So if we work with Beth's original question (see my last post), then the chances would be 8/4096?
Right? I hope so.
Because next I want to discuss how our biases towards certain players enter into the equation. I mean, There are four players to choose from at each position, but there is not an equal chance of any of the four being picked. It's not like flipping coins, where the probability of either outcome is the same.
For example: Duncan is one of four players, but he might have a 75% chance of being picked, while with Nash and Marbury in the mix, Tony Parker has maybe a 15-20% chance of being picked.
If I said "what are the chances of two people (in a group of two, for now) picking a team with players A, B, C, D, E, and F (Name the 6 players with the lowest V. AVG), the chances would be greater than one in 16 million.
But if I said "what are the chances of two people picking a team with Duncan or Stoudamire, Marbury or Nash, Nowitzki or Marion, Finley, Ginobili, or McGrady and any one of four centers and four coaches, the chances would be much less.
So to answer Beth's question, it's really more like 2 x 2 x 2 x 3 x 4 x 4, or 384-1 divided by the group size of 8, or about 1/48.
But because some of the above combos are impossible because of the cap (No, I am not going to run through all 384 to find out how many there are), lets call it about 1 chance in 24.
