I found this problem on brilliant.org. I get this kind of doubt a lot, in similar problems. (This is my 1st question here, please go easy on me)
An ant stands on point A of the lattice below.
Every minute, the ant selects an adjacent point uniformly at random and moves to it. For example, if the ant is on point B, then it moves to the points A, C, or G with probability 1/3 each.
What is the probability that the insect moves to point D in exactly 2 moves from the point shown?
If I find the product of probabilities (A->G->D), the answer I get is 1/3*1/6 = 1/18
But if I try to count the events, I get the probability to go from A to D in 2 steps as 1/12 (successful events/total events)
the 12 events being:
{A->B->C , A ->B->G, A->B->A , A ->F->E, A->F->A , A ->F->G, A->G->C , A ->G->A, A->G->B , A ->G->E, A->G->F , A ->G->D, }
The 1st answer is correct, but how? What are these 18 possible paths?
You have to fix your $4$th event from $A \rightarrow F \rightarrow D$ to $A \rightarrow F \rightarrow E$ obviously.
That said, your $12$ events do not have the same probability. The first $6$ on the list have probability $1/9$, the other $6$ have probability $1/18$, because they pass by $G$. You want the probability of the last event $A \rightarrow G \rightarrow D$, which can be expressed as:
$$\frac{1}{18}=\frac{\frac{1}{18}}{6\frac{1}{9}+6\frac{1}{18}} \neq \frac{1}{6+6} = \frac{1}{12}$$