I figured out the 4096 x 4096 problem, DB.

For that to be the correct answer, you need three people involved. I pick a roster. The odds of CC picking the same roster is 1 x 4096. The odds of SP picking the same roster as the two of us is 1 x 4096 x 4096.

The way to phrase the question so it involves only two people would be something like this, I think:

What are the odds of two people picking the same predetermined roster? Or to put it another way, what are the odds of two players picking a roster composed of players A, B, C, D, E, and F, as opposed to any six random players?

But here's something else we didn't factor in which greatly reduces the odds: The number of players in our group.

In a group of two, the odds are 4096-1. In a group of eight, however, the odds are much lower. I mean in a group of 3000-4000, the odds of two people picking the same random roster are actually very low, because in a group of 4097, it would be an absolute certainty that at least two people would have the same picks.

So what do you do? Divide the 4096 by the size of the group?

(Note to FS: Spare me the comments about "Get a Life" or "Too Much Time On My Hands" please )


"Difficult....not impossible"