# Resulting distribution when repeatedly sampling without replacement from weighted elements

I have a model where I need to sample from $m$ elements without replacement, where each element exists exactly once. The element have different weights and the probability of an element being drawn is equal to the fraction of the element's weight relative to the sum of the weights of the remaining elements. In R a single draw would simply be:

m <- 4                  # number fo elements
w <- c(.4,.3,.2,.1)     # weights
n <- 3                  # number of draws
sample(seq_along(w), size = n, prob=w, replace=FALSE)

> [1]  3  1  2


Simulating many draws and calculating the probability of occurrence of each element we get

v <- replicate(1e5, sample(seq_along(w), size = n, prob=w, replace=FALSE))
p.hat <- table(v) / sum(table(v))
p.hat

>
1         2         3         4
0.3075767 0.2902900 0.2526967 0.1494367


What is the name of the resulting distribution and how could I calculate the results above without simulation?

PS. I stumbled across Wallenius' noncentral hypergeometric distribution but my knowledge in this field is too limited as to decide if this distribution is given or not.

• Are you asking about the specific situation $m = 4$ and $n = 3$?Your population is size $m$. Sampling without replacement, you can draw at most $n = m$ elements. – Steve Samuels May 10 '16 at 20:06
• For a specific $m$ and $n$ would be fine. I would like to know the name of the resulting distribution, as I suspect that it might have been derived analytically somewhere, i.e. for the general case. – Mark Heckmann May 11 '16 at 9:46

I would not bother trying to come up with a name, or find this in the books. This is an example of a sampling distribution in finite population sampling, and that's about it. You can probably describe all the samples explicitly for a population this small, and derive all the probabilities (again, knowing exactly what your sampling scheme is), so you don't have to rely on sample() simulation.