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Let $T_{n,m}(p)=\sum_{j=0}^{m-1} \binom{n+j-1}{j}p^j$. In answering a recent question, I discovered the identity

$$ (1-p)^n T_{n,m}(p)+ p^m T_{m,n}(1-p)=1 \tag{1} $$

Is there a nice combinatorial proof or interpretation of (1) ? ($p$ looks a lot like the probablity of something in this formula, doesn't it.) I've got a feeling that this question is a duplicate of an already existing question, but my search was unsuccessful.

UPDATE : as explained in orangeskid's answer to the abovelinked question, (1) can be nicely derived by expanding Newton's binomial in $(p+(1-p))^{n+m-1}=1$. But this is still more computational than really combinatoric/probabilistic, I'm still waiting for a more direct interpretation of $T_{n,m}(p)$ and (1).

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If we take $p$ as the probability of success of a sequence of i.i.d. Bernoulli trials, then

$$ (1 - p)^nT_{n, m}(p) = \sum_{j=0}^{m-1} \binom {n + j - 1} {j}p^j(1-p)^n$$

is the probability that having less than $m$ successes before the $n$-th failures. Similarly,

$$ p^mT_{m, n}(1 - p) = \sum_{j=0}^{n-1} \binom {m + j - 1} {j}(1-p)^jp^m$$

is the probability that having less than $n$ failures before the $m$-th successes.

These two are complementary events - you either reach the $m$-th success first, or the $n$-th failure first, and you cannot have them together in the same trial. So their probabilities sum up to $1$.

P.S. These are well known negative binomial CDFs.

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  • $\begingroup$ Could you elaborate a little more ? Your interpretations are not obvious from the formulas. For example, what does the $j$ index represent in each situation ? $\endgroup$ – Ewan Delanoy Mar 15 '18 at 11:58
  • $\begingroup$ It depends on $j$ is in the exponent of $p$ or $1 - p$. For example, for the first term, each summand is a pmf of a negative Binomial, which can be interpreted as the probability of having exactly $j$ success before the $n$-th failure. To see this, write the summand as $\displaystyle \binom {n + j - 1} {j} p^j (1-p)^{n-1} \times (1 - p)$. Note that the $n$-th failure must be located at the $n+j$-th trial. Then the number of ways to arrange $j$ successes and $n-1$ failures in the first $n+j-1$ trials is the leading binomial coefficient, and hence the summand is derived. $\endgroup$ – BGM Mar 15 '18 at 15:23

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