best way to explain it was shown below:QuoteOriginally posted by: beero1000...Assume there is a 'best number' A. That is, all the geniuses recognize that picking A will give them the best shot at the prize, and therefore they all pick it. So that's 89 who choose A. If the closest integers to (4/5)A or (5/4)A are pickable and isn't A, then I can choose one and win the prize, while everyone else loses. That means A can't possibly be the best choice. The only possible best number is one where the closest choosable integer to (4/5)A or (5/4)A is A itself. For example, if for some reason all the geniuses think 10 is the best bet, then I can choose 8 and expect to win (I'm a genius too!). The geniuses should know this and therefore won't choose 10. The same reasoning invalidates all of the numbers but 1. No one person can steal the prize if everyone else chooses 1, so assuming that none of the 90 communicate a plan, its not worth it for any person to play anything but 1. It's the prisoner's dilemma. also assuming we only want the title of 'winner' (it doesn't matter how many people win, as long as I do), then everyone should conclude that everyone will conclude that 1 is the best bet, and when we all play 1, we all win.

Sorry about the blank.Had to backtrack and think a bit more.

Last edited by quantyst on December 4th, 2008, 11:00 pm, edited 1 time in total.

This was a contest conducted by my options pricing prof in my fin math program. Apparently it is actually a very famous problem and is more "behavioral" than quant. I think this was a part of a Financial Times competition, which found many people guessing the least answer( 1 or 0), a few people guessing the average of average, or average of average of average, and so on. I agree that 1 is probably the stable equilibrium, but there are supposedly a ton of behavior case studies on this kind of stuff. Our competition was to find 75% of the mean, and you will be surprised, the final answer was 18. There were lots of people who selected 1.The answers were annonymous, but there was one entry of a 100, suggesting that some "market manipulation" was happening, which is what I suspect will happen if geniuses play this game

Last edited by vxs on December 13th, 2008, 11:00 pm, edited 1 time in total.

- Traden4Alpha
**Posts:**23951**Joined:**

Actually, both 1 and 2 are stable solutions in the sense that one can't win by choosing a different number if everyone else picks 1 or if everyone else picks 2. This result implies we need a tiebreaker. I can think of four tiebreakers:1. Simplicity: 1 is the first number and so many people might think it is "more natural" as a choice.2. Contextual information: If the company is named Two Guys and an Algorithm LLC, then 2 might be the right answer. Or people in a Cantonese-speaking region might prefer "2" because that number is a good number and sounds like the work "easy."3. Broken Symmetry: Consider the possible distribution of choices of 1s and 2s, the calculation of the average, and the role of the round-up rule (e.g., that 1.5 rounds to 2). Of the 91 possible values for the fraction of "1"s in the population of 90 choices, 50.55% of them favor choosing a "2". Round-up breaks the symmetry between the two choices. In fact, the closer one thinks the distribution will be 50-50, the greater the chance of winning with a "2". 4. Robustness to mistakes: If one believes that some of the 90 "geniuses" are are not that smart (e.g., those losers in sales who couldn't integrate x*dx if their lives depended on it), then we might anticipate a few mistaken answers that are much higher than 1. The answer "2" would be more robust to rare mistakes. Of course, if one expects a non-negligible fraction of mistakes, then answers higher than 2 may be warranted (see vxs's post for an example of that).Factors 3 and 4 suggest a choice of "2", factor 1 suggests a choice of 1, and factor 2 depends on context. EDIT: "2" isn't stable. It looks like I should have read the problem statement more closely (or drank my coffee before posting)!

Last edited by Traden4Alpha on December 13th, 2008, 11:00 pm, edited 1 time in total.

Traden4Alpha, note the problem differs from the original version that the winning number is closest to either 4/5 or 5/4 of the mean.as beero1000 stated in his original post, if everyone else picks 2, one wins by choosing 3. (i don't agree that 1 is the answer though:P)----- ----- ----- ----- -----vxs, thank you for posting the distribution!

Last edited by wileysw on December 14th, 2008, 11:00 pm, edited 1 time in total.

Distribution:11111222235561010101011121313141414151515171717171818191919202021212325252525262627282828282933333436363638384242434750505563646772100

It's clear in vxs's case --- where the winner was the submission closest to 75% of the average --- whoever picked 100 was not expecting to win. In fact, I'd guess that everyone who picked above 40 did not understand the problem (or did not care about winning, which is the same thing). When I was an undergrad, I participated in some economics experiments like vxs's, not two-sided as in this problem, and recall that the winning guess was usually about two steps from the middle number --- something like 50 * 0.75^2 = 27. I was naive and picked 1.A generation passes, and vxs's cohort thinks (one or) two more steps ahead. I note that with 70+ students, the individual guess of 100 moved the mean up by 1 point.

You should read about the original (FT?) problem in my FAQs book. People think about this problem in a way similar to how they think about the magician question in my blog. Some treat it as an exercise in game theory, and look for the 'correct' answer, others try to anticipate what others are anticipating (think of the Keynes quote here). The former will always lose, the the winner will be from the latter group!@rcyeh: Not understanding the problem and not caring about winning are very different! You really should read about what happened when my students and I entered the FT competition!P