HMMT 二月 2026 · 冲刺赛 · 第 36 题
HMMT February 2026 — Guts Round — Problem 36
题目详情
- [20] Sebastian is going for a walk in the coordinate plane. He starts at the origin facing in the positive x direction. Each minute, he takes a step forward, then randomly chooses one of the three axial directions other than the opposite of his current orientation. Sebastian stops walking once he returns to a point he has already visited. Estimate the expected number of steps Sebastian walks. 2 − 0 . 3( E − A ) Submit a real number E . If the correct answer is A , you will receive ⌊ 20 . 05 e ⌋ points. ©2026 HMMT
解析
- [20] Sebastian is going for a walk in the coordinate plane. He starts at the origin facing in the positive x direction. Each minute, he takes a step forward, then randomly chooses one of the three axial directions other than the opposite of his current orientation. Sebastian stops walking once he returns to a point he has already visited. Estimate the expected number of steps Sebastian walks. 2 − 0 . 3( E − A ) Submit a real number E . If the correct answer is A , you will receive ⌊ 20 . 05 e ⌋ points. Proposed by: Jacob Paltrowitz Answer: ≈ 14 . 0844 Solution: The simplest way for Sebastian to stop walking is for him to walk in a 1 × 1 square. At any time step after three steps, the probability that his last 3 steps cause him to walk in a square (and thus 2 necessarily self-intersect) is . So, we can crudely upper bound the length of the walk by assuming 27 that this is the only reason why Sebastian will ever stop. If we assume that the survival of the random walk is independent of previous steps, we can approximate the probability of lasting exactly n steps as ( ) ( ) n − 4 25 2 . 27 27 This gives a corresponding expected value of ( ) ( ) ∞ n − 4 ∑ 25 2 n = 16 . 5 27 27 n =4 which is good for 3 points. To improve this bound, we consider the next smallest way for Sebastian to self intersect. This would involve him walking along the edge of a 2 × 1 rectangle. At any given time n > 6 , this occurs with 6 2 probability = . So, we can update our probability estimate of lasting exactly n (for n > 6 ) to be 243 81 ( ) ( ) ( ) n − 4 n − 6 25 79 2 2
- . 27 81 27 81 Plugging back in, this gives an expected value of ( ) ( ) ( ) ∞ ( n − 4) ( n − 6) ∑ 2 25 2 25 79 2 2