Probability of reaching a path Let's say we have a coordinate plane with only the top right quadrant (i.e. $x \geq 0$ and $y \geq 0)$. If we want to reach an arbitrary point $(m,n)$, then there are $\binom{m+n}{n} = \binom{m+n}{m}$ ways to reach that point. If for each step, I am equally likely to move up or right, what is the probability that it ever passes through the point $(n, m)$? 
I am having trouble computing this. With my current understanding, the shortest path consists of $m+n$ steps. If I have an equal chance to go up or right, then that's a $0.5$ probability. The new probability would then become $\binom{0.5m+0.5n}{n}$. However, this is incorrect. How do I approach this problem?
Edit: We can only move in the positive $x$ and $y$ direction. 
 A: You didn't explicitly say so, but I take it that you are only allowed to take steps in the positive $x$ and $y$ directions (otherwise this is a totally different problem).
First notice that if you ever reach $(m,n)$, it must be on your $(m+n)$th step. That happens if and only if you have taken $m$ total steps in the positive $x$ direction and $n$ total steps in the positive $y$ direction -- it doesn't matter in which order you take these steps, though. After your $(m+n)$th step, you must have taken the correct number in each direction.
If each direction has probability $0.5$ for each step, this is equivalent to asking for the probability of having exactly $m$ heads among $(m+n)$ tosses of a fair coin. This probability is
$$\frac{\binom{m+n}{m}}{2^{m+n}}.$$
A: If you're ever going to reach $(m,n)$ starting from $(0,0)$, you'll do so after exactly $m+n$ steps.  There are $2^{m+n}$ equally probable ways to choose your first $m+n$ steps.  As you noted, ${{m+n}\choose{m}}$ of these ways take you to the point $(m,n)$.  So the probability of reaching that point is
$$
P = \frac{1}{2^{m+n}}{{m+n}\choose{m}}=\frac{(m+n)!}{2^{m+n} m! n!}.
$$
