Probability of $x$ trials given $k$ successes

Question

I am looking for a probability distribution that calculates the probability of $x$ total Bernoulli trials given fixed $k$ successes.

I have looked into negative binomial distribution: $$ P(X=x) = \binom{x-1}{k-1} p^k (1-p)^{x-k} $$ where:

• $x$ is the number of Bernoulli trials (not fixed)
• $k$ is the number of successes (fixed)
• $p$ is the probability of success (fixed)

But I've read that this particular negative binomial distribution calculates the probability of $x$ trials UNTIL the $k$th success. I might be wrong, but this is not the same as what I am looking for, which is the probability of $x$ trials, given $k$ successes.

If they are the same, can someone explain how they are equivalent? The Wikipedia page is confusing to me.

Answer 1

Your question is relevant to the concept of conditional probability. Recall the Bayes' theorem $$P(X=x|k) = \frac{P(X=x)P(k|X=x)}{P(k)}=\frac{P(X=x)P(k|X=x)}{\sum_i P(X=i)P(k|X=i)}$$

We can back out the desired distribution LHS from RHS. In particular we make use of $P(k|X)$, which is the binomial distribution.

Without conditioning on $k$ and without a given prior distribution of $X$, the number of trials $X$ can be assumed to follow a uniform distribution. Thus, $P(X=i)$ is a constant and the equation can be reduced to $$P(X=x|k) = \frac{P(k|X=x)}{\sum_i P(k|X=i)}$$

First of all, it is obvious that $X\geq k$.

For every $X\geq k$, $P(k|X=x) = C^x_kp^k(1-p)^{x-k}$.

Therefore,

$$P(X=x|k) = \frac{C^x_kp^k(1-p)^{x-k}}{\sum_{i=k}^\infty C^i_kp^k(1-p)^{i-k}}$$ $$=\frac{C^x_k(1-p)^{x-k}}{\sum_{i=k}^\infty C^i_k(1-p)^{i-k}}$$ $$=\underline{\underline{C^x_k(1-p)^{x-k}p^{k+1}}} (*)$$ ($*$)For the sum of the infinite series, please refer to the comment section below.

Now, compare this result with the negative binomial distribution $C^{x-1}_{k-1}p^k(1-p)^{x-k}$. As a matter of fact, they highly resemble each other. The differences lie in the binomial coefficient and the number of success trials:

$$C^x_k \text{ vs } C^{x-1}_{k-1}$$ $$p^{k+1} \text{ vs } p^k$$

The intuitive explanation is as follows.

The negative binomial is a distribution of the trials until the $k$th success. Therefore, the last trial must yield a success, which means that the permutation is applied to the first ($x$-1) trials only that contains ($k$-1) successes.

On the other hand, the desired distribution does not have this restriction, so the permutation includes one more successful trial.