# Probability that $2$ dice selected from $3$ rolled dice will have sum $7$ [closed]

If I roll $$3$$ dice ($$6$$-sided) and then proceed to choose $$2$$ of them (discarding the 3rd), what is the probability of the total of those $$2$$ dice totalling $$7$$?

• How do you choose them? Are you selecting in favor of getting a 7? Against? Highest two, lowest two, Max and min, random? Commented Jul 5, 2019 at 12:24
• simply choosing the 2 you want to get the number you want Commented Jul 5, 2019 at 12:57
• Well, then it is just the number of cases where one die is a $1$ and other a $6$, or a $2$ and a $5$ or a $3$ and a $4$, divided by the total number of combinations with $3$ dice. Do you know how to do this?
– Rai
Commented Jul 5, 2019 at 13:02
• the probability of rolling a combo 7 with 2 dice is (6/36), there are 18 combinations with 3 dice and 108 total combinations, but that just gives the same probability (18/108), no? Commented Jul 5, 2019 at 13:30
• Say the dice are blue, green, and red. Then each outcome can be represented in the form $(b, g, r)$, so there are $6^3 = 216$ possible outcomes. What you need to count is the number of outcomes in which $b + g = 7$ or $b + r = 7$ or $g + r = 7$. Commented Jul 5, 2019 at 13:37

In Excel, I used the following formulas.

In cell A1: =1+MOD(ROW()-1,6)

In cell B1: =1+MOD(INT((ROW()-1)/6),6)

In cell C1: =1+MOD(INT((ROW()-1)/36),6)

In cell D1: =OR(A1+B1=7,A1+C1=7,B1+C1=7)

I copied these down 216 rows. This gives me every possible outcome of the roll of 3 dice, along with a TRUE/FALSE for if it satisfies the condition that at least one pair will add to 7. There are 90 such combinations out of 216.

In cell E1, I put: =COUNTIF(D1:D216,TRUE)

$$\dfrac{90}{216} = \dfrac{5}{12}$$

To calculate this combinatorially, we can count the number of ways three dice can achieve the desired result of at least one pair adding to 7 and divide by the total number of ways to roll three dice.

You can have 3 distinct numbers or two numbers with one number repeated.

For three distinct numbers, two of them must be one of the six pairs that add to seven and the last can be any of the four remaining numbers. Then we choose two of the numbers to be the pair that adds to seven: $$6\times 4\times 3 = 72$$

For two distinct numbers with one repeated, we choose one of the pairs of numbers, then we choose two of the dice to be one of the numbers: $$6\times 3 = 18$$

Total: 90 ways to achieve three dice where at least one pair adds to 7 out of 216 total.

• this is great, thanks. For someone who uses excel a lot, I dont know why i didnt consider using it to help me out as you did. great stuff. thanks again. Commented Jul 5, 2019 at 13:58

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.

The question isn't clear about how do we decide which 2 dice we pick.

1. If we pick the two uniformly random: every pick has 1/3 probability and since the choice of the dice is independent in the roll we get that for every choice the probilirt is 1/6. Overall ,since all the events are simetric, we get $$\frac{1}{6}$$.

2. If we choose as we wish, assuming that If there are a pair which sum to 7 we will pick them: denote by $$A$$ the event that the is a pair whose sum is 7. We first roll only two dice (arbitrary chosen beforehand) and denote by $$B$$ the event that their sum is 7. And by $$C$$ the event that that both are the same. by the law of total probability $$Pr(A) = Pr(A|B) Pr(B) + Pr(A|\bar{B}\cap C) Pr(\bar{B} \cap C) + Pr(A|\bar{B}\cap \bar{C}) Pr(\bar{B} \cap \bar{C} ) = 1*\frac{1}{6} + \frac{2}{6} * \frac{4}{6} + \frac{1}{6} * \frac{1}{6} = \frac{15}{36} =\frac{5}{12}.$$

When we used the fact that if the first pair didn't sum up to 7 the third dice can "complete" each of the first to into sum of 7 by exactly one option and hence $$Pr(\bar{B} ) = 2/6$$.

As the probability of rolling the same number stwice is $$1/6$$, the probability of rolling two different numbers which don't sum up to 7 is $$5/6 - 1/6$$.

• In the second case, are you accounting for the possibility that the first two dice have the same face up? Commented Jul 5, 2019 at 16:08
• You're right I missed that. Fixed
– Meni
Commented Jul 6, 2019 at 18:04