This problem is similar to a problem from "Heard on the Street". The only modification is making it zero-sum. I apologize if it's old already.Say A and B are playing a marble game. Each of them has 2 marbles Red and Blue. Each round, A and B may decide showing one of their marbles and ifR, R appears, A wins $3, B loses $3B, B appears, A wins $1, B loses $1R, B appears, B wins $2, A loses $2B, R appears, B wins $2, A loses $2Suppose you are risk-neutral, do you wanna be A or B?

Both A and B have expected outcome of zero, but A has maximum loss of 2, while B has maximum loss of 3, representing a positive skew for A, and therefore, A would be preferred.

Hey bilbo1408, I suggest you reconsider your answer. Remember each run, A and B can decide to show R or B. It's not necessarily true that they pick it completely random. QuoteOriginally posted by: bilbo1408Both A and B have expected outcome of zero, but A has maximum loss of 2, while B has maximum loss of 3, representing a positive skew for A, and therefore, A would be preferred.

With optimum mixed strategy, it appears that A is at a disadvantage and has an expected gain of -1/8.So would rather be B.

If A and B act simulataneously --Prob(A shows Red): pProb(B shows Red): q Then the expected value for A :8pq - 3p -3q + 1On differentiating This has maxima at p = q = 3/8Giving the exp for A to be -1/8. And +1/8 to be B.----------------------------------EDIT: My question is on reaction time. Assuming A is not aware of the so called optimal mixing strategy. Can you fool/hustle him?You play with suboptimal q for some time, make him feel like he's winning this game and then pounce on him -- before he realizes the equilibrium point your avg, winning will prbbly be higher than 1/8. At most you lose a friend.

Ok...I didn't catch that part, and I have reconsidered. How about this answer:If I chose to be player A, I would chose red each time so that I either have +3 or -2 for expected return of +.5If I chose to be player B, I would chose blue each time so that I either have +2 or -1 for expected return of +.5Therefore, my choice would be B because of the same expected return, but with less variance.

QuoteOriginally posted by: AdvaitaOn differentiating This has maxima at p = q = 3/8Not correct. Actually it is not a maximum. It is a saddle point.

LOL

Yeppe.Actually we don't need to assume anything.In my opinion, the crux is not q = 3/8 is maximal or not. You can't assume A obets our order. I think what makes it a solid argument is,8pq -3p -3q + 1 = p(8q -3) -3q + 1When q = 3/8, p disappears, meaning A has no control.QuoteOriginally posted by: AdvaitaIf A and B act simulataneously --Prob(A shows Red): pProb(B shows Red): q Then the expected value for A :8pq - 3p -3q + 1On differentiating This has maxima at p = q = 3/8Giving the exp for A to be -1/8. And +1/8 to be B.----------------------------------EDIT: My question is on reaction time. Assuming A is not aware of the so called optimal mixing strategy. Can you fool/hustle him?You play with suboptimal q for some time, make him feel like he's winning this game and then pounce on him -- before he realizes the equilibrium point your avg, winning will prbbly be higher than 1/8. At most you lose a friend.

Ahhh....got it.

If both A and B show together (not in a sequence), If A shows a red ball, his possible gain is: 3*(1/2) + (-2)*(1/2) = 3/2 - 1 = 1/2If A shows a blue ball, his possible gain is: 1*(1/2) + (-2)*(1/2) = 1/2 - 1 = -1/2If B shows a red ball, his possible gain is: 2*(1/2) + (-3)*(1/2) = 1 - 3/2 = -1/2If B shows a blue ball, his possible gain is: 2*(1/2) + (-1)*(1/2) = 1 - 1/2 = 1/2So, both A and B can earn the same maximum amount if they show the balls together.QuoteOriginally posted by: pk14This problem is similar to a problem from "Heard on the Street". The only modification is making it zero-sum. I apologize if it's old already.Say A and B are playing a marble game. Each of them has 2 marbles Red and Blue. Each round, A and B may decide showing one of their marbles and ifR, R appears, A wins $3, B loses $3B, B appears, A wins $1, B loses $1R, B appears, B wins $2, A loses $2B, R appears, B wins $2, A loses $2Suppose you are risk-neutral, do you wanna be A or B?

If you want to answer this question game-theoretically, you have to assume that your opponent will play optimally, i.e. unexploitably. A well-known result in game theory is that A's optimal strategy occurs when B is indifferent between his two strategic options. Build the table where A's options are down the side, B's options are across the top, and the payoffs are an ordered pair (A,B):....... R ...... BR (3,-3) (-2,2)B (-2,2) (1,-1)Let p the probability of A playing Red. B has two strategic options, play Red or play Blue:EV(B plays Red) = -3p + 2(1-p) = 2-5pEV(B plays Blue) = 2p - (1-p) = 3p-1A wants to make B indifferent to his strategic choices. Thus:2-5p = 3p-1p = 3/8Similarly, let q be the probability of B playing Red. Now B wants to make A indifferent to playing Red or Blue:EV(A plays Red) = 3q - 2(1-q) = 5q-2EV(A plays Blue) = (1-p) - 2q = 1-3q5q-2 = 1-3qq = 3/8Thus the optimal strategy for both A and B is to play Red 3/8 of the time. The EV of the game, played optimally by both, is that B wins 1/8 dollars per game.