[jimdmcevoy]

Well after a bit of thinking I think I am right and you are right (except for that last conclusion), I reckon once the guy picks 3 out of a hat, he can then correctly state:

SUM over i of P(3 given N=i)*P(N=i) from i =1 to 6 is equal to

SUM over i of P(3 given N=i)*P(N=i) from i =7 to infinity

As in he can put restrictions on the probability distribution of N just from knowing that x=3.

I reckon my theory(well it's not 'my' theory) predicts this result. My theory predicts a probability density distribution of the last person being born as {0 if y<x, x/y^2 if y>x} where y is number of number of total people. Besides the problem with continous and descreteness, I think this pretty much says that there is a x/y^2 chance that N=y, as in P(N=i)=x/i^2.

I got this from the culmulative probabity function of {0 if y<x, 1-x/y if y>x) where y is total people. This function saying that there is a 1-x/y chance that The last person born will be assigned a number less than y.

So anyway, P(N=i)=x/i^2 and P(3 given N=i)={1/i for i>=3 , 0 for i<3}

When I calulate both sums for large N, the first one is 3 times the second one and I'm not sure why. This is true for discrete or continuous calulations (for the discrete sum I think you need the Reiman Zeta function or something) I've made a mistake somewhere along the line, I'm not sure if it's a logical or mathematical mistake, I'll check it all over when I have some time.