Actually, DB (jump in anywhere), I think the answer is pretty easy once you know the number of possible combinations.
Now, I'm not about to figure that out (If I take Marbury, I can't take Bryant, etc., etc.), bit for the sake of argument let's say this:
18 teams are playing tonight, so there are 18 x 18 ways to pick a backcourt, or 324 possible combinations (we'll just use starters--no back-ups). Let's assumes that half of them, or 162 are not possible because of cap restrictions.
Now we can "back into" the answer (jump in anytime, DB).
If there are 162 possible combos, we would need 163 people playing to guarantee that at least two would have the same backcourt. So if 162 people were playing, the chances of two of them having the same picks would be 161/162, or 99.4% (jump in anywhere, DB). That means there would be a .6% chance of all 162 picking a different backcourt.
So, with 5 players, the chances of at least two of us having the same picks would be 4/162, or 2.47%.
And the chances of none of us picking the same combination would be 97.53%
I'm pretty certain that mathematically, that answer is correct. However, it doesn't allow for combinations that would never get picked because you would be so far below the cap that it would be ridiculous to pick those two guys together.
For example, who would combine Utah's DeShawn Stevenson (cap value 6.9, VGA 9.3) with Miami's Rafer Alston (9.5, 9.7)?
It's a mathematically doable combo under the cap restrictions, but picking those two with Brand, Garnett, and Dampier would still leave you about 20 points below the cap, and with Marbury, Cassell, Bryant, etc. available for at least one of those spots (and you would still be under the cap), why combine Stevenson & Alston?
Anyway, I have a feeling that there were no more than about 48 combinations that had an actual chance of being selected, so the chances of all five of us having a different backcourt would be 44/48, or 91.67%
