jimrandomh comments on Conditioning on Observers - Less Wrong

6 Post author: Jonathan_Lee 11 May 2010 05:15AM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Comments (118)

You are viewing a single comment's thread. Show more comments above.

Comment author: neq1 11 May 2010 04:16:30PM 0 points [-]

"Of course P(W) isn't bound within [0,1]"

Of course! (?) You derived P(W) using probability laws, i.e., solving for it in this equation: P(H)=P(H|W)P(W), where P(H)=1/2 and P(H|W)=1/3. These are probabilities. And your 1/3 solution proves there is an error.

If two variables have correlation of 1, I think you could argue that they are the same (they contain the same quantitative information, at least).

On your argumentation, you would assert confidently to that the coin is fair beforehand, yet also assert that the conditional probability that you wake on Monday depends on the coin flip, when in either branch you are woken then with probability 1.

No. You will wake on Monday with probability one. But, on a randomly selected awakening, it is more likely that it's Monday&Heads than Monday&Tails, because you are on the Heads path on 50% of experiments

Comment author: jimrandomh 11 May 2010 05:09:33PM *  0 points [-]

P(H)=P(H|W)P(W), where P(H)=1/2 and P(H|W)=1/3

No, P(H)=P(H|W)P(W) is incorrect because the W in P(H|W) is different than the W in P(W): the former is a probability distribution over a set of three events, while the latter is a boolean. Using the former definition, as a probability distribution, P(W) isn't meaningful at all, it's just a type error.