Wow, thanks Tim!

Assuming the same values from your post (2 premises each as likely as .7), except this time we add Argument C with the same number of premises and the same probability for each premise, would this be the correct mathematical representation:

~A1 & A2 & ~B1 & B2 & ~C1 & C2

~A1 & A2 & ~B1 & B2 & C1 & ~C2

~A1 & A2 & B1 & ~B2 & ~C1 & C2

~A1 & A2 & B1 &~ B2 & C1 & ~C2

Assuming the same values from your post (2 premises each as likely as .7), except this time we add Argument C with the same number of premises and the same probability for each premise, would this be the correct mathematical representation:

~A1 & A2 & ~B1 & B2 & ~C1 & C2

~A1 & A2 & ~B1 & B2 & C1 & ~C2

~A1 & A2 & B1 & ~B2 & ~C1 & C2

~A1 & A2 & B1 &~ B2 & C1 & ~C2

A1 & ~A2 & ~B1 & B2 & ~C1 & C2

A1 & ~A2 & ~B1 & B2 & C1 & ~C2

A1 & ~A2 & B1 & ~B2 & ~C1 & C2

A1 & ~A2 & B1 &~ B2 & C1 & ~C2

8(.3x.7x.3x.7x.3x.7)= .074088

~A1 & ~A2 & ~B1 & B2 & C1 & ~C2

~A1 &~ A2 & ~B1 & B2 & ~C1 & C2

~A1 & ~A2 & B1 & ~B2 & ~C1 & C2

~A1 & ~A2 & B1 & ~B2 & C1 & ~C2

~A1 & A2 & ~B1 & ~B2 & ~C1 & C2

~A1 & A2 & ~B1 & ~B2 & C1 & ~C2

A1 & ~A2 & ~B1 & ~B2 & ~C1 & C2

A1 & ~A2 & ~B1 & ~B2 & C1 & ~C2

~A1 & A2 & ~B1 & B2 &~ C1 & ~C2

~A1 & A2 & B1 & ~B2 &~ C1 & ~C2

A1 & ~A2 & ~B1 & B2 & ~C1 &~ C2

A1 & ~A2 & B1 & ~B2 &~ C1 & ~C2

12 (.3x.3x.3x.7x.3x.7)= .047628

~A1 &~ A2 & ~B1 & ~B2 & ~C1 & C2

~A1 &~ A2 & ~B1 & ~B2 & C1 & ~C2

~A1 & ~A2 & ~B1 & B2 & ~C1 & ~C2

~A1 &~ A2 & B1 & ~B2 & ~C1 & ~C2

~A1 & A2 & ~B1 & ~B2 & ~C1 & ~C2

A1 &~ A2 & ~B1 &~ B2 & ~C1 & ~C2

6 (.3x.3x.3x.3x.3x.7) = .010206

~A1 &~ A2 & ~B1 & ~B2 &~ C1 &~ C2

1 (.3x.3x.3x.3x.3x.3) = .000729

Kevin

6/01/2011 5:28 PM Tim said... Kevin,

It's late at night and my eyes are blurring a bit, so I might have missed something, but it looks to me like you've done it right. Remember that this calculation is made simple by the constraints I listed. The independence assumption is critical; without it, the whole problem becomes more difficult (indeed, without further information of some kind as to the direction and degree of dependence, insoluble). So it's not a one-size-fits-all schema. But it's very useful to see how these limiting cases work out.

A1 & ~A2 & ~B1 & B2 & C1 & ~C2

A1 & ~A2 & B1 & ~B2 & ~C1 & C2

A1 & ~A2 & B1 &~ B2 & C1 & ~C2

8(.3x.7x.3x.7x.3x.7)= .074088

~A1 & ~A2 & ~B1 & B2 & C1 & ~C2

~A1 &~ A2 & ~B1 & B2 & ~C1 & C2

~A1 & ~A2 & B1 & ~B2 & ~C1 & C2

~A1 & ~A2 & B1 & ~B2 & C1 & ~C2

~A1 & A2 & ~B1 & ~B2 & ~C1 & C2

~A1 & A2 & ~B1 & ~B2 & C1 & ~C2

A1 & ~A2 & ~B1 & ~B2 & ~C1 & C2

A1 & ~A2 & ~B1 & ~B2 & C1 & ~C2

~A1 & A2 & ~B1 & B2 &~ C1 & ~C2

~A1 & A2 & B1 & ~B2 &~ C1 & ~C2

A1 & ~A2 & ~B1 & B2 & ~C1 &~ C2

A1 & ~A2 & B1 & ~B2 &~ C1 & ~C2

12 (.3x.3x.3x.7x.3x.7)= .047628

~A1 &~ A2 & ~B1 & ~B2 & ~C1 & C2

~A1 &~ A2 & ~B1 & ~B2 & C1 & ~C2

~A1 & ~A2 & ~B1 & B2 & ~C1 & ~C2

~A1 &~ A2 & B1 & ~B2 & ~C1 & ~C2

~A1 & A2 & ~B1 & ~B2 & ~C1 & ~C2

A1 &~ A2 & ~B1 &~ B2 & ~C1 & ~C2

6 (.3x.3x.3x.3x.3x.7) = .010206

~A1 &~ A2 & ~B1 & ~B2 &~ C1 &~ C2

1 (.3x.3x.3x.3x.3x.3) = .000729

*Maximum uncertainty for belief in God based on three arguments with 2 premises; with each conclusion being only .49 likely is (1-.132651)***Thanks,**

**= greater than or equal to .867349 percent.**Kevin

6/01/2011 5:28 PM Tim said... Kevin,

It's late at night and my eyes are blurring a bit, so I might have missed something, but it looks to me like you've done it right. Remember that this calculation is made simple by the constraints I listed. The independence assumption is critical; without it, the whole problem becomes more difficult (indeed, without further information of some kind as to the direction and degree of dependence, insoluble). So it's not a one-size-fits-all schema. But it's very useful to see how these limiting cases work out.