Can conditioning on a uniform distribution be ignored?

Question

Assume you have a box with 3 blue balls and 5 red balls. Player $A$ choose 3 balls uniformly at random. Player $B$ choose 2 balls uniformly at random from the balls player $A$ has chosen. What is the probability that the player $B$ chose one ball from each color?

I've done it with the law of total probability $$ \frac{\binom{3}{1}\binom{5}{2}}{\binom{8}{3}}\cdot\frac{1\cdot2}{3}+\frac{\binom{3}{2}\binom{5}{1}}{\binom{8}{3}}\cdot\frac{2\cdot1}{3} $$ However, it seems that one can claim that you can ignore the first Player and compute the probability of choosing one red and one blue ball. $$ \frac{5\cdot 3}{\binom{8}{2}} $$ If someone can justify it formally, it would be awesome. Is it always true, when conditioning on uniform distribution?

Answer 1

Yes. One way to choose $k$ balls uniformly at random without replacement from $n$ is to repeatedly remove a random ball until only $k$ is left. If we stop part-way through this process, when $\ell>k$ balls are left, we have a uniformly random selection of $\ell$ balls. So choosing $k$ balls from $n$ uniformly at random is the same as first choosing $\ell$ balls uniformly at random (by performing the first $n-\ell$ steps) and then choosing $k$ balls from the remining $\ell$ (by performing the last $\ell-k$ steps).

A more "formal" way to show this is that the probability of selecting a given set $S$ of $k$ balls from a uniform set of $\ell$ balls is $$\sum_{\substack{T\supset S\\|T|=\ell}}\frac{1}{\binom n\ell}\times\frac{1}{\binom\ell k}=\binom{n-k}{\ell-k}\times\frac{1}{\binom n\ell}\times\frac{1}{\binom\ell k},$$ since the first expression is the sum over all possible choices of $\ell$ balls of the probability of first choosing those $\ell$ and then those $k$, and then there are $\binom{n-k}{\ell-k}$ ways to expand $k$ balls to $\ell$. Now we could show that the RHS simplifies to $1/\binom nk$, but in fact we don't have to since it is the same for every $S$, so we must have a uniform distribution.

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It's important here that we are selecting balls without replacement. It is not true otherwise, since the fact that B is choosing from a smaller pool makes there more likely to be repeated balls.