Logic question: Ant walking a cube There is a cube and an ant is performing a random walk on the edges where it can select any of the 3 adjoining vertices with equal probability. What is the expected number of steps it needs till it reaches the diagonally opposite vertex?
 A: Here is what I thought. It is a Markov Chain Problem.
Mark the start point as $E_3$, here we define $E_n$ which means that it takes $n$ sides to reach the final point.
we have such relationship:
take one side, it will definitely go to point as $E_2$:
$$E_3 = E_2 + 1$$
the probability is 2/3 to $E_1$ and 1/3 back to $E_3$:
$$E_2 = 2/3 (E_1 + 1) + 1/3 (E_3 + 1)$$
similar to $E_1$:
$$E_1 = 1/3 + 2/3 (1 + E_2)$$
solve these equations, you will get $E_3$ = 10.
A: Call the set containing only the starting vertex $A$.  You can move to any of three vertices next.  Call the set of them $B$.  For the next step, you can go back to $A$, or you can move on to any of three new vertices.  Call the set of those vertices $C$.  Finally, call the set containing the goal vertex $D$.
Call the expected number of steps from $A$ to $D$ $E(AD)$ etc.
Consider $E(BD)$.  We can write an equation for $E(BD)$ by considering what happens if you start at $B$ and take two steps.  You could go to $C$ and then to $D$.  The probability of this is $2/3$ for the first step and $1/3$ for the second, or $2/9$ over all.  
You could also go to $C$ and back, or to $A$ and back.  Either way, your new expected number of steps to $D$ is the same as it was before, because you're back where you started.  The probability of this is $7/9$ because probabilities add to one.
This gives
$E(BD) = 2/9(2) + 7/9(2 + E(BD))$
which means
$E(BD) = 9$
It takes one step to go from $A$ to $B$, so
$E(AD) = 10$
A: Hint: there are three kinds of vertices besides the target itself:
1) those one step from the target
2) those two steps from the target
3) the vertex opposite to the target
Let u_i be the expected number of steps, starting at a vertex of type i, until it reaches the target.  Considering the possibilities for the first step from a vertex of type i, write
three equations expressing u_i in terms of the other u_j.  Then solve.
A: I've been challenged whith this exercice but with a slight difference.
The difference is that if I get vertex 8 (beginning from 1) then the ant dies. And the question is "What probabilities I have to die?"
I know from the above answers that the number of expected steps from 1 to 8 is 10, then it means that is more difficult to die on vertex 8 than in vertex for example 7, due to the expected number of steps to reach vertex 7 is smaller than in vertex 8 and that means that the probability to die is higher on vertex 7 because of is easier to reach.
But...I don't know exactly how to express it in terms of PROBABILITY.
Anyone could open my mind? Thanks
