When three dice are thrown, what is the probability of getting a sum of 7?

Let's say the dice are blue, green, and red. Each outcome can be represented by the ordered triple $(b, g, r)$. Since there are six choices for each entry, there are $6^3 = 216$ possible outcomes. Assuming the dice are fair, they are equally likely to occur.

Method 1: The favorable outcomes are those in which at least one pair of dice selected from the three rolled dice have sum $7$.

There are $\binom{3}{2} = 3$ ways to select a pair of dice. Say we select the blue die and the green die. There are six ways for that pair of dice to have a sum of $7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$, where each ordered pair is of the form $(b, g)$. For each such pair, there are six possible outcomes for the red die (or, in general, the other die), giving $3 \cdot 6 \cdot 6$ favorable outcomes.

However, we have counted each outcome in which there are two pairs of dice that have sum $7$ twice, once for each way we could have designated one of the pairs as the pair that has sum $7$. For this to occur, two of the dice must show the same number and the third die must show the difference between $7$ and the number that appears on the other two dice. There are $\binom{3}{2} = 3$ ways for exactly two dice to display the same number and $6$ numbers these two dice could display. For any such outcome, there is only one possible outcome for the third die. Hence, there are $3 \cdot 6$ such outcomes.

Thus, the number of favorable cases is $3 \cdot 6 \cdot 6 - 3 \cdot 6 = 108 - 18 = 90$.

Therefore, the probability that when three dice are rolled that a pair of dice may be selected with sum $7$ is $$\frac{3 \cdot 6 \cdot 6 - 3 \cdot 6}{6^3} = \frac{90}{216} = \frac{5}{12}$$ as InterstellarProbe found.

Method 2: The favorable outcomes are those in which at least one pair of dice selected from the three rolled dice have sum $7$.

There are two possibilities. Either each die shows a different outcomes or two of the three dice show the same outcome. Since $7$ is not a multiple of $3$, it is not possible for all three dice to show the same outcome.

Each die shows a different outcome: There are $\binom{3}{2} = 3$ ways for two of the dice to have sum $7$ and $6$ ways for those two dice to have sum $7$. There are four possible outcomes for the other die. Hence, there are $3 \cdot 6 \cdot 4$ such outcomes.

Two of the dice show the same outcome: There are $\binom{3}{2} = 3$ ways for exactly two of the three dice to show the same outcome and six possible outcomes those two dice could show. The other die must show the difference between $7$ and the outcome on those two dice. Hence, there are $3 \cdot 6$ such outcomes.

That yields a total of $3 \cdot 6 \cdot 4 + 3 \cdot 6 = 90$ favorable outcomes.

Hence, the probability that a pair of the three dice has sum $7$ is $$\frac{3 \cdot 6 \cdot 4 + 3 \cdot 6}{6^3} = \frac{90}{216} = \frac{5}{12}$$ in agreement with the answer we obtained above.

Probability of getting a sum of 8 with three dice is nothing but getting any of the outcomes (1,1,6),(1,6,1),(6,1,1),(1,2,5),(1,5,2),(2,1,5),(2,5,1),(5,1,2),(5,2,1),(1,3,4),(1,4,3),(3, 1,4),(3,4,1),(4,1,3),(4,3,1),(2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3), and (3,3,2). A sum of 21 outcomes. Hence, the probability is .

Probability of getting a sum of 9 with three dice is nothing but getting any of the outcomes (1,2,6),(1,6,2),(2,1,6),(2,6,1),(6,1,2),(6,2,1),(1,3,5),(1,5,3),(3,1,5),(3,5,1),(5,1,3),(5,3,1),(1,4,4),(4,1,4),(4,4,1),(2,2,5),(2,5,2),(5,2,2),(2,3,4),(2,4,3),(3,2,4),(3,4,2),(4,2,3),(4,3,2), and (3,3,3). A sum of 25 outcomes. Thus, the probability is .

Probability of getting a sum of 10 with three dice is nothing but getting any of the outcomes (1, 3 ,6),(1,6,3),(3,1,6),(3,6,1),(6,1,3),(6,3,1),(1,4,5),(1,5,4),(4,1,5),(4,5,1),(5,1,4),(5,4,1),(2,2,6),(2,6,2),(6,2,2),(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2),(2,4,4),(4,2,4),(4,4,2),(3,3,4),(3,4,3), and (4,3,3). A sum of 27 outcomes. Thus, the probability is .

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