2

Let $N$ be a positive integer. Also let $0<p<1$ and $q=1-p$. Then, the following holds:

$$\sum_{k=N}^{2N-1}\binom{k-1}{N-1}(p^Nq^{k-N}+q^Np^{k-N})=1 \tag{1}\label{1}.$$

One may derive (\ref{1}) "combinatorially" as follows: Suppose teams $A$ and $B$ play a series of $2N-1$ games, where the first team to win $N$ games wins the series (and no further games are played once this occurs). Team $A$ has a probability $p$ of winning an individual game against team $B$ (and team $B$ has probability $q=1-p$). Let $X$ be a random variable for the number of games played in the series. Clearly, $P(X=k)=0$ for $k=1,...,N-1$. For $k=N,...,2N-1$, we have $$P(X=k)=\binom{k-1}{N-1}(p^Nq^{k-N}+q^Np^{k-N}).$$ Justification: The binomial coefficient counts the ways to arrange the winning team's first $N-1$ wins among the first $k-1$ games (the last game is not counted since it must be a win). The other term accounts for either team winning, given the individual-game win probabilities.

Finally, summing over all possibilities $k$ must yield $1$ since this is a valid probability distribution.

Question: Is it possible to derive this result (meaning the identity in (\ref{1})) in a purely algebraic manner, i.e., without appealing to combinatorial reasoning?

Coffee_Table
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    This identity appeared at the following MSE link where a proof was given. (Needs some algebra to produce a matching problem statement.) – Marko Riedel Jun 09 '17 at 01:09
  • Thanks, @MarkoRiedel. Perhaps this MSE link is more relevant? – Coffee_Table Jun 09 '17 at 01:36
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    Yes this link is relevant also since it has only one upper limit / parameter for the two sums as in your problem statement, but the first link has the better proof. – Marko Riedel Jun 09 '17 at 01:46
  • @MarkoRiedel Thanks! I have two follow-up questions, if you don't mind: 1) Can you refer me to proofs of the identities used in the Egorychev method? I have no idea how hard the derivations are; my complex analysis skills are lacking... 2) Do you know if there exists a closed-form solution for the expectation $EX=\sum_{k=0}^{2N-1}kP(X=k)$ given the above form of $P(X=k)$? – Coffee_Table Jun 09 '17 at 14:10
  • For 1) these more or less follow by inspection from applying the Cauchy Residue Theorem to binomials and negative binomials. It is important to note that many of these proofs can be done without complex variables using formal power series only. The MSE user @MarkusScheuer has contributed a considerable number of proofs of this type to the site. These formal power series proofs are preferable because there are fewer constraints on the functions involved. For 2) I cannot guarantee it but I might get to it. – Marko Riedel Jun 09 '17 at 22:07
  • For the record: I applied the method from the two linked-to posts to the expectation and it did not simplify as when summing the probabilities. So it looks like the only possibility is the expectation for an unbiased coin ($p=q=1/2$). – Marko Riedel Jun 10 '17 at 00:30
  • @MarkoRiedel Thanks a lot for the information and resources; looking forward to studying these methods! And thanks for the simplification attempt; I will try my hand at the simpler case $p=q=1/2$. – Coffee_Table Jun 11 '17 at 14:38

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