Question

Suppose 4 dice are rolled, what is the probability that 1 number appears at least twice?

Answer 1

The probability is `13/18`

Explanation:

Let's number the dice with 1,2,3, and 4. We first count the number of ways a roll of the four dice does not have a number that appears at least twice. Whatever is on top of the first die, there are 5 ways to have a different number on die 2.

Then, assuming that we have one of those 5 outcomes, there are 4 ways to have a number on die 3 that is not the same as on dice 1 and 2. So, 20 ways for dice 1, 2, and 3 to have all different values.

Assuming we have one of these 20 outcomes, there are 3 ways for die 4 to have a different number than dice 1, 2, or 3. So, 60 ways altogether.

So, the probability of NOT having two numbers the same is `60/6^3 = 60/216`, as there are `6^3` different outcomes for rolling three six-sided dice.

The probability of the opposite, i.e. having at least two, equals 1 minus the above probability, so it is `1 - 60/216` = `(216-60)/216 = 156/216`=`13/18`.