PUMaC 2019 · 团队赛 · 第 2 题

PUMaC 2019 — Team Round — Problem 2

In a standard game of Rock–Paper–Scissors, two players repeatedly choose between rock, paper, and scissors, until they choose different options. Rock beats scissors, scissors beats paper, and paper beats rock. Nathan knows that on each turn, Richard randomly chooses paper with probability 33%, scissors with probability 44%, and rock with probability 23%. If Nathan plays optimally against Richard, the probability that Nathan wins is expressible as a/b where a and b are coprime positive integers. Find a + b .

解析

In a standard game of Rock–Paper–Scissors, two players repeatedly choose between rock, paper, and scissors, until they choose different options. Rock beats scissors, scissors beats paper, and paper beats rock. Nathan knows that on each turn, Richard randomly chooses paper with probability 33%, scissors with probability 44%, and rock with probability 23%. If Nathan plays optimally against Richard, the probability that Nathan wins is expressible as a/b where a and b are coprime positive integers. Find a + b . Proposed by: Nathan Bergman Answer: 89 4 33 33 Playing paper obviously sucks. Rock wins of the time. Scissors wins of the time. is 7 56 56 best so play scissors.