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Combinatorics Difficulty 6.2 National Olympiad Prove it United States

Problem:

Ash and Gary independently come up with their own lineups of 15 fire, grass, and water monsters. Then, the first monster of both lineups will fight, with fire beating grass, grass beating water, and water beating fire. The defeated monster is then substituted with the next one from their team's lineup; if there is a draw, both monsters get defeated.

Gary completes his lineup randomly, with each monster being equally likely to be any of the three types. Without seeing Gary's lineup, Ash chooses a lineup that maximizes the probability $p$ that his monsters are the last ones standing. Compute $p$ .

Solution

Solution:

First, we show Ash cannot do better. Notice there is a $\frac{2^{15}}{3^{15}}$ chance that Gary's $i$ -th monster ties or defeats Ash's $i$ -th monster for each $i$ . If this is the case, Ash cannot win, as Ash's $i$ -th monster will always be defeated by Gary's $i$ -th monster, if not sooner. Thus, Ash wins with probability at most $1-\frac{2^{15}}{3^{15}}$ . It remains to show this is achievable.

Ash uses the lineup fire-grass-water repeated 5 times. Then, none of Gary's monsters can defeat more than one monster in Ash's lineup, so Ash will win unless Gary manages to take down exactly one monster with each of his. In particular, this means the $i$ -th monster Gary has must tie or defeat Ash's $i$ -th monster, which occurs with $\frac{2}{3}$ chance with each $i$ . Thus this construction achieves the answer of $1-\frac{2^{15}}{3^{15}}$ .

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