(This article was originally published at R – Xi'an's Og, and syndicated at StatsBlogs.)

**A**n interesting question came up on X validated a few days ago: given a probability vector **p**=(p¹,…,p⁷), is there a way to pick 5 values in {1,…,7} *without replacement* and still preserve the probability repartition in the resulting sample? In other words, is there a sampling *without replacement* strategy that leads to

for i=1,…,7..? Unless those probabilities p¹,…,p⁷ are close enough to 1/7, this is simply impossible as 5 values out of 7 have to be sampled, which imposes some minimal frequency on some of the values.

Hence a generic question:

given a vector p of k probabilities (summing up to 1), what is the constraint on this vector and on the number n of elements of the population one can draw without replacement in order to achieve a expected frequency of np on the resulting vector? That is,

In the cases n=2,3, I managed to find and solve the system of equations satisfied by the sampling probability vector **q**, but I wondered if there exists a less pedestrian resolution. I then showed the problem to Robin Ryder while at CIRM for the Bayesian week and he quickly pointed out the answer by Brewer’s and Hanif’s book Sampling with unequal probabilities to this question, which *does not use sampling with replacement* with a fixed probability vector but instead modifies the remaining probabilities after each draw, as in the following R code:

kuh=(1:N)/sum((1:N)) #example of target smpl=sample((1:N),1,rep=FALSE,pro=kuh*(1-kuh)/(1-n*kuh)) for (i in 2:n) smpl=c(smpl,sample((1:N)[-smpl],1,rep=FALSE, pro=(kuh*(1-kuh)/(1-(n-i+1)*kuh))[-smpl])

Hence the question is not completely solved, since I am still uncertain whether or not there exists a sampling without replacement that achieves the target probability! But at least this shows there is only a solution when all probabilities are less than 1/n, n being the number of draws…

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