What's the probability a die irolled 4 times you will get only two kinds of faces?

Question

Let $A$ be the event "only $2$ different faces in $4$ rolls of a die."

At each roll there's $6$ possibilities, so:
$$\omega = 6\cdot 6\cdot 6\cdot 6$$
Considering that it can be only two kinds of faces, for the first one there's $6$ possibilities, for the second one $5$ possibilities, for the third and forth there's only $2$ possibilities (they need to repeat one of the two first ones) so:
$$P(A) = \frac{6\cdot 5\cdot 2\cdot 2}{6^4}$$ Is that right?

Answer 1

Your method undercounts the number of ways of getting two kinds of faces. You've assumed the second die roll is different from the first, when there is no reason to believe this.

There are $\binom{6}{2}$ pairs of faces. You can either (1) roll $3$ or one face and one of the other, or (2) roll $2$ of each face. There are $8$ ways to get (1), and $\binom{4}{2}$ ways to get (2). So the total is $$\binom{6}{2}\left(8+\binom{4}{2}\right)=210$$

For instance, if the two faces are $1$ and $2$, the 14 possibilities are:

$$\begin{align} 1112&(*)\\1121&(*)\\1211\\2111\\2221&(*)\\2212&(*)\\2122\\1222\\1122&(*)\\1212\\2112\\1221\\2121\\2211&(*) \end{align}$$

You have failed to count the $6$ items marked with (*), giving your (wrong) total $$\binom{6}{2}\cdot 8 = 120 = 6\cdot 5\cdot 2 \cdot 2$$

Answer 2

I would say there are $6\choose2$ ways to pick the two faces and $\frac{4!}{2!2!}$ ways to arrange those two pairs of numbers.