I have some doubts about my answer to this question, and I hope someone can help me figure it out.
Considering two consecutive, independent draws with replacement card by card, where you draw $n$ cards in the first draw and $m$ in the second, what is the probability to draw $x$ aces in the first draw knowing that $w$ aces were drawn in both draws?
I have tried and arrived at $\frac{4^x 48^{n-x}}{(w+1)4^w 48^{m+n-w}}$, following the logic that if $w$ aces are drawn, then there are $w+1$ cases: we could get $0$ aces on the first draw then $w$, $1$ ace then $w-1$, etc.
Could someone explain if this answer makes sense? If not, what is the right answer?