HMMT 十一月 2011 · 冲刺赛 · 第 27 题
HMMT November 2011 — Guts Round — Problem 27
题目详情
- [ 12 ] In-Young generates a string of B zeroes and ones using the following method: • First, she flips a fair coin. If it lands heads, her first digit will be a 0, and if it lands tails, her first digit will be a 1. • For each subsequent bit, she flips an unfair coin, which lands heads with probability A . If the coin lands heads, she writes down the number (zero or one) different from previous digit, while if the coin lands tails, she writes down the previous digit again. What is the expected value of the number of zeroes in her string? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TH 4 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND Round 10
解析
- [ 12 ] In-Young generates a string of B zeroes and ones using the following method: • First, she flips a fair coin. If it lands heads, her first digit will be a 0, and if it lands tails, her first digit will be a 1. • For each subsequent bit, she flips an unfair coin, which lands heads with probability A . If the coin lands heads, she writes down the number (zero or one) different from previous digit, while if the coin lands tails, she writes down the previous digit again. What is the expected value of the number of zeroes in her string? Answer: 2 Since each digit is dependent on the previous, and the first digit is random, we note that the probability that In Young obtains a particular string is the same probability as that she obtains the inverse string (i.e. that where the positions of the 0s and 1s are swapped). Consequently, we would expect that half of her digits are 0s, so that B C = . 2 If we solve our system of equations for A, B, C , we get that C = 2. Solution of the system of equations for Problems 25, 26, 27: Thus, we have the three equations B + C 2 − AC B A = , B = , C = ( B + 1)( C + 1)(2) 2 A 2 Plugging the last equation into the first two results in 3 B 4 − AB A = ⇒ B = ( B + 1)( B + 2)(2) 4 A Rearranging the second equation gives 4 4 4 AB = 4 − AB ⇒ AB = ⇒ A = 5 5 B Then, plugging this into the first equation gives 4 3 B = 5 B ( B + 1)( B + 2)(2) 2 2 ⇒ 15 B = 8 B + 24 B + 16 2 ⇒ 7 B − 24 B − 16 = 0 ⇒ (7 B + 4)( B − 4) = 0 1 Since we know that B > 0, we get that B = 4. Plugging this back in gives A = and C = 2. 5 Guts Round