Explanation needed for why my solution is incorrect. Here's the original problem:
A spider and a fly are sitting at opposite corners of a $5×5$ square grid, with the spider at the bottom left corner and the fly at the top right corner. The spider moves randomly right or up and the fly moves randomly left or down at the same rate. Find the probability that they meet at the same point on the grid.
The intended solution is $$\frac{\sum\limits_{k=0}^5{5\choose k}^2}{2^{10}}$$
As I spent more time thinking about this question, I thought of solving it another way. What I did is that I changed the sample space so that both the spider and the fly reach the end of the grid. This would yield a denominator of ${10\choose 5}^2$. For the numerator, instead of having ${5\choose k}^2$, it would be ${5 \choose k}^4$, since there is an additional ${5\choose k}$ ways for each of the spider and the fly to reach the end of the grid. In conclusion, the answer would become $$\frac{\sum\limits_{k=0}^5{5 \choose k}^4}{{10 \choose 5}^2}$$ 
However, this yields a different answer. Based on the exact values, it seems like I have over-counted. I still can't figure out where went wrong. Any help would be appreciated!
 A: Define these events:


*

*Let $E$ be the event that they meet.  

*Let $F$ be the event that the spider (moving Up or Right with equal prob each turn) reaches $(5,5)$.  

*Let $G$ be the event that the fly (moving Down or Left with equal prob each turn) reaches $(0,0)$.
Your first answer is $P(E)$.  Your second answer is $P(E \mid F \cap G)$.  Why would you expect them to be the same?  In fact, intuitively you would expect $P(E \mid F \cap G) > P(E)$, right?  Since events $F, G$ kind of "steer" them towards each other.
A: Presumably each takes exactly $5$ random steps along the edges of the squares. 
The answer to the original question is also $\frac{{10 \choose 5}}{2^{10}}$, the binomial probability that the number right moves by the spider plus the number of the left moves by the fly add up to $5$ out of $10$ (therefore also the up and down moves, and so they meet after five moves each)
A: Spider starts at $(0,0)$ and the fly at $(5,5)$. The first thing to notice is for the spider to get to the point $(a,b)$ in $n$ moves, the possibilities are ${n\choose a}$ because the spider has to go $a$ times to the right and $b$ times up, so it's a question of choosing which of those $n$ moves are to the right. Similarly for the fly. So if they both get to the same point, there's ${n\choose a}^2$ options.
Next, after $n$ moves, spider is located at $(k,n-k)$ and the fly at $(5-m,5-(n-m)$ where $0\leq k,m\leq n$. We want that point to be the same, so $k=5-m$ and $n-k=5-(n-m)$. This will give you $n=5$ (meeting happens exactly after $5$ moves) and $k+m=5$. So when $(k,m)=(0,5)$, they meet at point $(0,5)$. As we said, there's ${5\choose 0}^2$ ways for that happen. For the next point $(1,4)$, there's ${5\choose 1}^2$ ways, and so on.
