# Probability with 2 dice

1. Sep 27, 2006

### Tasaio

Suppose we make 20 tosses using 2 dice. What is the probability of scoring a two numbers that sum to 7?

My attempt

The sample space for a single toss of a single die is S = {1, 2, 3, 4, 5, 6}

For a single toss of both dice, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

The sample points with sum 7 are:
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

These are 6 possibilties out of 36.

So for *each toss* of the 2 dice, there is a 6/36 probability of scoring two numbers that sum to 7.

My question is, how to we work out what the probabilty would be for 20 tosses?

If we have 20 tosses, then there are a total of 36*20 = 720 possible sample points. But how many of those possibilities contain numbers that sum to 7?

2. Sep 27, 2006

### abode_x

think of this one: 20 tosses of a single coin. what is the chance, you get a head?
the easiest approach is to count the chance no head appears (all tails for 20 tosses) and subtract it to 1.

3. Sep 27, 2006

### Tasaio

Using your coin example, on each toss, there is a 0.5 probability that there is a head.

So for 20 tosses, we calculate:
0.5 * 0.5 * 0.5 * ... * 0.5 (20 times)

Let's try that for my question.

For *each toss*, there is a 30/36 chance that the numbers do not sum to 7.

So after 20 tosses, the probability is:
(30/36) * (30/36) * (30/36) *...*(30/36) (with 20 terms)

= (5/6) * (5/6) *...* (5/6)
= ~0.026084

1 - 0.026084 = 0.973916

So there is a ~0.97 probablity that one of the tosses contains numbers that sum to 7.

This probability is very high. Does it sound about right?