Problem 25 · 2022 AMC 8
Hard
Counting & Probability
careful-countingcasework
A cricket randomly hops between 4 leaves, on each turn hopping to one of the other 3 leaves with equal probability. After 4 hops, what is the probability that the cricket has returned to the leaf where it started?
Show answer
Answer: E — 7/27.
Show hints
Hint 1 of 2
Don't track which of the four leaves — the three non-start leaves all behave identically. Collapse the whole thing to one number: pn = the chance of being on the start leaf after n hops.
Still stuck? Show hint 2 →
Hint 2 of 2
Build a one-step rule. If you're on start, you must leave (so you got there only from a non-start leaf); from any non-start leaf, exactly 1 of the 3 hops returns to start. So pn+1 = (1 − pn) · 13.
Show solution
Approach: collapse 4 states to 1 by symmetry, then step a recursion
- Insight: the three leaves that aren't the start are interchangeable, so you never need to know which one the cricket is on — only whether it's home. Track a single number pn = P(on starting leaf after n hops).
- One-step rule: to be on start next hop, the cricket must currently be off-start (probability 1 − pn) and then pick the one returning hop out of 3. So pn+1 = (1 − pn) · 13.
- Iterate from p0 = 1: p1 = 0, p2 = 13, p3 = 29, p4 = (1 − 29) ÷ 3 = 79 ÷ 3 = 727.
- You'll see this again: when many states behave the same, merge them into one tracked quantity (here “home vs. away”) — a multi-state random walk becomes a single tidy recursion.
Another way — just count the paths (all are equally likely):
- Each hop is 1 of 3 equally likely choices, so the 4 hops give 34 = 81 equally likely paths. Count how many end back at the start.
- Such a return path must use the start as a “stepping point” an even number of times in the middle. Counting the closed length-4 walks gives 21 of them.
- Probability = 2181 = 727 — matching the recursion, a good cross-check.
Mark:
· log in to save