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HMMT 二月 2006 · 冲刺赛 · 第 42 题

HMMT February 2006 — Guts Round — Problem 42

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [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

  1. 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.