What happens if I do a fractional number of Bernoulli trials? I want to simulate $n$ independent Bernoulli trials with probability of success $p$, and calculate the probability that at least one of those trials succeeds. If multiple trials succeed, I don't particularly care how many. So I use this formula to calculate the overall probability of at least one success:
$$
1 - (1 - p)^n
$$
I then do a single Bernoulli trial with the above probability. That works quite well, as it allows me to aggregate many Bernoulli trials into a single meta-trial. If this meta-trial fails, I can later conduct another (once a meta-trial succeeds, I stop running trials altogether).
(I am aware that I could instead draw from a geometric distribution in advance, add up the values of $n$, and wait until I exceed the value drawn. But keeping track of the total value of $n$ is inconvenient, so I don't do this.)
Now, I'm beginning to think that I want to have "fractional trials" to deal with cases where an event is smeared out over time instead of happening in a single discrete lump. There are two "obvious" ways to do that:

*

*Just let $n$ be a positive real number (technically, a floating point number) and use the same formula as above.

*Use a Poisson distribution with $\lambda = np$, so that the resulting probability is $1 - e^{-np}$.

I understand, in general terms, what (2) is simulating: It's a Poisson process with rate $p$ and length or "duration" $n$. On average, a success happens every $\frac{1}{p}$ units of length, and we're asking whether any successes happen within the next $n$ units of length.
What I don't understand is (1). I can visualize it as a binomial distribution for integral $n$, but if $n$ is a real number, it's unclear to me what the probability actually represents, in a simple and intuitive way. I can see that:

*

*(1) is equivalent to (2) for large $n$ and small $p$ (by the Poisson limit theorem).

*For both (1) and (2), the probability of at least one success is not affected by combining or splitting meta-trials, when $p$ is held constant. So neither of them is "obviously wrong."

*For non-small $p$, (1) has a higher probability of a success than (2) (by playing around in Desmos). For small $p$, (1) is still bigger, but the difference is negligible. I believe that this corresponds to the fact that in (1), $p$ is a probability, and we approach a degenerate case as $p$ approaches 1, while in (2), $p$ is "just" a density factor and is not bounded above by $p = 1$.

*My understanding of the geometric distribution suggests that (1) should have a success on average every $\frac{1}{p}$ units of length. But that seems unlikely since (2) already has that property, and (1) is always greater than (2).

Ultimately, I need to choose between these two ways of modeling my problem, and I want to understand what (1) represents in order to make that decision.
What does the binomial distribution intuitively represent, if $n$ is not an integer?
 A: The comments link to a nice paper that gives the continuous analog of the binomial CMF:
$$ F_{n,p}(x) = \frac{B(x,n+1-x,p)}{B(x,n+1-x,0)}$$
Where
$$B(x,y,p):= \int_p^1 t^{x-1}(1-t)^{y-1}
\;\;x,y>0, \;p\in[0,1]$$
For interpretation, you could envision flipping $n+1$ coins with a parameter $p$, but instead of each coin $X_p$ being worth (i.e., having measure) $1$, each coin has a worth distributed on the range $[0,1]$ according to some distribution $\nu_{p}$ so we get something like this:
$$S_{n,p} = \sum_{i=1}^{n+1} X_{p,i} \sim F_{n,p} $$
Where

*

*$X_{p,i} \sim Y_pm_p$


*$Y_p \sim \text{Bernoulli}(p)$


*$m_p\sim\nu_p$


*$Y_p$ and $m_p$ are independent random variables
I don't know how to quickly derive the distribution $\nu_p$ for $n>1$ but heuristically, we can see an idea of this from the $n=1$ case (i.e., "continuous Bernoulli"):
$$P(X_p < x) = P(Y_p=0 \;\;\cup\;\; m_p<x ) = P(Y_p=0) + P(m_p<x) - P(Y_p=0)P(m_p<x)$$
Using known values, we have
$$P(X_p < x) = F_{p,1}(x) = (1-p) + F_{\nu_p}(x) - (1-p)F_{\nu_p}(x)$$
Simplifying, we get:
$$F_{p,1}(x) = (1-p) + [1 - (1-p)]F_{\nu_p}(x) = 1-p+pF_{\nu_p}(x) = 1-p[1-F_{\nu_p}(x)]$$
So, if we knew $F_{p,1}$ then we can get $F_{\nu_p}$:
$$F_{\nu_p} = 1-\frac{1-F_{p,1}}{p}$$
