HMMT 十一月 2018 · THM 赛 · 第 8 题
HMMT November 2018 — THM Round — Problem 8
题目详情
- Crisp All, a basketball player, is dropping dimes and nickels on a number line. Crisp drops a dime on every positive multiple of 10, and a nickel on every multiple of 5 that is not a multiple of 10. Crisp 2 then starts at 0. Every second, he has a chance of jumping from his current location x to x + 3, and 3 1 a chance of jumping from his current location x to x + 7. When Crisp jumps on either a dime or a 3 nickel, he stops jumping. What is the probability that Crisp stops on a dime ?
解析
- Crisp All, a basketball player, is dropping dimes and nickels on a number line. Crisp drops a dime on every positive multiple of 10, and a nickel on every multiple of 5 that is not a multiple of 10. Crisp 2 then starts at 0. Every second, he has a chance of jumping from his current location x to x + 3, and 3 1 a chance of jumping from his current location x to x + 7. When Crisp jumps on either a dime or a 3 nickel, he stops jumping. What is the probability that Crisp stops on a dime ? Proposed by: James Lin 20 Answer: 31 Let “a 3” mean a move in which Crisp moves from x to x + 3, and “a 7” mean a move in which Crisp moves from x to x + 7. Note that Crisp stops precisely the first time his number of 3’s and number of 7’s differs by a multiple of 5, and that he’ll stop on a dime if they differ by 0, and stop on a nickel if they differ by 5. This fact will be used without justification. We split into two cases: (a) Crisp begins with a 3. Rather than consider the integer Crisp is on, we’ll count the difference, n , between his number of 3’s and his number of 7’s. Each 3 increases n by 1, and each 7 decreases n by 1. Currently, n is 1. The probability of stopping on a dime, then, is the probability n reaches 0 before n reaches 5, where n starts at 1. Let a be the probability n reaches 0 first, given a current i position of i , for i = 1 , 2 , 3 , 4. We desire a . We have the system of linear equations 1 2 1 a = a + · 1 1 2 3 3 2 1 a = a + a 2 3 1 3 3 2 1 a = a + a 3 4 2 3 3 2 1 a = · 0 + a 4 3 3 3 15 From which we determine that a = . 1 31 (b) Crisp begins with a 7. Now, let m be the difference between his number of 7’s and his number of 3’s. Let b denote his probability of stopping on a dime, given his current position of m = i . We i desire b . We have the system of linear equations 1 1 2 b = b + · 1 1 2 3 3 1 2 b = b + b 2 3 1 3 3 1 2 b = b + b 3 4 2 3 3 1 2 b = · 0 + b 4 3 3 3 30 From which we determine that b = . 1 31 2 1 2 15 1 30 20 We conclude that the answer is a + b = · + · = . 1 1 3 3 3 31 3 31 31