In a box you have 13 white marbles and 15 black marbles. You also have 28 black marbles outside the box.

You remove two marbles, randomly, from the box. If they are of different colors, you put the white one back in the box. If they are the same color you put a black marble in the box.

Continue this until only one marble remains in the box.

What color is the last marble? Please explain your answer.

My gut feeling initially was that the last marble was going to be black because of superior numbers, but I think that the last marble will be white on the basis that as far as I can see there can never be an equal number of each colour in the box, meaning that there will always be at least one white one being replaced.

I agree with Lex's answer. It's very counter-intuitive.

Can't put the reason mathematically, but it can be proved empirically.

Consider:

1. If the first 6 draws are all WW, we will end up with 21 blacks and 1 white
2. If the next 10 draws are all BB, we will end up with 11 blacks and 1 white
3. It the next 5 draws are all BB, we will end up with 6 blacks and 1 white
4. It the next 3 draws are all BB, we will end up with 3 blacks and 1 white
5. It the next draw is BB, we will end up with 2 blacks and 1 white
5. It the next draw is BB, we will end up with 1 black and 1 white

This makes the last marble white

Now, because the question was asked in the first place, there has to be a solution, and that solution must fit all cases. Therefore the answer is white.

Well done, gentlemen! White marbles could only be taken out in pairs. Since there was an odd number of white marbles the last one has to be white.

Good thinking by Rog: I had wondered whether the rule would hold good in all circumstances, but of course the question implies a single rule.