HMMT 二月 2006 · 冲刺赛 · 第 42 题
HMMT February 2006 — Guts Round — Problem 42
题目详情
英文原题
- [18] Suppose hypothetically that a certain, very corrupt political entity in a universe holdsan election with two candidates, say A and B . A total of 5,825,043 votes are cast, but, in asudden rainstorm, all the ballots get soaked. Undaunted, the election officials decide to guesswhat the ballots say. Each ballot has a 51% chance of being deemed a vote for A , and a 49%
− Xchance of being deemed a vote for B . The probability that B will win is 10 . What is Xrounded to the nearest 10?
IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUNDth
解析
英文解析
- Suppose hypothetically that a certain, very corrupt political entity in a universe holdsan election with two candidates, say A and B . A total of 5,825,043 votes are cast,
but, in a sudden rainstorm, all the ballots get soaked. Undaunted, the election officialsdecide to guess what the ballots say. Each ballot has a 51% chance of being deemed avote for A , and a 49% chance of being deemed a vote for B . The probability that B − Xwill win is 10 . What is X rounded to the nearest 10?
Answer: 510
Solution: Let N = 2912521, so that the number of ballots cast is 2 N + 1. Let Pbe the probability that B wins, and let α = 51% and β = 49% and γ = β/α < 1. Wehave
( ) ( )
N N
∑ ∑
2 N + 1 2 N + 1 − X N − i N +1+ i N N +1 i
10 = P = α β = α β γ
N − i N − ii =0 i =0
(think of 2 i + 1 as representing B ’s margin of victory). Now
( ) ( )
2 N +1N
∑
2 2 N + 1 2 N + 1
i 2 N +1 < < γ < 2 ,
2 N + 1 N N − ii =0
So 15 − X = log P = N log α +( N +1) log β +(2 N +1) log 2 − = N log(2 α )+( N +1) log(2 β ) − ,
where 0 < < log(2 N + 1) < 7. With a calculator, we find that − X ≈ 25048 . 2 − 25554 . 2 − = − 506 . 0 − ,
so X ≈ 510.