# Statistics Casino Crap game

1. Dec 14, 2012

### Biosyn

1. The problem statement, all variables and given/known data

Roll a pair of six-sided dice. If the sum is 7 or 11, you win. If the sum is 2,3, or 12, you lose. If the sum is any other number, you roll again. In fact, you continue throwing the dice until you either roll that number again (win!) or roll a 7 (lose!).

Suppose you roll a sum of 8 on the first roll. Find the probability that you subsequently win the game, given that you rolled an 8 to start with.

2. Relevant equations

Conditional probability

3. The attempt at a solution

I'm not sure which way is correct.
First, I did: 5/36 because there are 5 ways to get a sum of 8.

Then I thought about the probability being 5/11. Because there are only two outcomes that I care about. An 8 or 7.
There are 5 ways to get an 8 and 6 ways to get a 7.

So... 5/(5+6) = 5/11

Would someone please explain the difference between the two attempts and explain why one or the other is wrong?

2. Dec 14, 2012

### awkward

In your first try, the probability of rolling a 7 doesn't enter into the calculation at all. Doesn't that seem wrong to you?

Your second try gives the correct answer, but it's not clear to me that you know why.

3. Dec 14, 2012

### Biosyn

I was thinking about only finding the probability of getting an 8 to win. I didn't think finding the probability of getting a 7 would matter.

4. Dec 14, 2012

### Ray Vickson

There are two ways to look at it: (1) the straightforward, but technically lengthy way; and (2) the short, but tricky way.

Let's look at (1). The relevant outcomes are '7' (prob. = p_7 = 7/36), '8' (prob. = p_8= 5/36) and 'o' = 'other' (prob. = p_o = 1- p_7 - p_8 = 24/36). You keep tossing until a 7 or 8 appears; you win on '8' and lose on '7'. You can win on the first toss (i.e., 'first' after getting your original '8' = first toss of the rest of the game), or you can win on the second toss, or on the 3rd toss,.... or on the nth toss... . The probability you win on the nth toss p_o^(n-1)*p_8 for n = 1,2,3,.... So
$$P\{\text{win}\} = p_8 \sum_{n=1}^{\infty} p_o^{n-1} = p_8/(1 - p_o) = p_8/(p_7 + p_8).$$
Method (2). Since '7' and '8' are the only _relevant_ outcomes, P{win} = P{8|7 or 8} = p_7/(p_7 + p_8). Can you see the logic here?

5. Dec 14, 2012

### Biosyn

Where is this formula from?
I understand it much better now. Thank you.

6. Dec 14, 2012

### Ray Vickson

P{win} = P{win on toss 1} + P{win on toss 2} + P{win on toss 3} + ... .