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Suppose that there’s an election! Two candidates, Sherlock and Moriarty, are running for office. Suppose that Sherlock receives 8 votes and Moriarty receives 7 votes, and that these votes are being counted up one-by-one to create a running total. What is the probability that Sherlock is never behind in this running total? In general, if Sherlock got s votes and Moriarty got m votes, what is this probability? Can anyone help me explain this problem? There seems to be a bunch of different cases here for which I have not figured out what the best approach is.

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I give the answer for the special case of $s=m=n$. Considering an opening bracket a vote for Sherlock and a closing bracket a vote for Moriarty, you're actually looking for the number of valid strings with $n$ pairs of brackets, which equals the $n^{th}$ Catalan number, i.e., $C_n$. Thus, the probability you're looking after in this special case equals: $$\frac{C_n \times n! \times n!}{(2n)!}.$$

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