Basic Probability Question, Ant and lattice, Question from Brilliant.org

Question

I know this seems to be basic here, but I need your help,

So this is the question i found on brilliant.org

A bug stands on a random point of the lattice below. Each point is equally likely to be the starting point.

Every minute, the bug selects an adjacent point at random and moves to it. Each adjacent point is equally likely to be chosen. For example, if the bug is on point B, then each probability to move to the points A, C, or G is 1/3

What is the probability that the bug reaches point A in 2 moves or less? Each point is equally likely to be the bug's starting point. Also, assume starting at A will "reach" the point in 0 moves.

The answer was as below

i have some questions here, first why did it ignore the probability of A moving to B then C for example? why it only considered that it's there, or that would be considered only if we calculated all probabilities, i don't understand this point

Most importantly ( and sorry for writing a lot )

i retried the method with simple actions to make sure it's working so assuming it should be reached in zero moves only, so it's clear that if it started at A , then P is 1, dividing by 7, will be 1/7 which is True

but when i try with reaching in 1 or less move, by their method it should be ( 1 (for starting in A) + 1/3 B to A + 1/6 G to A + 1/3 F to A ) = 11/6 then i divide by 7 , then it should be 11/42

in the same time, probability of reaching A by 1 or less move, is by number of ways to reach A divided by all moves

A can be reached in 4 ways as i mentioned above, and total moves are ( 7 for starting and not moving ) + 6 *3 ( each point except of G has 3 moves) + 6 (G moves) = 4 / 31

I'm sure that i'm missing a lot, so anyone can help please

Answer 1

Why did [the author] ignore the probability of $A$ moving to $B$ then $C$ for example?

The author is considering the probability of favorable events. Moving from point $A$ to point $B$ and then to point $C$ is not a favorable event since you do not reach point $A$ in at most two moves.

Why is it only considered that [the bug] starts [at point $A$]?

Once the bug reaches point $A$, it has satisfied the requirements. If the bugs starts at point $A$, it is done. Therefore, we do not need to consider the sequences $A \to B \to A$, $A \to F \to A$, or $A \to G \to A$.

I retried the method with simple action actions to make sure it's working. So, assuming it should be reached in zero moves, so it's clear that if it started at $A$, then $P$ is $1$. Dividing by $7$, will be $1/7$, which is true.

Correct.

When I try with reaching in $1$ or less moves, by their method, it should be $$(1 (\text{for starting at $A$}) + \frac{1}{3} (\text{from $B$ to $A$}) + \frac{1}{6} (\text{from $G$ to $A$}) + \frac{1}{3} (\text{from $F$ to $A$}) = \frac{11}{6}$$ Dividing by $7$, then it should be $11/42$.

Also correct.

[At] the same time, [the] probability of reaching $A$ by $1$ or less moves is [the] number of ways of reaching $A$ divided by all moves.

False.

These moves are not equally likely. There are three possible moves from each vertex except $G$, from which six moves can be made. Your method would only work if each of the moves you counted were equally likely to occur.

Here is another way to express the result you obtained by your first method: \begin{align*} \Pr(\text{at most one move}) & = \Pr(A) + \Pr(B)\Pr(B \to A) + \Pr(F)\Pr(F \to A) + \Pr(G)\Pr(G \to A)\\ & = \frac{1}{7} + \frac{1}{7} \cdot \frac{1}{3} + \frac{1}{7} \cdot \frac{1}{3} + \frac{1}{7} \cdot \frac{1}{6}\\ & = \frac{1}{7}\left(1 + \frac{1}{3} + \frac{1}{3} + \frac{1}{6}\right)\\ & = \frac{11}{42} \end{align*}

Notice that one of your four successful cases occurs with probability $1/7$ (starting at $A$), two of them occur with probability $1/21$ (starting at $B$ or $F$ and then moving to $A$), and one occurs with probability $1/42$ (starting at $G$ and then moving to $A$).