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My approach to this is by counting the number of arrangements of needed wins in a sequence of games. If it's a best of 7 match, an example of an arrangement where a player won is WLWWLWL. So by arranging the 4 W's and 3 L's, I can get the number of ways to win a best of 7 match.

Then, in general, let $A$ be the set of ways to win a best of $n$ match, then $|A|=$${n}\choose{\lfloor n/2 \rfloor + 1}$ .

Is this correct?

marshblocker
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  • Not correct, because once someone has won $4$ times, no more matches are played. – DreiCleaner Jul 09 '20 at 15:40
  • @DreiCleaner yes that's true, but does that discard matches from set $A$? I mean WWWW is still equal to WWWWLLL which is part of $A$ and there could be no other variation to that since ordering of similar letters does not matter. – marshblocker Jul 09 '20 at 15:48

1 Answers1

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You are right, and there are $\binom{7}{4}=35$ ways to arrange $4$ Ws and $3$ Ls.

I would add that this only counts the ways for the series to end in favor of one of the teams. Meaning, if the teams are the Wombats and the Lemurs, then $4$ Ws and $3$ Ls ensures victory for the Wombats. There are similarly $35$ other ways to arrange $4$ Ls and $3$ Ws (if you want to count all possible series where either team wins).

DreiCleaner
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