HMMT 二月 2026 · 冲刺赛 · 第 24 题
HMMT February 2026 — Guts Round — Problem 24
题目详情
- [12] Two mice and 100 pieces of cheese are uniformly and independently placed at random on the boundary of a circle. Each mouse walks to the piece of cheese closest to it, with ties broken independently at random. Compute the probability that the two mice walk to the same piece of cheese. ©2026 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2026, February 14, 2026 — GUTS ROUND Organization Team Team ID#
解析
- [12] Two mice and 100 pieces of cheese are uniformly and independently placed at random on the boundary of a circle. Each mouse walks to the piece of cheese closest to it, with ties broken indepen- dently at random. Compute the probability that the two mice walk to the same piece of cheese. Proposed by: Sebastian Attlan 3 Answer: 202 Solution 1: The key claim is the following. ©2026 HMMT Claim 1. We may assume without loss of generality that each mouse picks the nearest piece of cheese 1 clockwise of it with probability and the nearest piece of cheese counterclockwise of it with probability 2 1 . 2 Proof. Label the mice M and M . Let the nearest pieces of cheese clockwise and counterclockwise of 1 2 ′ M be C and C , and let the nearest pieces of cheese clockwise and counterclockwise of M be C and 1 1 2 2 1 ′ ′ ′ ′ ′ ′ C . Let M lie on arc C C such that arcs C M and M C have the same measure, and let M lie on 1 1 1 2 1 1 1 1 2 ′ ′ arc C C such that arcs C M and M C have the same measure. 2 2 2 2 2 2 ′ C 1 C 2 M 2 ′ M 1 ′ M 2 M ′ 1 C 2 C 1 ′ ′ Notice that the maps induced by sending M to M and/or sending M to M are bijections on the 1 2 1 2 ′ set of all possible placements of mice and cheese. Therefore, the mice positions ( M , M ) , ( M , M ) , 1 2 2 1 ′ ′ ′ ( M , M ) , and ( M , M ) are all equally likely. Since exactly one of M and M is closer to C than 1 2 1 1 2 1 1 ′ ′ ′ C , and exactly one of M and M is closer to C than C , the claim follows. 2 2 1 2 2 Notice that all orderings of the two mice and the 100 pieces of cheese along the circle are equally likely. Moreover, if there is more than one piece of cheese between the mice, then it is impossible for the mice to pick the same piece of cheese. Therefore, we have two cases. • Case 1: There are no pieces of cheese between the mice. There are 102 possible orderings where the mice are consecutive, so the probability of this case 102 2 is = . Now focus on the pieces of cheese surrounding the mice: C M M C . Using our 1 1 2 2 102 101 ( ) 2 1 1 1 1 key claim, there is a · = probability that both mice pick C , and (similarly) a probability 1 2 2 4 4 2 1 1 that both mice pick C . Thus, the overall contribution of this case is · = . 2 101 2 101 • Case 2: There is exactly one piece of cheese between the mice. There are 102 possible orderings where the mice are separated by one piece of cheese, so the 102 2 probability of this case is = . Now focus on the pieces of cheese surrounding the mice: 102 101 ( ) 2 1 1 1 C M C M C . Using our key claim, there is a · = probability that both mice pick C . 1 1 2 2 3 2 2 2 4 2 1 1 Thus, the overall contribution of this case is · = . 101 4 202 1 1 3 Our answer is + = . 101 202 202 Solution 2: Fix the position of the cheese that both mice come to, and then multiply by 100 later. We have two cases. Suppose that this cheese is distance x away from the first mouse and y away from the second mouse. Remembering to multiply by 2 later, we assume that the first mice is clockwise of the target cheese (as opposed to counterclockwise). Thus, we have to multiply by 200 later. Case 1: The two mice are on opposite sides of the cheese ©2026 HMMT Then all other 99 cheese must be in a segment of length (1 − 2 x − 2 y ) . In particular, the probability 1 99 that this happens (conditioned on fixed x and y ) is (1 − 2 x − 2 y ) if x + y ≤ and 0 otherwise. 2 Therefore, by integrating across all x and y , the answer is ∫ ∫ 1 / 2 1 / 2 − x 99 I = (1 − 2 x − 2 y ) dy dx. x =0 y =0 Evaluating this integral, we get that ∫ ∫ 1 / 2 t 99 I = (1 − 2 t ) dx dt (substitute t = x + y ) t =0 x =0 ∫ 1 / 2 99 = t (1 − 2 t ) dt t =0 ∫ 1 1 1 − u 99 = u · du (substitute u = 1 − 2 t ) 2 2 u =0 ∫ 1 1 99 100 = u − u du 4 0 ( ) ∣ 1 100 101 ∣ 1 u u ∣ = − ∣ 4 100 101 u =0 ( ) 1 1 1 1 = − = , 4 100 101 40400 1 so the probbability in this case is . 40400 Case 2: The two mice are on the same side of the cheese Without loss of generality, let x < y (and multiply by 2 later). Then the remaining 99 cheeses must be at least y away from the second mouse. Thus, all the cheeses must be in a segment of length 1 − 2 y . Therefore, the probability in this case is ∫ ∫ 1 / 2 y 99 (1 − 2 y ) dx dy. y =0 x =0 Evaluating the inner layer, this is equal to ∫ 1 / 2 99 y (1 − 2 y ) dy, y =0 1 which is an integral that also appears above, so it is . Remembering to multiply by 2 , the answer 40400 2 to this case is . 40400 Finally, adding two cases and remembering to multiply by 2 · 100 , the final answer is ( ) 1 2 3 2 · 100 · + = . 40400 40400 202