It turns out that the uniform distribution on the interval [1/4,3/4] does the trick. Martin Osborne and Carolyn Pitchik proved this back in 1982. Link to paper: http://cowles.econ.yale.edu/P/cd/d06a/d0628.pdf See Proposition 3 for the relevant statement. (I haven't checked their proof, but I did check that the given distribution has the desired properties.) Jakob Jonsson ________________________________________ Från: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] för Bill Thurston [wpthurston@mac.com] Skickat: den 13 juli 2011 04:18 Till: math-fun Kopia: Chung-chieh Shan; Dylan Thurston Ämne: Re: [math-fun] Guessing game I'm interested in the simpified situation: let's say that the guessers make sealed choices without communicating ahead of time. Nonetheless, as a secondary question, it would be interesting if you can describe a good collusion strategy whereby the colluders have probability each greater than 1/3. If there's a unique Nash equilibrium, all three players need to have the same distribution, but it's possible that there are asymmetric equilibrium distributions or even optimal distributions where the three players have three different distributions. But, I'd be happy to see the "best" solution where all three pick the same distribution. Bill On Jul 12, 2011, at 10:04 PM, Michael Kleber wrote:
Bill -- How do you deal with collusion? If Dylan and I agree to pool our winnings, surely we want to bracket your guess and box you out entirely, if we can.
--Michael
On Tue, Jul 12, 2011 at 9:56 PM, Bill Thurston <wpthurston@mac.com> wrote:
One of my cousins asked me this question: Suppose 3 people make secret guesses of a number, uniformly randomly chosen from 1--100. Whoever is closest wins. What is their "best" strategy?
I'd like to modify this to let the unknown number to be uniformly chosen real number in [0,1]. In the event of ties, let's say the winner is chosen uniformly from all best guesses (or the "prize" is distributed equally, I don't care). By "best", I want a Nash optimal probability distribution for the guesses of the the three participants --- that is, if each one knows the probabilities with which the others choose their guess, they can do no better than using the same distribution.
Note that with two guessers, a guess of .5 wins over everything except another guess of .5, when it ties.
For 3 guessers, it's not optimal for all three to pick .5, because .49 then wins against the other two. It's also not optimal to pick a number uniformly at random from the unit interval: if two of the players do that a guess of .5 has an expectation of 5/12.
Bill Thurston