Tagged: probability

Ant on a Cube
Posted by Puzzle Prime on October 16, 2018 at 11:42 pmSuppose there is an ant in one corner of a cube. Every second, the ant moves from one corner to an arbitrary neighboring one along one of the three available edges. What is the expected time until the ant comes back to the starting corner?
Puzzle Prime replied 1 year, 9 months ago 3 Members · 4 Replies 
4 Replies

I think it takes 4 times to come back to the initial position, because I consider cube as a square.
I like the question, it is bit tricky, maybe my answer to this question is wrong, but I am curious to know the correct one.

I think the question needs a bit more clarification… Imagine the cube looks like the one on the picture below, and the ant starts from the red node. After one move, the ant will end up in one of the green nodes. The question is, after how many moves on average the ant will be back in the red node.
In order to solve the problem, you will need to write a few equations. You can denote with G, Y, and B to average times to get to the red node if the ant starts from a green, a yellow, or a blue node respectively. For example, you can see that:
Y = B/3 + 2G/3 + 1
You will need to write a few more equations and then solve them in order to find the values of G, Y, and B. Let me know if you need any tips:)

Let us name any two corners based on their position relative to each other.
Consider a corner, say A.
It has 3 adjacent corners, 3 corners that are diagonally opposite but lies on the same side as A, 1 corner which is exactly at the opposite corner.
Let the expectation to reach the same corner be S, to reach the adjacent corner be A, to reach diagonally opposite corner on the same side be D, exactly opposite corner be P.
So,
S=1/3*(A+1)+1/3*(A+1)+1/3*(A+1)
So, S=A+1
A=1/3*(1)+1/3*(D+1)+1/3*(D+1)
So, A=2D/3+1
D=1/3*(A+1)+1/3*(A+1)+1/3*(P+1)
So, D=2A/3+P/3+1
P=1/3*(D+1)+1/3*(D+1)+1/3*(D+1)
So, P=D+1
Solving these, gives A=7,S=8,D=9,P=10.