So here is the problem. Maybe someone can help me. Thanks ahead of time anyone.
John and Judy play a game of flipping a quarter. If it comes up heads, John gets a point and if it comes up tails, Judy gets a point. They each bet $5 to play and the first person to get ten points wins the whole $10. At a point where John has 7 points and Judy has 5 points, the game is interrupted and they can't continue. John thinks he should win since he's ahead while Judy thinks she should win since she'd make a comeback. You are their friend, they agree to let you decide how to split up the money between them. Come up with a split that is most fair to both John and Judy.
Let Y be the random variable number of points Judy wins.
Pr(John wins from this position) = Pr(X = 3) times Pr(Y < 5).
Pr(Judy wins from this position) = Pr(X < 3) times Pr(Y = 5).
X and Y both follow binomial distributions.
Edit: Actually it's slightly more complicated than this I just realised but I have to rush off now. I'll have more to say later unless someone else gets in first.
I want to put this down as the answer but I'm not sure if this would constitute as a fair answer:
John- 7/12 x $10= $5.83
Judy- 5/12 x $10= $4.17
Probability of John winning: You need to calculate Pr(X = 3) times Pr(Y < 5) for n = 3, 4, 5, 6 and 7.
Probability of Judy winning: You need to calculate Pr(X < 3) times Pr(Y = 5) for n = 5, 6 and 7.