# Probability of Rolling a 5 or 7

1. Nov 17, 2007

### e(ho0n3

[SOLVED] Probability of Rolling a 5 or 7

1. The problem statement, all variables and given/known data
A pair of dice is rolled until a sum of either 5 or 7 appears. Find the probability that a 5 occurs first.

2. Relevant equations
Axioms and basic theorems of probability.

3. The attempt at a solution
There are 36 possible outcomes of a roll of a pair of dice, 4 of which result in a sum of 5 and 6 of which result in a sum of 7. Let A be event of rolling a 5 and B of rolling a 7. Then P(A) = 4/36 and P(B) = 6/36. The probability of rolling a 5 or 7 is P(A or B) = P(A) + P(B) = 10/36. The probability of not rolling a 5 or 7 is thus 1 - 10/36 = 26/36.

Define E(n) as the event that the nth roll results in a 5 and let p(n) be it's probability. If the nth roll resulted in a 5, then there were no previous rolls resulting in 5 or 7, so p(n) is thus equal to (26/36)^(n - 1) * 4/36.

The probability sought then is just the sum of p(n) for all legal values of n, which is an infinite sum. Is this right? I don't know what the sum is but it must definitely be less than or equal to 4/36.

2. Nov 17, 2007

### Dick

I don't think you actually need to sum the infinite series. If you get to the nth roll without rolling a 5 or a 7 and then you roll a 5 or a 7, then the ratio of the two probabilities is just the ratio of the probability of rolling a 5 to that of rolling a 7. So if you choose to write an infinite series for each of the events, the ratio of each term in one series is a fixed ratio of the term in the other series.

3. Nov 17, 2007

### e(ho0n3

Do you mean that P(5 occurs first) / P(7 occurs first) = P(rolling 5) / P(rolling 7)? That is certainly true so P(5 occurs first) = P(rolling 5) / P(rolling 7) * P(7 occurs first). However, that doesn't help much in computing the sought probability.

4. Nov 17, 2007

### Dick

Umm. I don't see the difference in the problems. You end up with either a 5 or a 7. There are no other possibilities.

5. Nov 17, 2007

### D H

Staff Emeritus
You can do the infinite series, and it isn't too hard to do that. Use this identity:

$$\sum_{n=0}^\infty x^n = \frac 1 {1-x}$$

The identity is true for all $x\;\in\;(-1,1)$

Before you do that, there is an even easier way to solve the problem, as suggested by Dick. Suppose you get the 5 or 7 on the first roll, or the tenth roll, or the one thousandth roll. All the previous rolls were neither a 5 or a 7. You can solve the problem simply by examining one single roll where you know a-priori that either a 5 or a 7 was rolled with a pair of fair dice. Given that information, what is the probability that a 5 was rolled?

6. Nov 17, 2007

### e(ho0n3

You're saying that P(5 was rolled first) = P(5 was rolled | 5 or 7 was rolled) right?

Question: what does the sample space S look like for this problem? If e is in S then e has the form (e_1, e_2, ..., e_n). Each e_i is of the form (x, y) where x is the value of the upper face of one die and y is that of the other. If i = n, then x + y equals 5 or 7; otherwise x + y does not equal 5 or 7. S will contain such an e for all positive integers n.

If I'm examining one single roll, then the sample space is reduced considerably, i.e. it only contains e's of the form (e_1). How does this work?

7. Nov 17, 2007

### D H

Staff Emeritus
There are 36 possible outcomes for rolling a pair of fair dice. How many of these outcomes sum to either 5 or 7?

8. Nov 17, 2007

### e(ho0n3

I calculated that in my first post as 10/36.

9. Nov 18, 2007

### D H

Staff Emeritus
No. That is the unconditional probability of rolling a 5 with a pair of fair dice. You did not answer my question, which was how many of the thirty six total outcomes sum to either a 5 or 7? Note: I am asking for an integer, not a probability. This is the sample space. Now, how many of these sum to 5? This you can convert to a probability.

10. Nov 18, 2007

### e(ho0n3

Oops. Sorry. That should be 10: (3,2), (2,3), (1,4), (4,1), (5,2), (2,5), (3,4), (4,3), (6,1), (1,6). Of these, the first 4 sum to 5. So you're saying the probability of getting a 5 first is 4/10? I guess it makes sense to ignore the previous rolls because they don't affect the outcome of the last roll; it's as if the last roll was my first roll.