Wei_Dai comments on False Majorities - Less Wrong

35 Post author: JamesAndrix 03 February 2010 06:43PM

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Comment author: Wei_Dai 03 February 2010 08:06:15PM 7 points [-]

I suggest linking to the previous discussion. In particular, Toby Ord had a counterargument, that I don't think you adequately dealt with. You wrote:

If the red supporter contends that all green and blue objects were lost in the color wars, while the blue supporter contends that all objects are fundamentally blue and besides the color wars never happened, then their opinions roughly cancel each other out. (Barring other reasons for me to view one as more rational than the other.)

I don't see why they cancel each other out. Why shouldn't you assign 1/3 probability to "all green and blue objects were lost in the color wars" and 1/3 probability to "all objects are fundamentally blue and besides the color wars never happened", in which case there's 2/3 probability that the object is not green?

Comment author: JamesAndrix 04 February 2010 01:38:57AM 1 point [-]

So Edited.

Cancel was too strong a word.

It depends on how green justifies it's position, and how that is taken by the other experts.

Suppose also that the Green expert disbelieves both the color wars and fundamental Blueness, and supports green for scientific reasons whose facts are not strongly disputed by the other two sides. 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.

The green expert has support from 1 or 2 other experts in every reason they hold. The red and blue experts have support from 0 or 1 other experts, and 2 in the case of the evidence for green.

The green expert is more authoritative, because more experts think he's not-crazy on more positions within the field. (that no one disputes green's basic logic is a bonus, even if The Funadmentalist Blues disavowed the 'so-called science', green would still have 1 corroborating opinion for all of their positions.)

It may be the case in this example that not-green would still be the weighted majority position, but at less than 2/3. I'm not sure how to do the math on this.

Comment author: Wei_Dai 04 February 2010 11:35:19AM *  1 point [-]

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.

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...

Comment author: wedrifid 04 February 2010 02:58:12PM 0 points [-]

It may be the case in this example that not-green would still be the weighted majority position, but at less than 2/3. I'm not sure how to do the math on this.

I would just like to say: Maths is hard work. Not because it is particularly difficult in this case but because I have filled up a page worth of typing with underscores and parenthesis and aren't even finished. So much boring detail!

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