I have a problem, where a die is rolled $n$ times and r.v. $X_k$ shows the number of rolls on which $k$ has showed $(k = 1, 2, ..., 6)$. I need to find the conditional distribution of $X_j | X_k = m: k ≠ j$ (the number of rolls on which $j$ has showed given that one of the other sides $k$ has showed $m$ times).
Is it correct to reduce the sample space to $5$ sides because we can ignore the $6$th one which we already got $m$ times? Using this approach, the answer will be $$P\{X_j = i\} = C_i^{n-m} * (1/5)^i * (4/5)^{n-m-i}$$where $i$ is the number of rolls.
Check your intuition by deriving the conditional probability. (It is correct!)
$X_j\sim B(n,1/6)~\forall j\in\{1,2,...,6\}$. $$\begin{align*}P(X_j=p|X_k=m)&=\frac{P(X_j=p\text{ and }X_k=m)}{P(X_k=m)}\\&=\frac{\binom nm\cdot\binom{n-m}p(1/6)^{m+p}(4/6)^{n-m-p}}{\binom nm(1/6)^m(5/6)^{n-m}}\\&=\binom{n-m}p\left(\frac14\right)^p\left(\frac45\right)^{n-m}\\&=\binom{n-m}p\left(\frac15\right)^p\left(\frac 45\right)^{n-m-p}\end{align*}$$