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, 2019

Combinatorics Difficulty 6.0 AIME, harder Prove it United States

Problem:

Reimu and Sanae play a game using 4 fair coins. Initially both sides of each coin are white. Starting with Reimu, they take turns to color one of the white sides either red or green. After all sides are colored, the 4 coins are tossed. If there are more red sides showing up, then Reimu wins, and if there are more green sides showing up, then Sanae wins. However, if there is an equal number of red sides and green sides, then neither of them wins. Given that both of them play optimally to maximize the probability of winning, what is the probability that Reimu wins?

Solution

Solution:

Clearly Reimu will always color a side red and Sanae will always color a side green, because their situation is never worse off when a side of a coin changes to their own color. Since the number of red-only coins is always equal to the number of green-only coins, no matter how Reimu and Sanae color the coins, they will have an equal probability of winning by symmetry, so instead they will cooperate to make sure that the probability of a tie is minimized, which is when all 4 coins have different colors on both sides (which can easily be achieved by Reimu coloring one side of a new coin red and Sanae immediately coloring the opposite side green).

Therefore, the probability of Reimu winning is
$\frac{\binom{4}{3} + \binom{4}{4}}{2^4} = \frac{5}{16}.$

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