HMMT 十一月 2023 · 团队赛 · 第 4 题
HMMT November 2023 — Team Round — Problem 4
题目详情
- [30] There are six empty slots corresponding to the digits of a six-digit number. Claire and William take turns rolling a standard six-sided die, with Claire going first. They alternate with each roll until they have each rolled three times. After a player rolls, they place the number from their die roll into a remaining empty slot of their choice. Claire wins if the resulting six-digit number is divisible by 6 , and William wins otherwise. If both players play optimally, compute the probability that Claire wins.
解析
- [30] There are six empty slots corresponding to the digits of a six-digit number. Claire and William take turns rolling a standard six-sided die, with Claire going first. They alternate with each roll until they have each rolled three times. After a player rolls, they place the number from their die roll into a remaining empty slot of their choice. Claire wins if the resulting six-digit number is divisible by 6, and William wins otherwise. If both players play optimally, compute the probability that Claire wins. Proposed by: Serena An 43 Answer: 192 Solution: A number being divisible by 6 is equivalent to the following two conditions: • the sum of the digits is divisible by 3 • the last digit is even 1 Regardless of Claire and William’s strategies, the first condition is satisfied with probability . So Claire 3 simply plays to maximize the chance of the last digit being even, while William plays to minimize this chance. In particular, clearly Claire’s strategy is to place an even digit in the last position if she ever rolls one (as long as the last slot is still empty), and to try to place odd digits anywhere else. William’s strategy is to place an odd digit in the last position if he ever rolls one (as long as the last slot is still empty), and to try to place even digits anywhere else. To compute the probability that last digit ends up even, we split the game into the following three cases: • If Claire rolls an even number before William rolls an odd number, then Claire immediately puts the even number in the last digit. • If William rolls an odd number before Claire rolls an even number, then William immediately puts the odd number in the last digit. • If William never rolls an odd number and Claire never rolls an even number, then since William goes last, he’s forced to place his even number in the last slot. The last digit ends up even in the first and third cases. The probability of the first case happening is 1 1 1
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- , depending on which turn Claire rolls her even number. The probability of the third case 3 5 2 2 2 1 is . So the probability the last digit is even is 6 2 1 1 1 1 43
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- = . 3 5 6 2 2 2 2 64 1 Finally we multiply by the chance that the sum of all the digits is divisible by 3 (this is independent 3 from the last-digit-even condition by e.g. Chinese Remainder Theorem), making our final answer 1 43 43 · = . 3 64 192
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