I roll a 6-sided die an unknown number of times. The number 1 occurs no times, what is the chance that the number six occurs exactly 10 times?
I got as far as figuring out that I need the the probability that the die was rolled $x$ times given that a number occurred exactly $n$ times, however I can't figure out how to create an expression for this.
Edit:
To make this absolutely clear, I will define the problem like this. We roll a die $x$ times, where $x$ is a uniformly distributed random number between $0$ and $n$, and count the number of times that each number appears, $n_1$ is the number of ones, $n_2$ is the number of twos, etc. We want to calculate the probability $P(n_1 = 0 | n_6 = 10)$. We can use the binomial theorem to find a value for $P(n_1 = 0 | x=y)$ and $P(n_6 = 0 | x=y)$ for any $y$. By adding these together for all $x$, it should be possible to to find that number. I believe that the limit as $n\rightarrow \infty$ exists, but I might be wrong.
Guide:
If a particular number $j$ occured exactly $n_j=n$ times out of $x$ times, then for the remaining $x-n$ times, the other number occur equally likely. We already know that for those tosses, the outcome can't be $j$, but for it take other values that is not equal to $j$, it occur with probability $\frac15$ each.
Let $X \sim Unif \{ 1, 6\}$, let $j, k \in \{ 1, \ldots, 6\}, j \ne k, $then we have
$$P(X=j|X \ne k)=\frac{P(X=j, X\ne k)}{P(X \ne k)}=\frac{P(X=j)}{1-P(X=k)}=\frac{\frac16}{\frac56}=\frac15$$