# If three six-sided fair dice are rolled, what is the probability that two dice show one number, and the remaining die shows another number? [closed]

If three six-sided fair dice are rolled, what is the probability that two dice show one number, and the remaining die shows another number?

I think the total of possible outcomes is $6^3= 216$, but I don't know how to apply the n C r in this case.

## closed as off-topic by José Carlos Santos, JonMark Perry, Hurkyl, Brian Borchers, user223391 Mar 25 '18 at 0:10

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• Where are you stuck? – Robert Z Mar 24 '18 at 9:10
• @RobertZ I'm trying to figure out how to even start with this question by using the combination method. – jasminvvian Mar 24 '18 at 9:34
• The entire statement of the question should be in the body of your question. Had I not noticed that the question was the continuation of a sentence started in the title, I would have just solved for the favorable cases. Please edit your question to tell us what you know and where you are stuck. For instance, how many possible outcomes are there? – N. F. Taussig Mar 24 '18 at 10:01
• @N.F.Taussig sorry about that, I'm a bit new on this site, still trying to get used to all the UI and such. I think the total possible outcome is 6^3= 216? but I don't know how to apply the n C r in this case – jasminvvian Mar 24 '18 at 10:14

Strategy: Assume the colors of the dice are blue, red, and green to make the dice distinguishable. To count the favorable cases:

1. Choose which two dice show the same outcome.
2. Choose which number those dice show.
3. Choose which of the remaining numbers the other die shows.

Finally, divide by the total number of possible outcomes, which you have correctly calculated.

• So I got 3 C 2 for choosing 2 dice out of three being the same number. Then I got 6 C 1 for the number of ways those number show, then finally I got 6 C 5 for the remaining numbers for the other die. divided by 216 I got 0.5 but the number doesn't seem right – jasminvvian Mar 24 '18 at 10:22
• You are correct that there are $\binom{3}{2}$ ways to select two of the three dice to show the same outcome and $\binom{6}{1}$ ways to select the number those dice show. However, in part 3, how many possible outcomes remain for the third die given that it is different from the outcome shown on the other two dice? Of these, how many can it show? – N. F. Taussig Mar 24 '18 at 10:23
• omg I think I got it. It should be 5 C 1 because one of the number was already picked out. You are a savior ! Thank you guys! – jasminvvian Mar 24 '18 at 10:26
• Yes, that is correct. The answer is $$\frac{\binom{3}{2}\binom{6}{1}\binom{5}{1}}{6^3}$$ – N. F. Taussig Mar 24 '18 at 10:28
• I like this strategy of thinking about these type of problems. Thank you for walking me through it – jasminvvian Mar 24 '18 at 10:29

There are $6\cdot5\cdot4=120$ ways of three different numbers and $6$ ways of three equal numbers. The remaining $216-126=90$ ways are favorable. The probability in question therefore is ${90\over216}={5\over12}$.

Hint, when you can't find a magical formula, start by counting the favourite outcomes. There are $6\cdot6\cdot6$ possible outcomes altogether. But favourite ones are:

• $d_1=d_2=1$ and $d_3\in\{2,3,4,5,6\}$. $5$ outcomes.
• $d_1=d_2=2$ and $d_3\in\{1,3,4,5,6\}$. $5$ outcomes.
• $...$
• $d_1=d_2=6$ and $d_3\in\{1,2,3,4,5\}$. $5$ outcomes.

Then count for

• $d_1=d_3=1$ and $d_2\in\{2,3,4,5,6\}$. $5$ outcomes.
• $d_1=d_3=2$ and $d_2\in\{1,3,4,5,6\}$. $5$ outcomes.
• $...$
• $d_1=d_3=6$ and $d_2\in\{1,2,3,4,5\}$. $5$ outcomes.

And one more ...

It should be easy to finish the exercise from here.

• thank you for showing me another way to think of this problem! – jasminvvian Mar 24 '18 at 10:27

First we need to select those two dies out of the three which show same number which can be done in 3C2 ways. Now for those two dice with same number the third dice can have five different results to satisfy the condition. So the answer will be (3C2*5*6)/63 = 5/12