Great Problems BruceZ. I have some comments below.


Question 1 :

I agree that the posted solution to #1 is correct, but it is not very enlightening. It would be very hard to extend the given solution to n=10000. I will post my solution to problem #1 separately.


Question 3 :

I agree that the posted solution would be correct if the names were replaced in the hat after each draw.


I disagree with your solution to #3. You claim that


f(2) = 1

f(n) = (n-1)f(n-1)


where f(n) is the number of ways n people can draw names without drawing their own. Obviously f(2) = 1. It is easy to write out the 6 permutations to check that f(3) = 2 = 2*1 which agrees with your formula. When I wrote out the 24 permuatations for f(4), I got f(4) = 9, which does not obey your formula.


Question 5 : I like your reasoning on this one. I got slightly different numbers. You state that the 3rd tallest person would be visible 1/4 of the time. I think that he would be visible 1/3 of the time. There seem to be 6 equally likely possibilities to me: 123, 132, 213, 231, 312, 321. In 1/3 of these possibilities the 1 can be seen. Similarly, I think the nth person can be seen with probability 1/n, so then, using your reasoning, the average number of people seen in an N person line would be


1+ 1/2 + 1/3 + 1/4 + ... + 1/N