Warp wrote:
> Invisible <voi### [at] devnull> wrote:
>> If we select one ball, the probability of it being ball #1 is 1/N.
> 
>   You are thinking it in the wrong way.
> 
>   The very first time you select one ball, the probability it's a ball you
> have never seen before is 1.
> 
>   The second time you select a ball, the probability it's a ball you have
> never seen before is (N-1)/N.
> 
>   With the third ball, the probability is (N-2)/N. And so on.
> 

It is very likely that, for the third pick of a substantial N balls that 
the probability is (N-2)/N. However, if the first two picks happened to 
be the same, the probability would still be (N-1)/N.

Example: Lets say 100 balls, labeled 1 to 100. For the first 99 picks, 
I've pulled out ball #1. That doesn't mean that, on the 100th draw, I 
have (100-99)/100 chance of pulling out a ball I haven't seen. In 
(N-K)/N, K needs to account for how many you've seen, not how many 
previous picks have been performed. Which means to solve for K on any 
draw at N=100, we have to solve this problem of how many picks it would 
take for N=99. For N=99, we'd have to solve for N=98. And so on.

I remember this problem from combinatorics, I just don't remember where 
my textbook is.

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