A puzzle: what happens in the blue-eyed islanders puzzle if we allow a small probability p for each each person to make a mistake each night? Assume that the probability of a mistake is the same for each and is common knowledge, the islanders are all ideal Bayesian reasoners, and let's say that 80% is the confidence cutoff past which someone leaves.

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I mean, if they're Bayesian reasoners then they see that everyone else has blue eyes and start to notice a pattern. If there's 100 islanders, each has 99 bits of evidence favoring "everyone has the same eye color" over "eye color is a fair coin".

So let's assume they have some perverse prior, and they can't update away from regarding eye color as a fair coin flip between blue and brown.

Next, I'm not sure what you mean by mistake. If the two actions are stay and go, then we might define mistake as doing the one of those opposite to what our Bayesian reasoner would do. Then, if there's a p probability of mistake, this is the same as a 2p probability of just acting randomly.

So now what I'm imagining is someone shows up and tells them that someone has blue eyes, and now each night a fraction p of the islanders think "I know where this is going, and I'm gonna get out before the rush. See ya!"

So I need a lemma - what happens in the original problem if N islanders decide to arbitrarily leave immediately? Does the timeline change?

If there are just three people and you hypothetically have brown eyes, the normal course is that you expect the other two to look at each other the first night, and then when they don't leave the second night you know you have blue eyes. If one person leaves immediately, and you have brown eyes, the remaining other person will think they're safe and not leave - and so neither of you end up leaving. And in the four person case, if you have brown eyes, everyone else follows the three-person logic and doesn't leave, etc etc.

So this "probability of mistake" thing is looking more like "probability of heroic self-sacrifice."

If p is small and not enough people leave to change the conclusion, then the only informative night is still the 99th, Bayesian reasoners or no. Then if you see the usual n*p people leave (n being the number remaining) you just need to compare the probabilities of np people leaving mistakenly, and n(1-p) staying mistakenly. With high probability, it was the smaller number making the mistake, so people leave on the 100th night if p is small.

Note that some mistakes are correctable - if you see 99 blue eyes, and two of them leave on the next boat, you know those were errors and don't change your behavior at all - you're still uncertain between 99 and 100 blue eyes (originally) on the island.

I think this means that the logic remains the same as long as the error rate is low enough that you can distinguish errors from group-similar actions. For instance, if so many leave that there are no other blue eyes visible by day 98, you will get no further evidence on whether your eyes are blue or not.

edit: on reading your comment more closely, this is mostly a repetition.