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Problem 1300

AIME late

Combinatorics Difficulty 5.5 Prove it HMMT · United States

On a game show, Merble will be presented with a series of 2013 marbles, each of which is either red or blue on the outside. Each time he sees a marble, he can either keep it or pass, but cannot return to a previous marble; he receives 3 points for keeping a red marble, loses 2 points for keeping a blue marble, and gains 0 points for passing. All distributions of colors are equally likely and Merble can only see the color of his current marble. If his goal is to end with exactly one point and he plays optimally, what is the probability that he fails?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

The probability that Merble fails is $\frac{1}{2^{2012}}$ .

First, we note that if all the marbles are red or all are blue, then it is impossible for Merble to win; we claim that he can guarantee himself a win in every other case. In particular, his strategy should be to keep the first red and first blue marble that he encounters, and to ignore all the others. Consequently, the probability that he cannot win is $\frac{2}{2^{2013}} = \frac{1}{2^{2012}}$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.