# How do you mechanically draw a sample of size n from N without replacement

Say I have a physical (or virtual) jar or N marbles and I wish to draw n marbles at random without replacement where the likelihood of each n marbles being drawn is 1/N ... how do I do that? Is that even possible?

I mean won't the probability for the 2nd marble in the sample change to 1/(N-1) and so on?

I remember comments on this in a statistical inference class -- would appreciate any comments/help.

• These kinds of sampling questions usually have two states.The chance a marble will be picked is 1/N, looking at the initial and final states only. Now, the chance any marble is the second, third, etc. to be picked is not 1/N. Jul 15, 2016 at 14:39
• Recursion is a useful aid: draw one marble with a chance of $1/N$ per marble. Now apply your solution to drawing $N-1$ marbles from the jar you are left with. Done! The truly hard part with a physical jar is drawing that first ball in a genuinely unbiased way.
– whuber
Jul 15, 2016 at 14:42

Since you mentioned virtual jar, I'd say it is possible given an infinite number of marbles in the jar. In which case, each sampling will represent a Bernoulli trial with p = 1/N.

• With an "infinite number" of marbles, exactly what would "$1/N$" equal??
– whuber
Jul 15, 2016 at 14:43

Let's say you number the marbles {1, ..., N} and you let mi be the ith marble you pick (i goes from 1 to n).

So the probability that the first marble is some marble a is: Pr(m1 = a) = 1/N

Now, turning to the second marble, we can say two different things:

Pr(m2 = a) = 1/N

BUT, if we look at the conditional probabilities:

Pr(m2 = a | m1 = b) = 1/(N-1) if a != b

Pr(m2 = a | m1 = b) = 0 if a = b

So, once you draw the first marble, each marble has a 1/(N-1) chance of being the second marble. But, if you haven't drawn any marbles yet, each marble has a 1/N chance of being the second marble.