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In a sequence of Bernoulli trials with success chance $p$, the probability that we will see $k$ failures before the $r$-th success is given by the negative binomial distribution (NBD) with pmf: $$ Pr(k) = \binom{k+r-1}{r-1} p^r (1-p)^k $$ But what if we had two sequences of simultaneous Bernoulli trials and we wanted to know the probability of how many trials we need, before we see at least $r_1$ successes in the first sequence and $r_2$ successes in the second, with success chances $p_1$ and $p_2$, respectively?

For simplicity, consider the case where $r_1 = r_2 = 1$. The NBD pmf simplifies to: $$ Pr(k) = p (1-p)^k $$ Let's distinguish two independent random variables $X$ and $Y$ with their respective pmf's: $$ Pr_x(k_x) = p_x (1-p_x)^{k_x} \quad \text{and} \quad Pr_y(k_y) = p_y(1-p_y)^{k_y}$$

If we perform a sequence of trials for both $X$ and $Y$ simultaneously, and it takes $n$ trials before we see at least one success in each sequence, then at least one sequence must have had its first success exactly after $n$ trials (or $n-1$ failures). Suppose that this was the $X$-sequence. Then the $Y$-sequence could also have had its first success after $n-1$ failures. The probability of this happening is: $$ Pr_x(n-1) \cdot Pr_y(n-1) $$ But the $Y$-sequence could also have seen its first success before the $n$-th trial. The probability of that happening is: $$ Pr_x(n-1) \cdot Pr_y(< n-1) = Pr_x(n-1) \cdot \Bigg(\sum_{j = 0}^{n-2} Pr_y(j)\Bigg) $$ Lastly, there is the case where the $Y$-sequence has its first success after exactly $n-1$ failures, while the $X$-sequence had its first success before the $n$-th trial. The probability of this occurring is: $$ Pr_x(< n-1) \cdot Pr_y(n-1) = \Bigg(\sum_{j = 0}^{n-2} Pr_x(j)\Bigg) \cdot Pr_y(n-1) $$ The summations in the two previous expressions are geometric. As long as we assume that $p_x, p_y \neq 0$, we can simplify them. Let $(q_x, q_y) := (1-p_x, 1-p_y)$ and by adding all three probabilities together we find a nice, symmetric expression: $$ Pr(n) = p_x q_x^{n-1} p_y q_y^{n-1} + p_x q_x^{n-1} (1 - q_y^{n-1}) + p_y q_y^{n-1} (1 - q_x^{n-1}) $$ I've verified this expression with a few experiments in Python. Is the constructive argument I've given above enough to prove that the pmf is correct?

So, what do we do in the more general case where we're looking for more than one success? I've already struggled quite a bit with this simple case. And more importantly, is this sort of distribution well-known and if so, what is it called?

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  • $\begingroup$ I just realised that the expression $Pr(n)$ is invalid for $n < 2$, since then the summations make no sense. $\endgroup$
    – beertje00
    Mar 22, 2021 at 13:57
  • $\begingroup$ If $n=1$ your expression gives $Pr(1) = p_x p_y $ which is correct if $n=1$ is the number of attempts, successful on both trials $\endgroup$
    – Henry
    Mar 22, 2021 at 14:10
  • $\begingroup$ Ah yes. It turns out to be true anyways. But the derivation is strictly speaking not correct for $n<2$, I think. $\endgroup$
    – beertje00
    Mar 22, 2021 at 14:26

1 Answer 1

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An alternative approach to your calculation for $r_1 = r_2 = 1$ is to say the probability of both being successful at least once by the $n$th attempt is $(1-q_x^{n})(1-q_y^{n})$ and of both being successful at least once by the $n-1$th attempt is $(1-q_x^{n-1})(1-q_y^{n-1})$ so the probability the second is initially successful after $n$ attempts or $n-1$ failures is the difference: $$(1-q_x^{n})(1-q_y^{n}) - (1-q_x^{n-1})(1-q_y^{n-1})$$

If you multiply this out and rearrange than you get your result.

For more general $r_1,r_2$ you could produce a similar expression involving regularised incomplete beta functions or something like

$$\left(1-\sum_{i=0}^{r_1-1}{n \choose i}p_x^{i}q_x^{n-i}\right)\left(1-\sum_{i=0}^{r_2-1}{n \choose i}p_y^{i}q_y^{n-i}\right) \\-\left(1-\sum_{i=0}^{r_1-1}{n-1 \choose i}p_x^{i}q_x^{n-1-i}\right)\left(1-\sum_{i=0}^{r_2-1}{n-1 \choose i}p_y^{i}q_y^{n-1-i}\right)$$

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