This question appears in several books written (or co-written) by Robert L. Scheaffer.
Two gamblers bet 1 dollar each on the successive tosses of a coin. Each has a bank of 6 dollars.
(a) What is the probability that they break even after six tosses?
(b) What is the probability that one player, say Jones, wins all the money on the tenth toss of the coin?
I have no problem with (a) and after much thought (and some help) I have gotten the "right" answer (which is in the book) for (b). But I have some doubts about the (b) answer. That answer is: $\frac{27}{1024}$, which is $$\frac{\binom{6}{2}\binom{2}{0} + \binom{6}{1}\binom{2}{1}}{1024}$$ The numerator makes sense to me because Jones must have two losses at least one of which must be during the first six tosses, and if only one is during the first six, the other must be during the seventh or eighth toss. But I have some doubts about the denominator, which counts all possible win/loss events for ten tosses. However, if (say) Jones wins on the sixth toss, the game ends, so why do we count the subsequent tosses, which I call "phantom" tosses?
Essentially the reason is that you must have equally likely events for the simple division of cases to work.
A sequence which ends after 6 turns such as Jones winning all these turns has probability $\frac{1}{2^6}$. If, however, we count all the possibilities for 10 turns (using your nicely phrased "phantom" turns) then we are in effect considering this to have probability $\frac{2^4}{2^{10}}$. There's no difference re. the maths.
P.S. It has just occurred to me that Fermat solved a gambling problem using just this way of considering events and the issue you have raised was questioned at the time.