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I derived a relationship between sequences drawn with and without replacement for an application in genetics. The proof is easy enough, but I would rather find a source than provide a derivation of a well-known result. But can't seem to find it.

Simplified problem setup: You have a stack of 52 card from which you draw uniformly at random with replacement until an arbitrary condition is met (say, you picked the ace of spades). This generates sequences with replacement. The order in which cards are first picked defines a random sequence without replacement.

An alternative way of generating the same distribution of our original sequences with replacement is as follows:

First draw this "order of first sampling" by generating a random permutation of the cards (i.e., shuffle the deck).

Then you pick cards from the top of the deck and keep a stack of previously drawn cards. suppose that by the time you completed the n th draw, the "previously picked cards" deck has c(n) cards. We then pick randomly from the "previously picked" pile with probability c(n)/52, and from the top of the original deck with probability 1-c(n)/52

Again, I don't want a proof that the alternative way is equivalent to the original, I am just wondering whether people know a name or reference for this.

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  • $\begingroup$ [ We then pick randomly from the "previously picked" pile with probability c(n)/52, and from the top of the original deck with probability 1-c(n)/52 ] - but this is clearly just the same as choosing a random card, right? At any rate, possibly your question would be better received at stats.stackexchange.com. $\endgroup$ Sep 22 '21 at 1:57
  • $\begingroup$ Yes, this is clearly the same! This equivalence, even though trivial, underlies a proof I have written for something less trivial (the look-down model, projecteuclid.org/journals/annals-of-probability/volume-24/…), so I'm asking around to see if people have seen this used or named before. Maybe it's just too trivial. $\endgroup$
    – user381021
    Sep 22 '21 at 11:39
  • $\begingroup$ For stats.stackexchange -- this feels morel ike probability than stats to me, but I suppose I can try. Thanks for the suggestion! $\endgroup$
    – user381021
    Sep 22 '21 at 11:42

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