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.
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.