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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.

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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.

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  • $\begingroup$ hmm... I'm not sure I see it, but I maybe understand your argument. Is this somehow so obvious that it doesn't require explanation, or is it not explained because this is an intro textbook? Thanks for your help, Peter. $\endgroup$ – DavidT Aug 31 '14 at 5:46

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