Probability question: names in hats

[robvalue]

But that's what I'm saying too.  So why shouldn't the probability that the last guy draws his own name be the product of everyone else successfully drawing any name but his?

If I want to know what the probability of drawing (in the dark) a pair of black socks in a drawer that contains 3 black and 2 red socks.  I can find this two ways.  Lets start by naming each individual sock b1, b2 and b3 for the black socks and r1 and r2 for the red ones.  Now list all the possible combinations (reversing the order wouldn't affect the outcome):

1-b1,b2
2-b1,b3
3-b1,r1
4-b1,r2
5-b2, b3
6-b2,r1
7-b2,r2
8-b3,r1
9-b3,r2
10-r1,r2

Only combinations 1, 2 and 5 include a pair of black socks so the P(pair of black socks) is 3/10

The other way to calculate the probability is to take the product of the probability the first draw is black and the probability the second draw is also black (after the first draw):

3/5 • 2/4 = 6/20 = 3/10

So I think multiplying the probability of that each person draws any name except person ten's should get it done.  Am I missing something?

Yeah, but the probability isn't fixed for each person. It depends on previous picks. I explained how even person 2's probability isn't fixed, but depends on what person 1 takes. It alters the pool of numbers he has, affecting the probability. It matters whether or not his number is still there, because it effectively disappears from the selection, giving less numbers, if it is there. So it requires more of a breakdown into possibilities.