HMMT 二月 2021 · 冲刺赛 · 第 35 题
HMMT February 2021 — Guts Round — Problem 35
题目详情
- [20] Geoff walks on the number line for 40 minutes, starting at the point 0. On the n th minute, he flips 1 a fair coin. If it comes up heads he walks in the positive direction and if it comes up tails he walks n 1 in the negative direction. Let p be the probability that he never leaves the interval [ − 2 , 2]. Estimate n 4 N = b 10 p c . ( ⌊ ⌋) ( ) 1 / 3 | E − N | An estimate of E will receive max 0 , 20 − 20 points. 160 2
解析
- [20] Geoff walks on the number line for 40 minutes, starting at the point 0. On the n th minute, he 1 flips a fair coin. If it comes up heads he walks in the positive direction and if it comes up tails he n 1 walks in the negative direction. Let p be the probability that he never leaves the interval [ − 2 , 2]. n 4 Estimate N = b 10 p c . ( ⌊ ⌋) ( ) 1 / 3 | E − N | An estimate of E will receive max 0 , 20 − 20 points. 160 Proposed by: Michael Ren Answer: 8101 Solution: To estimate it by hand, we’ll do casework on the most likely ways that Geoff will go past +2, and double the answer. If Geoff starts with one of the three sequences below, he will be past 2 or very close to 2: (+ , + , + , +) , (+ , + , + , − , + , +) , (+ , + , − , + , + , +) . 1 2 3 3 The probability of one of these happening is + = . This gives an estimate of p = , which 16 64 32 16 gives E = 8125 and earns 9 points. We can justify throwing out other starting sequences as follows. For example, suppose we start with 11 (+ , + , − , − ). At this point we are at . The variance of the rest of our random walk is 12 40 2 ∑ 1 π 1 1 1 < − 1 − − − < 0 . 25 . 2 n 6 4 9 16 n =5 13 So, the standard deviation of the rest of our walk is bounded by 0 . 5, which is much less than the 12 Geoff needs to go to get to +2. One can use similar estimates for other sequences to justify them as negligible. Furthermore, we can even use similar estimates to justify that if Geoff get close enough to +2, he is very likely to escape the interval [ − 2 , 2]. The exact value for p is 0 . 8101502670 ... , giving N = 8101. 2