[BruceZ]

[ QUOTE ]
The "Expected number of pairs picked = 10 * the chance that any given pair is picked" argument is correct.

The chance of picking one particular pair, however, is 18C4 / 20C6 = 3060 / 38760, rather than the number quoted by the other responders.

[/ QUOTE ]

Damn all those other responders, how could they all have missed that? [img]/images/graemlins/grin.gif[/img] I would have figured that out when I fixed the combinatorics in the original post, which I still plan to do later tonight.

Edit: Looks like you already found the combinatoric problem.