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Say I start off with N single cells in a sample. These are then allowed to grow untill there are x progeny from each individual cell. How many cells must I take so that I have at least one progeny from each the N original cells in the sample?

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  • $\begingroup$ As stated this is not a probability problem, but a straight mathematics/logic question. It's also so straightforward as to be routine bookwork - is this for some subject? It's related to "From a drawer containing N pairs of socks, how many socks do you need to take from the drawer (in the dark) to be certain to have at least one pair?" $\endgroup$
    – Glen_b
    May 26 '14 at 11:09
  • $\begingroup$ Really? I admit my sock drawer is somewhat muddled and would be grateful for more explanation. Actually, this is a real example where I scrape colonies off of one selective plate and want to plate a representative sample (ie containing at least one cell from each of the original colonies) on a new medium. $\endgroup$
    – David
    May 26 '14 at 14:36
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    $\begingroup$ You are not looking for a probability distribution (for instance, which is the probability that you sample all N cells taking y cells after growth?), but asking for the smallest y which gives you at least one of each N cells. In this case (N-1)x+1 cells would be the smallest value you need to take. If this is too expensive, you could define a threshold and look for the minimum number of cells such that the probability you are missing one of the N cells is smaller than your threshold... $\endgroup$
    – Giancarlo
    May 26 '14 at 17:00
  • $\begingroup$ David, if Giancarlo's (correct) answer doesn't cover what you really want to know, please refine your question to clarify what you do want. Since you haven't clarified further I'll now post the answer I've had sitting in the edit window since you posted, then I'll add the information about socks. $\endgroup$
    – Glen_b
    May 26 '14 at 21:39
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You have a total of $Nx$ progeny.

You could select $(N-1)x$ without having at least one progeny of every original member by having all progeny except those of one particular original cell.

So to be sure to get at least one of each, you need to select $(N-1)x+1$. Anything less can leave you missing one type.

As for the relationship to the relevant sock drawer problem, the problem works like this:

In your sock drawer, you have 2N socks from N distinct pairs, all as single socks. The light in your room isn't working, and you need to grab a pair of socks, so you plan to take some socks out to the light to find a pair from. What's the smallest number of socks you need to take to be sure of finding at least one pair?

(This is similar to your problem - you need to cover the case where you get one sock from every pair - and then take one additional sock, which would then match one of the already selected socks. That is, you need to take N socks in case you get get one from each pair, and then one more sock will be certain to form a pair. Any fewer could leave you with all single socks. The count of "once sock from every pair" is where it's very much like yours, but the form of problem is slightly different)

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    $\begingroup$ +1. An alternative understanding of the problem supposes the original $N$ cells are still in the population, requiring $N+(N-1)x+1$ cells to be chosen to assure at least one of each of the progeny is included. An answer that satisfies both readings is that you need to take enough cells so that no more than $x-1$ remain (for otherwise--clearly--the remaining cells could include all $x$ progeny of a single cell). $\endgroup$
    – whuber
    May 26 '14 at 21:54
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    $\begingroup$ +1 @whuber (and the last sentence would be worth an extra +1 on its own). I considered - but didn't pursue in my posted answer - the possibility of considering the originals still in the sample. I left it out for two reasons -- (i) I took a (possibly over-) literal meaning of the word 'progeny'; and (ii) that the problem was about cells - with cell division it's hard to call either of the two cells you end up with "the original"; as usually conceived, one cell splits into two daughter cells, which may then divide further. After the first division we no longer really have an original. $\endgroup$
    – Glen_b
    May 26 '14 at 22:28
  • $\begingroup$ You are right I was mixing up the probability of having of progeny from all the N original cells in a given sample with the simpler case here. Out of interest, how would one calculate the probability as mentioned above? $\endgroup$
    – David
    May 27 '14 at 9:11
  • $\begingroup$ Whuber gives an answer that works for either case in his second sentence. $\endgroup$
    – Glen_b
    May 27 '14 at 10:54

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