And P_G, P_R, P_G be the probability functions of the three experts:
P_G(G) = P_G(S) = 1
P_R(R) = P_R(W) = 1
P_B(B) = P_B(F) = 1
If we take P to be the average of the three probability functions, then P(G)=P(R)=P(B)=P(S)=P(W)=P(F)=1/3.
The Blue supporter acknowledges it would likely be green if not all things were blue, and the Red supporter the same if not for the color wars.
In that case, it would be something like this:
P_R(R) = P_R(W) = .99, P_R(G) = P_R(S) = .01
P_B(B) = P_B(F) = .99, P_B(G) = P_B(S) = .01
But if you take the average, P(G) still comes out pretty close to 1⁄3. In order to conclude that P(G)>1/2, I think we need to argue that taking the average of the 3 probability functions isn’t the right thing to do. I’m still trying to figure that one out...
Here is how I see the math. Let:
R = object is red
B = object is blue
G = object is green
W = color war
F = fundamentally blue
S = scientifically green
And P_G, P_R, P_G be the probability functions of the three experts:
P_G(G) = P_G(S) = 1
P_R(R) = P_R(W) = 1
P_B(B) = P_B(F) = 1
If we take P to be the average of the three probability functions, then P(G)=P(R)=P(B)=P(S)=P(W)=P(F)=1/3.
In that case, it would be something like this:
P_R(R) = P_R(W) = .99, P_R(G) = P_R(S) = .01
P_B(B) = P_B(F) = .99, P_B(G) = P_B(S) = .01
But if you take the average, P(G) still comes out pretty close to 1⁄3. In order to conclude that P(G)>1/2, I think we need to argue that taking the average of the 3 probability functions isn’t the right thing to do. I’m still trying to figure that one out...