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I am working through my college text book (mathmatical statistics freund/walpole) trying to refresh my stat skills. Its been a couple years... I would sure appreciate any pointers on an exercise question that has me stumped...

In a two -team playoff in some sport, the winner is the first team to win m games.

(a) Counting separately the number of play offs requiring m,m +1,...., 2m-1 games, show that the toatl number of different outcomes (sequences of wins and losses) is $$ 2\left[ {{m-1}\choose {m-1}} + {{m}\choose {m-1}} + ... + {{2m-1}\choose{m-1}}\right] $$ (end of question)

I drew this out as a path problem, wins along the X and wins losses along the Y. By inspection I can see that one team can win a best of 5 (m=3) in the following ways Team A has to win 3 games, but they can only loose 2 in the process...

$$ \left[ {{3}\choose {3}} + {{4}\choose {3}} + {{5}\choose{3}}\right] $$

The other team might win the series as well, so I would need to multiply the above by 2.

I can generalize this to

$$ 2\left[ {{m}\choose {m}} + {{m+1}\choose {m}} + ... + {{2m-1}\choose{m}}\right] $$

I have tried various substitutions but I can not seem to get to the form asked for. I notice they are using (m-1) in the bottom. For my working scenario, this would equate to

$$ 2\left[ {{2}\choose {2}} + {{3}\choose {2}} + {{4}\choose {2}} + {{5}\choose{2}}\right] $$

This only makes sense to me in the case of a full series, or a sweep since $$ {{5}\choose{3}} = {{5}\choose{2}} and {{3}\choose{3}} = {{2}\choose{2}} $$

Are they counting the ways for Team B to loose?

Any tips or pointers? I am sure I am missing something fundamental. Thanks in advance.

yiyi
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greg
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  • The given answer seems strange to me - it has (m+1) terms though it says it is considering separately m possibilities, and it disagrees (at least for m=3) with your answer which is correct as far as I can see. – aPaulT Dec 09 '13 at 15:45
  • What I said here is incorrect: as the answer below says, we've both counted, for example, AAABB as a five-game series when in fact it would end after three games and we've counted it there as AAA. – aPaulT Dec 09 '13 at 16:33

2 Answers2

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Unless I'm missing something, I don't think the proposed answer that the question provides is correct.

Let's enumerate the cases. Let $A$ and $B$ be wins for each team.

Then, for $m=3$:

Series that end in three games: $AAA$, $BBB$.

Series that end in four games: $AABA, ABAA, BAAA, BBAB, BABB, ABBB$.

Series that end in five games: $AABBA, AABBB, ABABA, ABABB, ABBAA, ABBAB, BBAAB, BBAAA, BABAB, BABAA, BAABB, BAABA$.

This is consistent with the following formula:

$$ 2\left[ {{m-1}\choose {m-1}} + {{m}\choose {m-1}} + ... + {{2m-2}\choose{m-1}}\right]. $$

If we look at your best-of-five case, the series which end in three games had two wins by one of the teams. This is different than saying that the series will end in three games if one of the teams wins the first two.

In other words, when counting the series that end in less than $2m-1$ games, one only needs to consider the combinations of wins and losses for the games prior to the series-ending game.

This is consistent with the formula originally proposed in the problem, but it goes one too far. I think it should stop at ${{4}\choose {2}}$ rather than ${{5}\choose{2}}$.

John
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  • John. Thanks for the input. Following your logic, why is your first term ${{m-1}\choose{m-1}}$ and not ${{m}\choose{m}}$.
    It looks to me like you have 3 choices and are specifying all of them. Similar question holds for the second term... (im really confused here...)     – greg Dec 10 '13 at 21:21
  • your answer works out for case m=4 as well. Oddly enough, the answer by your method, for both m=3 and m=4 gives the same permutation count as $2{{2m-1}\choose{m}}$. Which, by brute force, seems to be the right answer. my head hurts. – greg Dec 10 '13 at 21:24
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For the play-off requiring $m$ games, the number of possible sequences of wins is given by: $$n_m={m\choose{m}}={{m-1}\choose{m-1}}$$ For the play-off requiring $m+1$ games, the number of possible sequences of wins is given by: $$n_{m+1}={{m+1}\choose{m}}={{m}\choose{m}}+{{m}\choose{m-1}}={{m-1}\choose{m-1}}+{{m}\choose{m-1}}$$ ...

For the play-off requiring $2m-1$ games, the number of possible sequences of wins is given by: $$n_{2m-1}={{2m-1}\choose{m}}={{2m-2}\choose{m}}+{{2m-2}\choose{m-1}}\\=\left[{{m-1}\choose{m-1}}+{{m}\choose{m-1}}+...+{{2m-3}\choose{m-1}}\right]+{{2m-2}\choose{m-1}}$$

Using the same argument you used, we consider the possible sequences of losses (the other team wins) and that gives: $$n_{Total}=2\left[{{m-1}\choose{m-1}}+{{m}\choose{m-1}}+...+{{2m-2}\choose{m-1}} \right]$$

Osvaldo López Zúñiga
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