Isn't the probability that we'll be back at $A$ in 6 minutes just $\frac{5}{13}$ ?

Continuing the path that Mr. Rusczyk drew, for each of the five $1$ 's in the five-minute scenario, we will have two possibilities: $0$ or $2$ . This gives $5$ 0's and $5$ 2's in the extended diagram. However, for the three possibilities where we end up at $3$ in five moves, we have no choice but to go to $2$ . Hence there are $3$ more ways we can get to a 2, so tallying it up, we see the total number of paths to reach a 0 is $5$ , and the total number of paths to reach a 2 is $8$ .

Each of these paths is equally likely, since they are chosen at random. At the end of $6$ moves, we want to be back at A, so we want to be at Point 0 (zero away from where we started). Since there are $13$ total possibilities of places where we can end, and $5$ of them are successful, the probability is $\frac{5}{13}$ .

Is this right? Aren't the possible paths to take all equally likely? And there are $13$ possible paths, since there are $13$ possible final outcomes (we could either take one of the 5 paths to reach a 0, or one of the 8 paths to reach a 2). So since these paths are chosen at random, there are a total of $13$ possible outcomes and $5$ paths which result back in Point A after 6 moves, which means the probability is $\frac{5}{13}$ . Am I understanding this correctly, or is there something I am missing?

Also, is there a faster way to solve this problem than draw a flowchart like Mr. Rusczyk did above? That method may be really time-consuming, and it's more of a brute-force kind of method. Is there a faster or more creative way to solve this problem? If anybody has any suggestions, please let me know. Thank you!

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