Conditional Probability – Red and black marbles

Question

I saw this question on one of the app.

A wise man presents you with an urn containing an equal number of red and black marbles.

If you reach in and draw two marbles (uniformly and independently) at random from the urn, the probability that they're of matching color is necessarily less than ½. Here's the tricky part: if the colors of those first two marbles do match, and you reach in to grab two more, the probability of those next two being of matching color is exactly ½.

How can that be? What is the total number of balls in the urn — red plus black — before you take any out?

I am unable to work it out with equal number of initial red and black marbles.

Answer 1

Suppose there are $n$ red marbles and $n$ black ones. We choose a marble from the urn. There are $n-1$ matching marbles left among the $2n-1$ marbles in the urn, so the probability that the first two match is $${n-1\over2n-1}<\frac12$$

Now suppose the first two match. We may assume that there are $n-2$ black marbles and $n$ red marbles left. There are ${n-2\choose2}$ ways to choose two matching black marbles, ${n\choose2}$ ways to choose two matching red marbles, and ${2n-2\choose2}$ ways to choose two marbles. Can you finish it from here?