2022 AMC 8 Problems/Problem 25
Contents
 1 Problem
 2 Solution 1 (Casework)
 3 Solution 2 (Casework)
 4 Solution 3 (Complement)
 5 Solution 4 (Recursion)
 6 Solution 5 (Dynamic Programming)
 7 Solution 6 (Generating Function)
 8 Solution 7 (Also Generating Functions)
 9 Remark
 10 Video Solution by MathX (First understand the problem!!!)
 11 Video Solution(🚀Under 2 min🚀 Easy logic with all paths colorcoded ✨)
 12 Video Solution
 13 Video Solution by OmegaLearn
 14 Video Solution
 15 Video Solution
 16 Video Solution
 17 Video Solution
 18 Video Solution
 19 See Also
Problem
A cricket randomly hops between leaves, on each turn hopping to one of the other leaves with equal probability. After hops what is the probability that the cricket has returned to the leaf where it started?
Solution 1 (Casework)
Let denote the leaf where the cricket starts and denote one of the other leaves. Note that:
 If the cricket is at then the probability that it hops to next is
 If the cricket is at then the probability that it hops to next is
 If the cricket is at then the probability that it hops to next is
We apply casework to the possible paths of the cricket:

The probability for this case is

The probability for this case is
Together, the probability that the cricket returns to after hops is
~MRENTHUSIASM
Solution 2 (Casework)
We can label the leaves as shown below:
Carefully counting cases, we see that there are ways for the cricket to return to leaf after four hops if its first hop was to leaf :
By symmetry, we know that there are ways if the cricket's first hop was to leaf , and there are ways if the cricket's first hop was to leaf . So, there are ways in total for the cricket to return to leaf after four hops.
Since there are possible ways altogether for the cricket to hop to any other leaf four times, the answer is .
~mahaler
Solution 3 (Complement)
There are always three possible leaves to jump to every time the cricket decides to jump, so there is a total number of routes. Let denote the leaf cricket starts at, and be the other leaves. If we want the cricket to move to leaf for its last jump, the cricket cannot jump to leaf for its third jump. Also, considering that the cricket starts at leaf , he cannot jump to leaf for its first jump. Note that there are paths if the cricket moves to leaf for its third jump. Therefore, we can conclude that the total number of possible paths for the cricket to return to leaf after four jumps is , so the answer is .
Solution 4 (Recursion)
Denote to be the probability that the cricket would return back to the first point after hops. Then, we get the recursive formula because if the leaf is not on the target leaf, then there is a probability that it will make it back.
With this formula and the fact that (After one hop, the cricket can never be back to the target leaf.), we have so our answer is .
~wamofan
Solution 5 (Dynamic Programming)
Let denote the leaf cricket starts at, and be the other leaves, similar to Solution 2.
Let be the probability the cricket lands on after hops, be the probability the cricket lands on after crawling hops, etc.
Note that and For the probability that the cricket land on each leaf after hops is the sum of the probability the cricket land on other leaves after hops. So, we have It follows that
We construct the following table: Therefore, the answer is .
Solution 6 (Generating Function)
Assign the leaves to and modulo and let be the starting leaf. We then use generating functions with relation to the change of leaves. For example, from to would be a change of and from to would be a change of This generating function is equal to It is clear that we want the coefficients in the form of where is a positive integer. One application of roots of unity filter gives us a successful case count of
Therefore, the answer is
~sigma
Solution 7 (Also Generating Functions)
Let the leaves be and on the coordinate plane, with the cricket starting at . Then we write a generating function. We denote a change in the xvalue of the cricket, and similarly for . Then our generating function is and we wish to compute the number of terms in which the exponents of both x and y are even. To do this, we first square to get . Note that every term squared will give even powers for x and y, so that gives us Then every combination of and will also give us even powers for x and y, so that yields more terms, for a total of Now in total there possible sequences, so gives us the answer of
~littlefox_amc
Remark
This problem is a reduced version of 1985 AIME Problem 12, changing steps into steps.
This problem is also similar to 2003 AIME II Problem 13.
Video Solution by MathX (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=n9aPrcW_qLqFC8IF&t=5261
~MathX
Video Solution(🚀Under 2 min🚀 Easy logic with all paths colorcoded ✨)
~Education, the Study of Everything
Video Solution
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Video Solution by OmegaLearn
https://youtu.be/kE15Sy0B2Pk?t=633
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=85A6av3oqRo
~Mathematical Dexterity
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=2588
~Interstigation
Video Solution
https://www.youtube.com/watch?v=H1zxrkq6DKg
~David
Video Solution
https://youtu.be/0orAAUaLIO0?t=609
~STEMbreezy
Video Solution
~savannahsolver
See Also
2022 AMC 8 (Problems • Answer Key • Resources)  
Preceded by Problem 24 
Followed by Last Problem  
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25  
All AJHSME/AMC 8 Problems and Solutions 
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.