HMMT 十一月 2012 · THM 赛 · 第 8 题

HMMT November 2012 — THM Round — Problem 8

[ 4 ] In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeats rock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeats rock and scissors, as shown in the below diagram. As before, if two players choose the same move, then there is a draw. If three people each play a game of rock-paper-scissors-lizard-Spock at the same time by choosing one of the five moves at random, what is the probability that one player beats the other two? Scissors Spock Paper Lizard Rock

解析

[ 4 ] In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeats rock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeats rock and scissors, as shown in the below diagram. As before, if two players choose the same move, then there is a draw. If three people each play a game of rock-paper-scissors-lizard-Spock at the same time by choosing one of the five moves at random, what is the probability that one player beats the other two? Scissors Spock Paper Lizard Rock 12 Answer: Let the three players be A, B, C . Our answer will simply be the sum of the probability 25 that A beats both B and C , the probability that B beats both C and A , and the probability that C beats A and B , because these events are all mutually exclusive. By symmetry, these three probabilities are the same, so we only need to compute the probability that A beats both B and C . Given A ’s play, the probability that B ’s play loses to that of A is 2 / 5, and similarly for C . Thus, our answer is ( ) ( ) 2 2 12 3 · · = . 5 5 25