1. Originally Posted by Alan
SWW - you rock!

haha i did.
tried not no crack any ribs....

math lesson (from GMT+1). Just happen to have internet now.

Let's suppose that any given PM selects one of the hammocks with equal probability, 1/3, independent of the others.

The first PM selects one not already selected, with probability 1.
The chance that the second PM picks one that the first one did not is
(1/3 + 1/3) = 2/3.

The chance that the second PM selects one that the first one did not, AND
that the third one picks the one neither of the other two did is
(2/3)*(1/3) = 2/9.

This is the same kind of logic behind the "birthday paradox", in which one can show that the chance of some pair of people in a group of size 25 or so
having the same birthday is nearly 1/2.

Let me tell you about the Monty Hall problem sometime...

Grizz

Yep, that's exactly what I figured for the probability. That makes the odds, the ratio of the probability of the event and its complement, 2/9 : 7/9 which is 2:7.

3. Wow. 14 minutes and they are gone. Impressive.