I have been the question:
Independent trials, each resulting in a success with probability $p$ or a failure with probability $q = 1 - p$, are performed. We are interested in computing the probability that a run of $n$ consecutive successes occurs before a run of $m$ consecutive failures.
This is the given solution:
Let $E$ be the event that a run of $n$ consecutive successes occurs before a run of $m$ consecutive failures. To obtain $P(E)$, we start by conditioning on the outcome of the first trial. That is, letting $H$ denote the event that the first trial results in success, we obtain: $$P(E) = pP(E|H) + qP(E|H^c)$$
Now, given that the first trial was successful, one way we can get a run of $n$ successes before a run of $m$ failures would be to have the next $n - 1$ trials all result in successes. So, let us condition whether or not that occurs. That is, letting $F$ be the even that trials $2$ through $n$ all are successes, we obtain $$P(E|H) = P(E|FH)P(F|H) + P(E|F^cH)P(F^c|H)$$
On the one hand, clearly, $P(E|FH) = 1$; on the other hand, if the event $F^cH$ occurs, then the first trial would result in success, but there would be a failure some time during the next $n - 1$ trials. However, when this failure occurs, it would wipe out all of the previous success, and the situation would be exactly as if we started out with a failure. Hence, $$P(E|F^cH) = P(E|H^c)$$
Now I understand why the failure wipes out the entire work, but I don't get why the last equation holds. In particular, $P(E|F^cH)$ means the probability of success given that the last $n - 1$ trials have at least one failure. I don't see how that can reduce to just given the information that the last trial was a failure. For example, if the sequence ends with $...FSSS$, we also have the information that the last $3$ are successes.
Can someone please explain this? I know there is a solution here but I don't get it.
Thanks
When you say "the probability of success given that the last $n-1$ trials have at least one failure" I think this is an example of the confusion, because this isn't quite what you're conditioning on.
It's just saying "there is a failure that occurs when you are going for the next $n-1$ flips, and therefore we never made it through this entire event - we failed and are now in this other event space where we are starting over.
We're not "going through with the rest of the $n-1$ flips" because once we hit that failure we are restarting, so we're not talking about a situation where "the last three flips could have been successes" for example.
In other words, you are thinking of it like "We flip a coin $n-1$ times and see if there was a failure in there somewhere" when really it is more like "Flipping a coin repeatedly and not making it to $n-1$ flips because we failed at some point."