alex_zag_al comments on What is Evidence? - Less Wrong

60 Post author: Eliezer_Yudkowsky 22 September 2007 06:43AM

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Comment author: Gray_Area 22 September 2007 09:15:37AM 3 points [-]

Why not just say e is evidence for X if P(X) is not equal to P(X|e)?

Incidentally, I don't really see the difference between probabilistic dependence (as above) and entanglement. Entanglement is dependence in the quantum setting.

Comment author: alex_zag_al 16 September 2012 01:44:18AM *  0 points [-]

That definition does not always coincide with what is described in the article; something can be evidence even if P(X|e) = P(X).

Imagine that two cards from a shuffled deck are placed face-down on a table, one on the left and one on the right. Omega has promised to put a monument on the moon iff they are the same color.

Omega looks at the left card, and then the right, and then disappears in a puff of smoke.

What he does when he's out of sight is entangled with the identity of the card on the right. Change the card to one of a different color and, all else being equal, Omega's action changes.

But, if you flip over the card on the right and see that it's red, that doesn't change the degree to which you expect to see the monument when you look through your telescope. P(monument|right card is red) = P(monument) = 25/51

It does change your conditional beliefs, though, such as what the world would be like if the left card turned out to also be red: P(monument|left is red & right is red) > P(monument|left is red)

Comment author: moshez 31 December 2012 11:40:50AM 0 points [-]

Of course e can be evidence even if P(X|e)=P(X) -- it just cannot be evidence for X. It can be evidence for Y if P(Y|e)>P(Y), and this is exactly the case you describe. If Y is "there is a monument and left is red or there is no monument and left is black", then e is (infinite, if Omega is truthful with probability 1) evidence for Y, even though it is 0 evidence for X.

Similarly, you watching your shoelace untied is zero evidence for my shoelaces...