January 17th, 2008, 5:28 pm
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?