Dice role: What is the probability to observe 2 times 1 and 2 times 5 with the outcome of a fifth die role being unknown? I tried to solve the following exercise: Given a dice with $P(X=2) = P(X=4) = P(X=5) = \frac{2}{15}$ and $P(X=1) = P(X=6) = P(X=3) = \frac{2}{10}$. What is the probability to observe 2 times 1 and 2 times 5 with the outcome of a fifth die role being unknown? Tip: Use marginalization! I tried to use the multinomial probability mass function and marginalize: So I got: \begin{equation*} \begin{split} &\sum^{6}_{i=1} \frac{5}{2! \cdot 2! \cdot x_i!} \cdot (\frac{2}{10})^2 \cdot (\frac{2}{15})^2 \cdot p(x_i)^1\\\ =& 0.0213 \end{split} \end{equation*} and as a result obtained that the probability is 2,13%. But the real solution should be 1,6%. Can you give me an advice what I did wrong?

You were mostly okay, except for the cases where you have three ones or three fives. (Also note the multinomial coefficient in the other cases .)

$$\begin{equation*} \begin{split} &\sum_{i\in\\{2,3,4,6\\}} \tfrac{5!}{2! \cdot 2! \cdot 1!}~ (\tfrac{2}{10})^2~(\tfrac{2}{15})^2 ~\mathsf P(X{=}i)^1 ~+~\tfrac{5!}{3!\cdot 2!}~(\tfrac{2}{10})^3~(\tfrac{2}{15})^2+\tfrac{5!}{3!\cdot 2!}~(\tfrac{2}{10})^2~(\tfrac{2}{15})^3 \\\\[1ex] =& \tfrac{5!}{2! \cdot 2! \cdot 1!} ~ (\tfrac{2}{10})^2 ~ (\tfrac{2}{15})^2 ~(\tfrac{2}{15}+\tfrac{2}{10}+\tfrac{2}{15}+\tfrac{2}{10}) ~+~\tfrac{5!}{3!\cdot 2!}~(\tfrac{2}{10})^2~(\tfrac{2}{15})^2~(\tfrac{2}{10}+\tfrac{2}{15})\\\\[1ex] =& 0.016\dot{\overline{592}} \end{split} \end{equation*}$$