# You roll two balanced dice one time. What is the probability that you obtain either a sum of 8 or the same number on each of the two dice?

Dec 17, 2014

A probability can be calculated as:

$\left(\text{Desired events")/("All possible events}\right)$

Let's start by calculating the denominator first, i.e. all possible events. When you roll two dice, there are 36 possible combinations that might come up. Each dice can be any of the numbers 1 to 6. So $6 \cdot 6 = 36$. You could make a list to convince yourself, i.e. {1,1}, {1,2}, {1,3},...,{6,6}.

Now let's look at the numerator, i.e. all possible desired events. We desire either a sum of 8 or the same number on each of the two dice.

• It's relatively easy to count the number of ways you can get the same number on both dice. It's exactly six, i.e. {1,1}, {2,2},...,{6,6}.
• Now let's look at the different ways we can get a sum of 8. These are: {2,6}, {3,5}, {4,4}, {5,3} and {6,2}. This is a total of five.

So, in total we have 11 ways to get either a sum of 8 or the same number on both dice. However, note that the option {4,4} appears in both sets, so we can't count this twice. Removing that double-counting, we end with 10 ways to get what we 'desire'.

So, our final answer is $\frac{10}{36} = \frac{5}{18}$.