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Two players A and B plays a series of $2n$ games. Each game can result in either a loss or win for A. Find the total number of ways in which A can win this series of games. (All games are to be played)

I'm feeling a bit confused with this one. How to solve?

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Does the series run until one of the sides reaches $n+1$ wins, or does it run all the way through (until all $2n$ games are done)? – barak manos Jan 14 at 14:46
I'm not sure @barakmanos I just wrote what my book says.But I find the former more practical (personally) – ZOZ Jan 14 at 14:47
If it runs until all games are done, then the answer is simply $\sum\limits_{k=n+1}^{2n}\binom{2n}{k}$. – barak manos Jan 14 at 14:48
Since it says all games are to be played, I think all 2n games are played. – SchrodingersCat Jan 14 at 14:48
@barakmanos: It says "all games are to be played" so that seems to mean "run all the way through". – Eli Rose Jan 14 at 14:48
up vote 6 down vote accepted

Since $2n$ games are going to be played each string of $2n$ games can be described by the games which A wins. Now A wins $k$ out of $2n$ games in $\dbinom{2n}{k}$ ways (order matters here when we count like this), so the total number of outcomes is equal to $$\sum_{k=0}^{2n}\dbinom{2n}{k}=(1+1)^{2n}=2^{2n}=4^n$$ by the Binomial Theorem. If A wins exactly $n$ games, then we have a tie. So there are $\dbinom{2n}{n}$ ties. Now due to symmetry, A wins exactly half of the rest of the strings, so the answer is $$\frac{\text{#total outcomes$-\#$ties}}{2}=\frac{4^n-\dbinom{2n}{n}}{2}=2^{2n-1}-\dbinom{2n-1}{n}$$

Plug in some values for $n$ to verify that this works. For example:

  1. $n=1$. Then there are $4^1$ outcomes: $ww, wl, lw, ll$ (where $w$ stands for a win of A and $l$ for a loss). In $1$ of those A wins.
  2. $n=2$. Then there are $4^2=16$ outcomes: $wwww, wwwl, wwlw, \dots$ and A wins in $$\frac{4^2-\dbinom{4}{2}}{2}=\frac{16-6}{2}=5$$ of them. B wins in other $5$ (by symmetry) and in $6$ outcomes we have a tie. You can check this by writing out all $16$ strings.
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Very nice.But will @Barack Manos's answer give the same result? – ZOZ Jan 14 at 15:07
@SanchayanDutta Yes. – Jimmy R. Jan 14 at 15:10
@SanchayanDutta: It must -- this is just a closed form for the sum barak manos gives. – Eli Rose Jan 14 at 15:10
@barakmanos No, for $n=1$ it gives $1$ to me. But you are correct in the following sense: The sum as I have it (with result $4^n$) counts the outcomes $wl$ and $lw$ as different, so order matters (but I say that order does not matter). So, there are $4^1$ strings: $wl, ww, ll, lw$, but in the end $1$ way for A to win, so this comes out correctly. Note also, that $\sum_{k=0}^{2n}\dbinom{2n}{k}=\dbinom{2n}{n}+2\sum_{k=n+1}^{2n}\dbinom{2n}{k}$ so that your approach and mine give exactly the same result. – Jimmy R. Jan 14 at 15:21

If the series runs until all $2n$ games are played, then the answer is:


You simply need to sum up the number of ways in which A can win:

  • $n+1$ games out of $2n$ games
  • $n+2$ games out of $2n$ games
  • $\dots$
  • $2n$ games out of $2n$ games
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So that is $2^{2n}-2^{n}$ ? – ZOZ Jan 14 at 15:08
@SanchayanDutta: No, why would it be? – barak manos Jan 14 at 15:12
Can you tell me how to sum that series? I was trying to use $C(n, 0) + C(n, 1) + C(n, 2) + ... + C(n, n-1), + C(n, n) = 2^{n}$ and $C(2n, 0) + C(2n, 1) + C(2n, 2) + ... + C(2n, 2n-1), + C(2n, 2n) = 2^{2n}$ and subtracted the first from second..where am I going wrong? – ZOZ Jan 14 at 15:16
@SanchayanDutta: $C(n,\dots)$ is where you're going wrong. This term has nothing to do with $C(2n,\dots)$. It's like taking Pascal Triangle, summing up a row of $2n$ terms, then a row of $n$ terms, then subtracting one from the other. In the formula above, you need to sum up a row of $2n$ terms, and then subtract the first $n-1$ terms from the result. – barak manos Jan 14 at 15:17
Oh right...but how to sum that then? – ZOZ Jan 14 at 15:21

Since all games are to be played you want to count how many combinations of $0100 \ldots 0011$ there are (where $0 \to A$ loses and $1 \to A$ wins). The formal result is $$\sum\limits_{k=n+1}^{2n} (\text{combination where there are $k$ ones}) = \sum\limits_{k=n+1}^{2n}\binom{2n}{k}$$

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if you seek the number of combinations to arrange A and B (for which players who wins), the number of ways if all games are played is (2n)!

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You are supposed to determine in how many outcomes player A wins at least $n + 1$ of the $2n$ games. – N. F. Taussig Jan 14 at 14:54

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