# Thread: Help with Probability flipping a quarter!!!!

1. ## Help with Probability flipping a quarter!!!!

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.

2. Originally Posted by 2y4life
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 3 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.
Calculate the probability $p_{John}$ that John would have won from this position.

Calculate the probability $p_{Judy}$ that Judy would have won from this position.

Then give John $\text{\}10 \, p_{John}$ and give Judy $\text{\}10 \, p_{Judy}$.

3. Originally Posted by mr fantastic
Calculate the probability $p_{John}$ that John would have won from this position.

Calculate the probability $p_{Judy}$ that Judy would have won from this position.

Then give John $\text{\}10 p_{John}$ and give Judy $\text{\}10 p_{Judy}$.
Yea...but is there a more simple way of finding that out? I tried and there are too many possibilities. John could flip another head and then flip 5 tails in a row and Judy would win or vice versa.

4. Originally Posted by 2y4life
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.
This question is historically interesting ..... read this to see why.

5. Originally Posted by 2y4life
Yea...but is there a more simple way of finding that out? I tried and there are too many possibilities. John could flip another head and then flip 5 tails in a row and Judy would win or vice versa.
Let X be the random variable number of points John wins.
Let Y be the random variable number of points Judy wins.

Then:

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.

6. Originally Posted by mr fantastic
This question is historically interesting ..... read this to see why.
Yea, this was an extra credit problem for us to take home and figure out and my TA said that this problem came from two mathematicians named Blaise and Pierre.

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

7. Originally Posted by mr fantastic
Let X be the random variable number of points John wins.
Let Y be the random variable number of points Judy wins.

Then:

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.
You only need to work out one of John winning or Judy winning (you know why, right).

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.