HMMT 二月 2019 · 冲刺赛 · 第 13 题

HMMT February 2019 — Guts Round — Problem 13

[ 7 ] Reimu has 2019 coins C , C , . . . , C , one of which is fake, though they look identical to each 0 1 2018 other (so each of them is equally likely to be fake). She has a machine that takes any two coins and picks one that is not fake. If both coins are not fake, the machine picks one uniformly at random. For each i = 1 , 2 , . . . , 1009, she puts C and C into the machine once, and machine picks C . What is the 0 i i probability that C is fake? 0

解析

[ 7 ] Reimu has 2019 coins C , C , . . . , C , one of which is fake, though they look identical to each 0 1 2018 other (so each of them is equally likely to be fake). She has a machine that takes any two coins and picks one that is not fake. If both coins are not fake, the machine picks one uniformly at random. For each i = 1 , 2 , . . . , 1009, she puts C and C into the machine once, and machine picks C . What is the 0 i i probability that C is fake? 0 Proposed by: Yuan Yao 1009 2 Answer: 1009 2 +1009 Let E denote the event that C is fake, and let F denote the event that the machine picks C over C for 0 i 0 P ( E ∩ F ) all i = 1 , 2 , ... 1009 . By the definition of conditional probability, P ( E | F ) = . Since E implies F, P ( F ) 1 P ( E ∩ F ) = P ( E ) = . Now we want to compute P ( F ) . If C is fake, F is guaranteed to happen. If C 0 i 2019 is fake for some 1 ≤ i ≤ 1009 , then F is impossible. Finally, if C is fake for some 1010 ≤ i ≤ 2018 , then i − 1009 1 F occurs with probability 2 , since there is a probability for each machine decision. Therefore, 2 1009 1009 1 1009 1009 − 1009 2 +1009 2 P ( F ) = · 1 + · 0 + · 2 = . Therefore, P ( E | F ) = . 1009 1009 2019 2019 2019 2019 · 2 2 +1009