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We toss a coin $n + m$ times. When we are done, there have been $n$ heads and $m$ tails, with $n > m$. All sequences of results are equally likely. For example, if $n = 2$ and $m = 1$, then $P(HHT) = P(HTH) = P(THH) = \frac{1}{3}$.

Now what is the probability that during the course of the experiment, the number of heads we've seen so far is always strictly larger than the number of tails we've seen so far?

My first thought was to define $A_i$ to be the event that we've always have more heads than tails up to the end of the $i$th toss, and calculate $P(\bigcap_{i=1}^{n+m}A_i)$, but then I got stuck.

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    So this is all conditional on the end result having $m$ heads and $n$ tails? i.e you want the probability that the substring of tosses 1 : k has more heads than tails given that the total string has n heads and m tails? – 2017-01-26
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    @spaceisdarkgreen Yes, that's right. – 2017-01-26
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    (my only hang up is that this is not really an 'experiment' one could do since we must look into the future, so I wasn't sure ) – 2017-01-26
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    This is known as the ballot theorem. See here: https://en.wikipedia.org/wiki/Bertrand%27s_ballot_theorem or Feller's probability book. – 2017-01-26
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    @carmichael561 Yeah that's exactly what I'm looking for. Thanks! – 2017-01-26

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