Probability of a run of $n$ successes before a run of $m$ failures I'm doing a self-review of probability, working my way through Ross' Introduction to Probability.  I don't understand the final claim of the example below (from chapter 3, example 5c):

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.

Following the given solution, but paraphrasing, let $E$ be the event that a run of $n$ consecutive successes occurs before a run of $m$ consecutive failures; let $H$ be the event that the first trial succeeds, and condition on the first trial.
$$ ￼P(E) = p P(E|H) + q P(E|H^C) $$
Now, let $F$ be the event that trials 2 through $n$ all are successes, so:
$$
\begin{align*}
P(E|H) & = P(E|FH)P(F|H) + P(E|F^CH)P(F^C|H) \\\
       &= 1 \cdot p^{n-1} + P(E|F^CH) (1 - p^{n-1})
\end{align*}
$$
My question, now, is how to calculate $P(E|F^CH)$.  Ross writes

[...] if the event $F^CH$ occurs, then the first trial would result in a 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 successes, and the situation would be exactly as if we started out with a failure. Hence, $P(E|F^CH) = P(E|H^C)$.

Um... what?  I don't follow this, at all.  I've googled a bit and other books simply repeat this paragraph almost verbatim.  If anyone can explain more clearly, I would be grateful.
Note this question was also asked here: Run of $N$ successes before run of $k$ failures .  I read the link and the referenced book but it says virtually the same thing as Ross.
 A: If a failure occurs after a success, the situation is indeed the same as starting with
 a failure. You can see this as follows : 
Suppose, you a have sequence that gives the  desired event. 
Now, just add the trials before the failure to this sequence to get a sequence also 
 giving the desired event. 
Vice versa, you can delete the trials to get the shorter sequence. 
So there is a $1-1$-correspondence to the sequences, so the probability for the
 desired event must be the same.
A: Ross's phrasing is poor and is just not correct as stated. I'm teaching from this book right now and ran into this issue this evening. Unfortunately Ross' phrasing seems to have infected the rest of the sources on this problem.
Ross was originally published in 1976. The current edition (10th) says:

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.

Tapas K. Chandra and Dipak Chatterjee's A First Course in Probability from 2001 included the same example, very likely derived from Ross, on p.71:

Independent trials, each resulting in a success with probability $p$ or a failure with probability $q = 1-p$, are performed. Find the probability that a run of $n$ consecutive successes occurs before a run of $m$ consecutive failures.

This MSE thread uses very similar phrasing and the accepted answer points to Chandra--Chatterjee:

What is the probability that a run of $N$ consecutive successes will occur before a run of $k$ consecutive failures when each trial has a probability $p$ of success and $q=1−p$ of failure?

In any case, let's use Ross's formulation. The issues with his formulation:

*

*It's vague about how many trials are performed. If it means $n+m$ trials, the only option will be $S^n F^m$, for a probability of $p^n q^m$, which is not interesting and not what he computes. Fixing some number $N$ of trials leads to a more interesting calculation, but now it's not clear if the consecutive failures must be immediately preceded by consecutive successes, or if there can be a gap; in any case, Ross means neither of these alternatives and intends arbitrarily many trials.

*Ross says "run", but does not mean it. Earlier he used the term to refer to maximal sequences of successes or failures. Here he merely intends sequences of consecutive successes or failures.

*Most importantly, Ross means "first $m$ consecutive failures". If $p, q > 0$, then any infinite string from $\{S, F\}$ will contain infinitely many length-$m$ consecutive subsequences $F^m$ with probability $1$, and at least one of these will be preceded by $S^n$ with probability $1$, trivializing the problem.

I will propose the following phrasing to address these issues:

An infinite sequence of 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 we see $n$ consecutive successes before we first see $m$ consecutive failures.

(One could make this more finite by stopping the experiment after we first see $m$ consecutive failures, though there's still a probability $0$ case of getting an infinite number of trials.)
