First, I will show an extension of simple random sample without replacement, and then put forward the question.


From Wiki, a simple random sample is a subset of individuals (a sample) chosen from a larger set (a population), where each individual is chosen randomly and entirely by chance, such that each individual has the same probability of being chosen at any stage during the sampling process.

Now, suppose that each individual in the population has a different weight representing the chance of being sampled.

Take the following example:

The population is {1, 2, 3, 4}, and the relative weights of the individual 1, 2, 3, 4 are 1, 2, 3, 4 for concise. Then if a sample {1, 2} is taken from the population without replacement. Then the possibility of the sample {1, 2} should be calculated by:

$\frac{1}{1 + 2 + 3 + 4} * \frac{2}{2 + 3 + 4} + \frac{2}{1 + 2 + 3 + 4} * \frac{1}{1 + 3 + 4}$.


My question is how to give out a general expression of the possibility for a sample taken in the above way? If there doesn't exist an closed form, how to approximate it?


Extended comment:

I don't know if it helps with your particular application, but this kind of sampling is exactly implemented in the sample function of R statistical software, using the prob argument.

For your example, it would be sample(1:4, 2, prob=1:4).

A million draws made to these specifications give a 1 and a 2 (in either order) 4.74% of the time. This agrees to two places with your computation $(2/90 + 2/80) = 0.04722222.$

 m = 10^6;  hit = logical(m)
 for (i in 1:m) {
  x = sample(1:4, 2, prob=1:4)
  hit[i] = (sum(x)==3) }
 ## 0.047389

 2/90 + 2/80
 ## 0.04722222

Notes: (a) In R, 1:4 (consecutive integers) is shorthand for the vector c(1,2,3,4). (b) The 4-vector for the argument prob need not sum to unity because R normalizes it before use. (c) See R documentation and a reference on the sample function by typing ? sample in the session window. (d) In symbols, the sum 1 + 3 + 4 might be written $\sum_{1, i\ne 2}^4 i.$


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