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?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 ${90\over216}={5\over12}$.

Answer 4

First we need to select those two dies out of the three which show same number which can be done in $^3C_2$ 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 $$\frac{^3C_2\times5\times6}{6^3} = \frac{5}{12}$$