2.4 Adding Dice

Example 1 What is the probability of the sum of two dice getting a particular value, say, 7?

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In this case, we simply outline every single possibility, and count the fractions. In a more complex case we may need to find a better method of counting, but the idea will be the same.

Note

All possible results from rolling two dice:

sum	(die 1,die 2)
2	(1,1)
3	(1,2),(2,1)
4	(3,1),(1,3),(2,2)
5	(1,4),(4,1),(3,2),(2,3)
6	(1,5),(5,1),(4,2),(2,4),(3,3)
7	(1,6),(6,1),(5,2),(2,5),(4,3),(3,4)
8	(3,5),(5,3),(6,2),(2,6),(4,4)
9	(5,4),(4,5),(3,6),(6,3)
10	(4,6),(6,4),(5,5)
11	(6,5),(5,6)
12	(6,6)
	(36 arrangments total)

We find immediately that the probability of getting a sum of 7 is the largest, because there are more arrangements of the two dice which yield a sum of 7 than for any other sum. Each probability of a particular sum is just the number of arrangements to get that particular sum divided by the total number of arrangements of a two dice (i.e. 36).

$\begin{matrix} P (2) = \frac{1}{36} = 0.028 \\ P (3) = \frac{2}{36} = 0.055 \\ P (4) = \frac{3}{36} = 0.083 \\ P (5) = \frac{4}{36} = 0.111 \\ P (6) = \frac{5}{36} = 0.139 \end{matrix}$

$\begin{matrix} P (8) = \frac{5}{36} = 0.139 \\ P (9) = \frac{4}{36} = 0.111 \\ P (10) = \frac{3}{36} = 0.083 \\ P (11) = \frac{2}{36} = 0.055 \\ P (12) = \frac{1}{36} = 0.028 \end{matrix}$

Example 2 What is the probability of rolling a sum more than 7 with two dice?

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In our notation this is

$P open parenthesis 8 or 9 or 10 or 11 or 12 close parenthesis$

which are all exclusive events, so we use the Sum Rule for exclusive events (Equation 1.11) and obtain

$\begin{matrix} P (8 or 9 or 10 or 11 or 12) = P (8) + P (9) + P (10) + P (11) + P (12) \\ = 0.139 + 0.111 + 0.083 + 0.055 + 0.028 \\ = 0.416 \end{matrix}$

Adapted from Statistical Inference for Everyone, by Brian Blais (Bryant University), licensed under CC BY-SA 4.0 (dual-licensed under the GNU FDL 1.2 or later; this adaptation uses the CC BY-SA grant). Changes were made; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.