What's the probability of going to (6, 4) without ever visiting (3, 4)?

Question

The ant starts at (0, 0). With each move it goes either up or right by one unit distance. After 10 moves, what is the probability that the ant settles down at point (6, 4) without ever visiting the point (3, 4)?

My solution:

Since the ant has to make 10 moves moving either up or right, the amount of possible routes is 2^10 = 1024.

There are 10!/(6!4!) = 210 possible ways to reach (6, 4) in 10 moves.

Hence, the probability of going to (6, 4) is 210/1024 ≈ 0.2051.

Since the ant is only able to move either up or right, it has to make 7 moves (3 right, 4 up) to get to (3, 4).

There are 2^7 = 128 possible routes to take in 7 moves.

And so, there are 7!/(3!4!) = 35 possible ways to reach (3, 4) in 7 moves.

So, the probability of going to (3, 4) in 7 moves should be 35/128 ≈ 0.2734.

The probability of not going to (3, 4) is 1 - 0.2734 = 0.7266.

With that, the probability of going to (6, 4) are 0.2051 and the probability of not going to (3, 4) is 0.7266.

Since both of these events should occur to satisfy the condition, my answer would be 0.2051 * 0.7266 ≈ 0.149.

However, I suppose my solution is wrong because if the ant goes to (7, 0) or (2, 5) in the first 7 moves, there's no way for it to reach (6, 4).

Is there a way to fix my solution or have I gone absolutely the wrong way?

Answer 1

You were very close before you began on the probabilities! You know the number of ways the ant can get to $(6, 4)$, and you know the number of ways the ant can get to $(3, 4)$. From there you are only two small steps away from calculating the number of ways to get to $(6, 4)$ through $(3, 4)$.

To do that you first need to calculate the number of ways to get from $(3, 4)$ to $(6, 4)$. And once you have that, you know the number of ways to get to $(3, 4)$ and the number of ways to get from $(3, 4)$ to $(6, 4)$, so the number of ways to get to $(6, 4)$ through $(3, 4)$ is the product of these.

Once you have the number of ways to get to $(6, 4)$ through $(3, 4)$, and the number of ways to get to $(6, 4)$ in total, it is easy to calculate the number of ways to get to $(6, 4)$ without going through $(3, 4)$. And once you have that, you can easily find the probability of the ant actually choosing such a route.

Also, I feel I have to point out that you are vastly overcounting the number of places the ant can end up. After 10 steps, there are only 11 possible places it can be. However, there are indeed 1024 different routes the ant can take to get to one of those 11 points.

Answer 2

If in general for nonnegative integers $r,u$ the event of visiting $\left(r,u\right)$ is denoted as $V_{r,u}$ then to be found is: $$P\left(V_{6,4}\cap V_{3,4}^{\complement}\right)=P\left(V_{6,4}\right)-P\left(V_{6,4}\cap V_{3,4}\right)=P\left(V_{6,4}\right)-P\left(V_{3,4}\right)P\left(V_{6,4}\mid V_{3,4}\right)=$$$$P\left(V_{6,4}\right)-P\left(V_{3,4}\right)P\left(V_{3,0}\right)$$

So if we can find an expression for $P\left(V_{r,u}\right)$ then we are ready.

You already found correctly that $P(V_{6,4})=\binom{10}6\times2^{-10}$ so I guess you know how to do that.