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!