If 4 dice are rolled, what is the probability that three of the top faces show the same odd number and the other top face shows an even number?

Question

I am currently struggling to solve the question in the title:

If 4 dice are rolled, what is the probability that three of the top faces show the same odd number and the other top face shows an even number?

I thought the answer was $\frac{3*3}{6^4}$, instead it is $\frac{4*3*3}{6^4}$.

Can someone explain the reason for that?

Answer 1

There's no problem in your logic, except the fact that you are assuming that the dice are ordered, i.e. there is a first, second, third, and last dice, and in the end you are saying that the even number must come on the last dice. Indeed, the even number can come on any of the dice, and there are four of them, so the actual answer is your answer multiplied by four, which is the case.

Answer 2

It is: $$4(P(1\cap 1\cap 1\cap E)+P(3\cap 3\cap 3\cap E)+P(5\cap 5\cap 5\cap E))=$$ $$4\left(\frac{1}{6^3}\cdot \frac{1}{2}+\frac{1}{6^3}\cdot \frac{1}{2}+\frac{1}{6^3}\cdot \frac{1}{2}\right)=$$ $$4\cdot \frac{3\cdot 3}{6^4}$$ Note: $P(1\cap 1\cap 1\cap E)=P(1\cap 1\cap E\cap 1)=P(1\cap E\cap 1\cap 1)=P(E\cap 1\cap 1\cap 1).$