[durtyrute]

Originally Posted by Rain Man

So if I look at those groups I referred to earlier, let's assume the following:

In Group 1, Manning, Rivers, McNabb, and Whiny Manning are franchise QBs. That's a stretch with Rivers and Whiny, but let's give it to them. So 4 of the 14 are franchise QBs, meaning you had a 28 percent chance of success.

In Group 2, Roethlisberger is a franchise QB. 1 in 7 chance, or 14 percent. We'll ignore Cutler and Culpepper to be conserve, though one could make a case that they're equivalent to the whiny Manning.

In Group 3, Rodgers is a franchise QB. 1 in 6 chance, or 17 percent.

In Group 4, Brees is a franchise QB. 1 in 7 chance or 14 percent.

So you take a top 10 pick and you have a 28 percent chance of success.

If you draft a QB from Group 2, Group 3, and 2 from Group 4, the odds of getting a franchise QB are:

1 - [6/7*5/6*6/7*6/7) = 48%

Would you rather have a 48% chance of winning or a 28% chance of winning?

I like where you are going, but you can't do the math like that. It would be a 28% chance vs a 14 or 17% chance. You can't add up the percentages unless you pick a qb in each of those rounds.