TITLE: Find the probability that the thirteenth spade will appear before the thirteenth diamond?
QUESTION [8 upvotes]: The $52$ cards of an ordinary deck of cards are placed successively one after the other from left to right. Find the probability that the thirteenth spade will appear before the thirteenth diamond? 
This appears to be a hard problem since counting the favorable cases are hard to count. Clearly, the position of the $13$th diamond must be between $26$ and $N$. Now, I tried to count for specific values, but I find this difficult, perhaps there is a better way?  
Note:
I checked the answer and the favorable cases are supposed to be:
$$\sum_{k=26}^{52}(n-1)!(52-n)!\binom{26}{n-26}\cdot 13.$$
So I guess this ugly sum divided by $(52)!$ is equal to $1/2,$ right?
PS:
Question:

Answer:

REPLY [2 votes]: In case if you are curious how to simplify the summation directly to $\frac12$:
\begin{align}
\sum_{n=26}^{52}(n-1)!(52-n)!\binom{26}{n-26}\cdot 13
=&{13\cdot 26!}\cdot 25!\cdot  \sum_{n=26}^{52}\binom{n-1}{25}
\\\stackrel{\text{H.S.}}=&13\cdot 26!\cdot 25!\cdot \binom{52}{26}
\\=&52!\cdot \frac12
\end{align}
In $\stackrel{\text{H.S.}}=$ we use the Hockey Stick identty.