# malthrin comments on Uncertainty - Less Wrong

19 29 November 2011 11:12PM

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Comment author: 30 November 2011 08:47:39PM *  1 point [-]

Can you elaborate on the calculation for S? I think it should be this, but I'm not confident in my math.

Comment author: 30 November 2011 10:38:20PM 1 point [-]

Yours was correct; editing the post. I skipped a step and that made my previous answer wrong.

Comment author: 11 December 2011 11:29:32PM 0 points [-]

Maybe I'm missing something obvious here, but I'm unsure how to calculate P(S). I'd appreciate it if someone could post an explanation.

Comment author: 12 December 2011 05:00:30PM *  2 points [-]

Sure. S results from HH or from TT, so we'll calculate those independently and add them together at the end. We'll do that by this equation: P(p=x|S) = P(p=x|HH) * P(H) + P(p=x|TT) * P(T).

We start out with a uniform prior: P(p=x) = 1. After observing one H, by Bayes' rule, P(p=x|H) = P(H|p=x) * P(p=x) / P(H). P(H|p=x) is just x. Our prior is 1. P(H) is our prior, multiplied by x, integrated from 0 to 1. That's 1/2. So P(p=x|H) = x*1/(1/2) = 2x.

Apply the same process again for the second H. Bayes' rule: P(p=x|HH) = P(H|p=x,H) * P(p=x|H) / P(H|H). The first term is still just x. The second term is our updated belief, 2x. The denominator is our updated belief, multiplied by x, integrated from 0 to 1. That's 2/3 this time. So P(p=x|HH) = x*2x/(2/3) = 3x^2.

Calculating tails is similar, except we update with 1-x instead of x. So our belief goes from 1, to 2-2x, to 3x^2-6x+3. Then substitute both of these into the original equation: (3/2)(x^2) + (3/2)(x^2 - 2x + 1). From there it's just a bit of algebra to get it into the form I linked to.