rolling $4$ dice probability question I thought I understood the $4$ dice rolling probabilities until considering the sum of all possible results.
Probability of getting different results while rolling 4 standard dices:

*

*all $4$ dice have one value - $1/216$


*having $2$ different values:
$\dfrac{4C2 \cdot  6 \cdot 5}{6^4} = \dfrac{30}{216}$


*having $3$ different values:

$4C2 \cdot 6 \cdot 5 \cdot 4= \dfrac{120}{216}$


*having $4$ different values:

$\dfrac{6 \cdot 5 \cdot 4 \cdot 3}{6^4} = \dfrac{60}{216}$
Those are all the possible results so the probabilities should sum to $1$. But it's missing $5/216$??? Where is it? Which one of the above cases is wrong?
After some thinking - the second case above seems wrong. What is the probability of getting $2$ different values when rolling $4$ dice? $\dfrac{6 \cdot 5}{6^4} = \dfrac{5}{216}$ seems wrong.
 A: You made an error when considering the case with two different values.  There are two possibilities:

*

*One value appears three times and another value appears once.

*Two values each appear twice.

One value appears three times and another value appears once:  There are six ways to select the value that appears three times, $\binom{4}{3}$ ways to select the three dice on which that value appears, and five ways to select the value that appears on the other die, giving
$$\binom{6}{1}\binom{4}{3}\binom{5}{1}$$
such cases.
Two values each appear twice:  There are $\binom{6}{2}$ ways to select the two values which each appear twice and $\binom{4}{2}$ ways to select the two dice on which the larger of those values appear, giving
$$\binom{6}{2}\binom{4}{2}$$
such cases.
Hence, the probability that exactly two different values appear on the four dice is
$$\frac{\dbinom{6}{1}\dbinom{4}{3}\dbinom{5}{1} + \dbinom{6}{2}\dbinom{4}{2}}{6^4} = \frac{120 + 90}{1296} = \frac{210}{1296} = \frac{35}{216}$$
