Imagine I offer you to play the following game. I will pick at random some number between 0 and 100 and I will offer to pay you that amount. You can accept the offeror reject it. If you reject it, I will pick another number between 0 and 100 at random. Again, I will offer to pay you that amount. If you reject it again, I will make you a third and final offer (again choosing a number at random from 0 to 100). You MUST accept this final offer. Should you play the game if I insist the ante be $65? If you do, what should your strategy be?cheers

Assume continuous distributions and acceptance bounds p and q for the first two rounds. The expected payoff (between 0 and 1) is then(1-p) * (1+p) / 2 + p * (1-q) * (1+q) / 2 + pq / 2The maximum is reached for p = 5/8, q = 1/2, which gives an expected payoff of 89/128 = 0.6953125.

Same outcome, just more clarification of what is going on.G1 = The value of the game at first throw and X1 = the outcome of throw one, etc. With a uniform distribution the contingent probabilities are just the average of the contingent range.E(G3) = E(X3) = 0.5 (i.e. on second throw you don't accept unless value higher than 0.5)E(G2) = P(X2>E(G3)) * E(X2 | X2>E(G3)) + P(X2<E(G3) * E(G3) = 1/2 x (1 + 1/2) /2 + 1/2 x 1/2 = 5/8E(G1) = P(X1>E(G2)) * E(X1 | X2>E(G2)) + P(X1<E(X2) * E(X2) = (1 - 5/8) x (1 + 5/8) / 2 + 5/8 x 5/8 = 89/128You continue unless you throw an outcome higher than the expected value of the remaining game.

typo fixed above

here is a related problem (recently asked on a monthly puzzle website):the setting is similar, but you have to pay a fixed amount x (0<=x<=100, note x is nonrandom) to enter each round of the game. you can play the game infinite many times till you decide to cash out, your final payoff is the maximum of all the random numbers you got minus the entry fees you paid. (so instead of an american-style payoff, it's a lookback-like payoff)what is your best strategy? what is your expected payoff as a function of x? (note you don't have to participate the game at the first place, and your payoff is 0)

The answer converges to 100% when you are able to play infinitely. Was that the question?

QuoteOriginally posted by: ZiggyThe answer converges to 100% when you are able to play infinitely. Was that the question?That's only true if x = 0. If x > 0, after 100/x rounds your payoff will definitely be negative.

The expected value of the maximum of n iid uniform random variables on [0,100] should be 100 * n / ( n+1). Hence you should choose n such thatn * ( 100 / (n+1) - x )becomes maximal.

Agreed, as the fee is paid every turn. I missed that point earlier.

QuoteOriginally posted by: cm27874The expected value of the maximum of n iid uniform random variables on [0,100] should be 100 * n / ( n+1). Hence you should choose n such thatn * ( 100 / (n+1) - x )becomes maximal.the strategy above is not optimal for 0<x<50. it's a little tricky to see what went wrong for this argument, since it gives correct result for the trivial cases: x=0 (n->infty and expected return->100) and x>50 (n=0 and expected return is 0). e.g., if x=25, it reaches maximum when n=1 and gives expected return of 25, but the correct answer is ~29.3.a big hint: the optimal strategy does not fix n. the criteria to stop playing is similar to what of the original problem.

QuoteOriginally posted by: cm27874The expected value of the maximum of n iid uniform random variables on [0,100] should be 100 * n / ( n+1). Hence you should choose n such thatn * ( 100 / (n+1) - x )becomes maximal.I am just a beginner in probability but I have a concern about this solution. The expected maximum 100*n/(n+1) is true only if no round has been played yet. Once the first round is played, then that expected maximum is not that value any more, isn't it ?Here is my solution for this problem. Call the expected value of the game F. Consider what happens after the first round. If you stop after the first round, the payoff is X1 - x. If you continue, then the game basically repeats itself, only that you have lost x paying for the first round's fee. Thus the expected payoff for continuing is F - x. The optimal stopping strategy is then to stop if X1 >= F. Consequently, we have the equation F = (100-F)*(100+F)/200 + F*F/100 - x => F = 100 - 10*sqrt(2x).

snapshooter, your answer is correct.(just for the sake of clarity, let me rewrite your equation as F = F/100*(F-x) + (100-F)/100*[(100+F)/2-x], where the 1st term is for the case that you got a number lower than F and continue. the 2nd term is that you got a number higher than F and cash out. here is the original problem.)