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Two teams play best of 7, and when one team wins $4$ games, the series $ends$.

How many different win or loss scenarios are possible in this case?

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  • $\begingroup$ I would start by writing out all the possibilities for a visual understanding. i.e. WWWW, WWWLW and so on $\endgroup$ – Brandon Dec 12 '16 at 3:51
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While you can always list it, apart from being lengthy, it is also quite error prone !

Even if the series ends, we can extend it notionally to $7$ matches, with $\xcancel{x}$ against unplayed ones.

Player $A$ must win $4$ matches to win the series, so s(he) has $\binom74$ ways to win, ditto for player $B$,

thus there can be a total of $2\times \binom74 = 70$ scenarios.

You could confirm the formula by listing for, say, best of $3$ or best of $5$ matches.

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If Team A wins $4$ games and Team B wins $k$ with $0 \le n \le 3$ then, as Team A must win the last game, there are $\displaystyle {k+3 \choose 3}$ possible patterns

So if Team A wins $4$ games then there are $\displaystyle \sum_{k=0}^3{k+3 \choose 3}$ possible patterns

So by symmetry, overall there are $\displaystyle 2 \sum_{k=0}^3{n+3 \choose 3}$ possible patterns

So overall there are $2(1+4+10+20)=70$ possible patterns

It is not a coincidence that this is $\displaystyle {2\times 4 \choose 4}$, and indeed for a "first to win $n$ games" or "best of $2n-1$ games" type match, there are $\displaystyle {2 n \choose n}$ possible patterns

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