Probability of flipping a quarter

Hey guys, I'm new here but hopefully someone can help me with this question:

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.

Now I know that the probability of flipping heads or tails is 1/2 or 50%. But there are so many possibilities and that's where I'm stuck. I think that John will win because he's ahead by more than twice as many as Judy but I have a hard time figuring out the exact probability of him winning three more times. Hopefully someone can help me out

2y4life said:

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. Come up with a split that is most fair to both John and Judy.

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This problem is the oldest in probability theory. It dates back to at least the \(\displaystyle 12^{th}\) century. It goes by the name the problem of points. How does one divide a prize in an interrupted game? Do a web search for details. I will outline the solution.

In the game you have been given, the two would have flip the coin seven more times to absolutely determine a winner. That means \(\displaystyle 2^7 = 128\) outcomes; John wins in 99 of those outcomes while Judy wins in only 29.
So a fair division would be $7.73 to John and $2.27 to Judy.