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) = x1/(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) = x2x/(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.
Can you elaborate on the calculation for S? I think it should be this, but I’m not confident in my math.
Yours was correct; editing the post. I skipped a step and that made my previous answer wrong.
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.
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) = x1/(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) = x2x/(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.