Because if you write down the Bayes’ Rule calculation, that’s not the ratio that appears in it.
HTTTHHHTHTHTTHHTTHTHTTHTHHHTHHTTTHTH. Using Bayes’ Rule, what are the odds I actually got that sequence, as opposed to randomly typing letters? (If you miss my point: You’re misusing Bayes’ Rule in this argument.)
Nope. They both mean: for large n, for a fraction of sequences that tends to 1 as n → infinity, that’s what happens.
If Alice cheats 100% of the time, your formula produces probabilities greater than 1 for any n less than infinity, which I’m reasonably certain doesn’t happen.
Using Bayes’ Rule, what are the odds I actually got that sequence, as opposed to randomly typing letters?
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’re misusing Bayes’ Rule in this argument.
I think you may be misunderstanding my argument.
If Alice cheats 100% of the time, your formula produces probabilities greater than 1
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?
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.
Because if you write down the Bayes’ Rule calculation, that’s not the ratio that appears in it.
Nope. They both mean: for large n, for a fraction of sequences that tends to 1 as n → infinity, that’s what happens.
HTTTHHHTHTHTTHHTTHTHTTHTHHHTHHTTTHTH. Using Bayes’ Rule, what are the odds I actually got that sequence, as opposed to randomly typing letters? (If you miss my point: You’re misusing Bayes’ Rule in this argument.)
If Alice cheats 100% of the time, your formula produces probabilities greater than 1 for any n less than infinity, which I’m reasonably certain doesn’t happen.
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.
I think you may be misunderstanding my argument.
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?
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.