Pretending for the sake of argument that I don’t see any regularities in your sequence that I wouldn’t expect from genuinely random coin flips (it actually looks to me more human-generated, but with only n=36 I’m not very confident of that): the odds are pretty much the same as the prior odds that you’d actually flip a coin 36 times rather than just writing down random-looking Hs and Ts.
You said something interesting there, and then skipped right past it. That’s the substance of the question. You don’t get to ignore those regularities; they do, in fact, affect the probabilities. Saying that they don’t appear in the ratio of Bayes’ Rule is… well, misusing Bayes’ Rule to discard meaningful evidence.
The only formula I wrote down was “2^-n times Pr(Alice cheated)” and those probabilities are definitely not greater than 1. Would you care to be more explicit?
2^(-n) approaches 1 as n approaches infinity, but for any finite n, is greater than 1. Multiply that by a probability of 1, and you get a probability greater than 1. [ETA: Gyah. It’s been too long since I’ve done exponents (literally, ten years since I’ve done anything interesting). You’re right, I’m confusing negative exponents with division in exponents.]
Saying that they don’t occur in the ratio of Bayes’ Rule [...]
But I didn’t say that. I didn’t say anything even slightly like that.
This is at least partly my fault because I was too lazy to write everything out explicitly. Let me do so now; perhaps it will clarify. Suppose X is some long random-looking sequence of n heads and tails.
Odds(Alice cheated : Alice flipped honestly | result was X) = Odds(Alice cheated : Alice flipped honestly) . Odds(result was X | Alice cheated : Alice flipped honestly).
The second factor on the RHS is, switching from my eccentric but hopefully clear notation to actual probability ratios, Pr(result was X | Alice cheated) / Pr(result was X | Alice flipped honestly).
So those two probabilities are the ones you have to look at, not Pr(Alice cheated) and Pr(result was X | Alice flipped honestly). But the latter is what you were comparing when you wrote
it’s -always- more likely that Alice cheated than that this particular sequence came up by chance.
which is why I said that was the wrong comparison.
You said something interesting there, and then skipped right past it. That’s the substance of the question. You don’t get to ignore those regularities; they do, in fact, affect the probabilities. Saying that they don’t appear in the ratio of Bayes’ Rule is… well, misusing Bayes’ Rule to discard meaningful evidence.
2^(-n) approaches 1 as n approaches infinity, but for any finite n, is greater than 1. Multiply that by a probability of 1, and you get a probability greater than 1. [ETA: Gyah. It’s been too long since I’ve done exponents (literally, ten years since I’ve done anything interesting). You’re right, I’m confusing negative exponents with division in exponents.]
But I didn’t say that. I didn’t say anything even slightly like that.
This is at least partly my fault because I was too lazy to write everything out explicitly. Let me do so now; perhaps it will clarify. Suppose X is some long random-looking sequence of n heads and tails.
Odds(Alice cheated : Alice flipped honestly | result was X) = Odds(Alice cheated : Alice flipped honestly) . Odds(result was X | Alice cheated : Alice flipped honestly).
The second factor on the RHS is, switching from my eccentric but hopefully clear notation to actual probability ratios, Pr(result was X | Alice cheated) / Pr(result was X | Alice flipped honestly).
So those two probabilities are the ones you have to look at, not Pr(Alice cheated) and Pr(result was X | Alice flipped honestly). But the latter is what you were comparing when you wrote
which is why I said that was the wrong comparison.