Whoa. Useful. 0o Thankyou.Expand full comment
 James, to end up with a 39% posterior on the coin being heads-weighted, must have seen four heads and one tail:P(four heads and one tail| heads-weighted) = (0.75^4 ∙ 0.25^1) = 0.079. P(four heads and one tail | fair) = 0.031. P(heads-weighted | five heads) = (0.2∙0.079)/(0.2∙0.079 + 0.8∙0.031) = 0.39, which is the posterior belief James reportsI think most of these numbers should strictly speaking be multiplied by five. The end result is the same, but we should say e.g.P(four heads and one tail| heads-weighted) = (0.75^4 ∙ 0.25^1 ∙ 5) = .396I mention this only in case anyone else saw the maths and thought "huh, that's not right". Agree with the end product, though, and great post.Expand full comment
 Since people are asking for help, I'll take the liberty of asking for help on the problem of the cab fare and the extra twenty.Expand full comment
 Great post! I would like to read more posts like this one dependent on math and less fiction.Expand full comment