HMMT 十一月 2020 · 团队赛 · 第 4 题

HMMT November 2020 — Team Round — Problem 4

[35] Marisa has two identical cubical dice labeled with the numbers { 1 , 2 , 3 , 4 , 5 , 6 } . However, the two dice are not fair, meaning that they can land on each face with different probability. Marisa rolls the two dice and calculates their sum. Given that the sum is 2 with probability 0 . 04, and 12 with probability 0 . 01, the maximum possible probability of the sum being 7 is p . Compute b 100 p c .

解析

[35] Marisa has two identical cubical dice labeled with the numbers { 1 , 2 , 3 , 4 , 5 , 6 } . However, the two dice are not fair, meaning that they can land on each face with different probability. Marisa rolls the two dice and calculates their sum. Given that the sum is 2 with probability 0 . 04, and 12 with probability 0 . 01, the maximum possible probability of the sum being 7 is p . Compute b 100 p c . Proposed by: Shengtong Zhang Answer: 28 Solution: Let p be the probability that the dice lands on the number i . The problem gives that i 2 2 p = 0 . 04 , p = 0 . 01, so we have 1 6 p = 0 . 2 , p = 0 . 1 , p + p + p + p = 0 . 7 . 1 6 2 3 4 5 We are asked to maximize 2( p p + p p + p p ) = 2(0 . 02 + p p + p p ) . 1 6 2 5 3 4 2 5 3 4 2 2 y x Let x = p + p and y = p + p . Then by AM-GM, p p ≤ , p p ≤ . Also, 2 5 3 4 2 5 3 4 4 4 2 2 2 2 2 2 x + y x + 2 xy + y ( x + y ) 0 . 7 ≤ = = = 0 . 1225 . 4 4 4 4 Hence, 2( p p + p p + p p ) ≤ 2(0 . 02 + 0 . 1225) = 0 . 285 , 1 6 2 5 3 4 where equality holds if p = p = 0 . 35 , p = p = 0. 2 5 3 4 Thus, we conclude that p = 0 . 285 and b 100 p c = 28.