If you roll four fair dice, what is the probability that you will end up with two 1's and two 3's?

Question

Can someone help me with this? I'm completely stumped on how to go about it and no one I know can help. I just need to be shown how to do it.

Answer 1

The probability of an event is the number of "successful outcomes" over the number of possible outcomes. Since you have correctly stated that there are $6^4$ potential outcomes, let's try to count the number of outcomes that are "successful", i.e. contain two 1's and two 3's.

If we roll four dice, the "successful" outcomes are

• $3,3,1,1$
• $3,1,3,1$
• $1,3,3,1$
• $3,1,1,3$
• $1,3,1,3$
• $1,1,3,3$

That is, there are $6$ successful outcomes out of the $6^4$ possibilities.

If we wanted to approach the above systematically, we could note that any valid possibility can be created by selecting $2$ out of the $4$ dice to land on $1$, and setting the other dice to land on $3$. Thus, there would be $\binom{4}{2} = \frac{4!}{2!2!}=6$ successful outcomes.

Answer 2

Hint: You are correct that the total outcomes are $1296$. The number of successful outcomes is the number of ways of listing two $1$'s and two $3$'s in a row. You can do that by hand, or find the number of ways to choose which two of the four positions have $1$'s in them.