A kangaroo is jumping around on the corners of the square ABCD. The kangaroo starts at corner A. In each jump it moves to one of the two adjacent corners, with a probability of exactly 1/2 for each. The kangaroo stops when it has visited all corners at least once.

What's the probability that the kangaroo finally stops in corner C?

  • $\begingroup$ Just to clarify, corner C is diagonally opposite corner A? $\endgroup$ Commented Apr 16, 2016 at 15:23

2 Answers 2


The probability is


Explanation :

The first moves are symmetrical, so let's consider that the kangaroo moves to B. Let the probability that the walk ends in D be $p$.
As the kangaroo is in B, the walk ends in D if he goes to C. If he goes back to A, then he has to jump back to B, at which point the probability of ending in D is $p$ again. Therefore $p = \frac12 + \frac{p}{4}$, or $p = \frac23$. As we're looking for the probability of ending in C, it should be equal to $1-p$.

This is not a coincidence. Generalizing for a polygon with any number of vertices:

For every vertex, there is a point in which the kangaroo first reaches an adjacent corner. Then that vertex is the last if and only if the kangaroo proceeds to reach the vertex by going around the long way. The probability of that happening is the same for every vertex, so the probability that the walk ends in every vertex is the same (except for the sad starting vertex).


The probability that the kangaroo stops in corner B is equal to the probability that it hits C before it hits B. As the kangaroo performs a symmetric random walk on B-A-D-C, this probability is 1/3.

Then by symmetry, P(stops in C) = 1 - 2*P(stops in B) = 1/3.


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