# Can conditioning on a uniform distribution be ignored?

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?

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.
• The first analogy is true because to $k$ u.a.r. has probability $\binom{n}{k}^{-1}$ and removing $n-k$ balls u.a.r. has probability $\binom{n}{n-k}^{-1}$. As $\binom{n}{n-k}=\binom{n}{k}$ those two processes have the same distribution. Thanks Commented Aug 12 at 8:30