Probability of Two Samples Containing Overlapping Data Quick question about sampling a data population that contains non-overlapping data: if I take two samples of 50 from a population of 100,000 unique data points, how would I calculate the probability of the two samples containing a shared data point?
As well, how do I determine the population size required for the probability of overlap to be below a certain point (0.05%) for $n=50$?
Thanks.
 A: Since this looks like bookwork - and you clearly even know which distribution you're supposed to use - I'll avoid doing it directly in those terms, and just outline the basic logic.
Imagine you have a big urn with 100,000 grey balls in it (maybe one in the shape of a small skip). Take your first sample, painting those 50 balls red and return them.
Now you draw the second sample. Easy question: What is the chance you draw no red balls?
(1) quick approximation: removing a few grey balls won't change the probability much, so we can approximate it by doing it as if it were sampling with replacement.
P(0 red balls) = $(0.9995)^{50}$ = 0.975304
So the chance of one or more red balls in the second sample is about 2.5%
This is handy because it gives us a reasonableness check; the real answer should be close but a little smaller.
(2) Exact - sampling without replacement.
P(0 red balls) = $(1-\frac{50}{100000})\times(1-\frac{50}{99999})\times(1-\frac{50}{99998})\times \ldots\times(1-\frac{50}{99951})$ 
The answer is very similar to the earlier approximation (as expected, is close but a little smaller). Given the hints here already, you should be able to cast it directly into hypergeometric terms ... though starting by doing it from first principles may be a better way to start.

how do I determine the population size required for the probability of overlap to be below a certain point (0.05%) for n=50?

A calculation almost identical to the one above. You could get a very close idea from the approximate calculation I did. To do it more exactly, write the inequality for the probability in terms of factorials (hypergeometric), use Stirlings approximation, cancel out everything you reasonably can, solve to get a lower bound for n.
A: ref to Markatou, M., et al. (2005). "Analysis of Variance of Cross-Validation Estimators of the Generalization Error." J. Mach. Learn. Res. 6: 1127-1168.
it is a hypergeometric distribution.
