How does this basic probability question solution make sense? I am just having some trouble understanding why it is valid to answer the following question in a certain way.
Suppose we have iid Bernoulli trials, with $p=0.7$,
We do trials until we have 4 success or 4 fail, if we get 4 success first, we win, else we lose
what is the probability that we win.
Now what I dont understand is my, one a work page I had seen, the professor had wrote that we can solve it by noting that it is binomial with 7 trials, and we win if we have 4 or more success. But I dont understand why this is valid because how can we have more then 4 successes, ie , that game is over at that point.
I would think it would be 
=
$$P[S=4, F=0]+P[S=4, F=1] + P[S=4, F=2]+ P[S=4, F=3]$$
But mostly I am wondering about that solution and how it is valid to say they are equivalent even though it does not make sense to have more than 4 successes or failures
Thanks
 A: The key is to observe the following: after $7$ trials,  you are guaranteed that the game has finished (indeed, there is not way to have seen both fewer than $4$ success and fewer than $4$ failures in $7$ trials).
Moreover, in $7$ trials there cannot be both at least $4$ successes and $4$ failures.
So once you have that, the game becomes equivalent to doing $7$ trials, and looking at which of the successes or the failures happened at least $4$ times. This tells you the outcome of the game.
To generalize: try to replace $4$ by another number, say $n$, in the definition of the game; and see what would be the corresponding Binomial to look at.
A: 
Now what I dont understand is my, one a work page I had seen, the professor had wrote that we can solve it by noting that it is binomial with 7 trials, and we win if we have 4 or more success. But I dont understand why this is valid because how can we have more then 4 successes, ie , that game is over at that point.

A winner will be declared when one player has four successes before the other does.   The players could continue playing a whole seven games without affecting their probability for winning.   However, imagining they do so may make it easier to calculate.
In seven games exactly one player will have four or more successes while the other must have three or less.   That means that one player will have reached four successes before the other does so.   The events "have at least four successes in seven games" and "have four successes before the other player" are the same event and hence will have the same probability.
$$\binom 74 p^4(1-p)^3+\binom 75 p^5(1-p)^2+\binom 76 p^6(1-p)+p^7$$

I would think it would be 
  $=
P[S=4,F=0]+P[S=4,F=1]+P[S=4,F=2]+P[S=4,F=3]$

Sure, but with the additional criteria that the last game played must be the fourth success.   That is:
$$\binom 33 p^4+\binom43 p^4(1-p)+\binom 53p^4(1-p)^2 +\binom63 p^4(1-p)^3$$
These two expressions are actually equal.   Both evaluate to: $~p^4 (35-84p+70p^2-20 p^3)~$
A: Your professor not taking 7 trials. He is taking all cases with 0, 1, 2, 3 failures in between successful trials.
