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Seth_Goldin comments on Open Thread: January 2010 - Less Wrong
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Mike Gibson has a great and interesting question. How would Bayesian methodology address this? Might this be an information cascade?
Yes, that would be an information cascade.
In the toy problem in the link, as long as we know the rule that people use to write down their guesses (e.g., write down the hypothesis with maximum posterior probability; if 50-50, write down what the last person wrote,) at each stage we can treat the previous sequence as a latent variable about which we have partial information. The solution is straightforward to set up.
My intuition is that if you assume everyone before you has written down the correct most likely answer based on the sequence they observe (and using the same assumption) then you fairly quickly reach a point where additional people's guesses add no new information. Can anyone confirm or refute that and save me trying to do the math?
If the tiebreak strategy is "agree with the previous person's guess", then you reach that point immediately. The first person's draw determines everyone's guess: If the second person's draw is the same as the first, then of course they agree, and if not then they're at a 50/50 posterior and thus also agree.
If the tiebreak strategy is "write down your own draw (i.e. maximize the information given to subsequent players)", then information can be collected only so long as the number of each color drawn remains tied or +/-1. As soon as one color is ahead by 2 draws, all future draws are ignored and the guesses so far suffice to determine everyone else's guess.
If the draws are with replacement, then the probability that what you get locked into is the right guess is 4/5. (Assume WLOG that the urn is primarily white. Consider two draws: WW is 4/9 and determines the right answer; RR is 1/9 and determines the wrong answer; WR or RW have no net likelihood change, so recurse.)
If the draws are without replacement, then it's... 80.6%. (Very close to 4/5 since with very high probability you'll run into a cascade one way or the other before the non-replacement changes the ball proportions much.)
Otoh, "tally all the votes at the end of the pulling, and that determines the group’s Urn choice" is an entirely different question, and doesn't have the same strategy as maximizing your individual chance of correctness.
Nice - I hadn't gotten so far as analyzing the other tiebreak policy.
"Prior information" in this kind of problem includes a bunch of rather unlikely assumptions, such as that every player is maximally rational and that the rules of the game reward picking the true choice of urn.
Unfortunately there is no reason to prefer one tiebreak policy over the other. Does it make the problem more determinate if we assume the game scores per Bayesian Truth Serum, that is, you get more points for a contrarian choice that happens to be right ?
Since the total evidence you can get from examining all previous guesses (assuming conventional strategy and rewards as before) gives you only a 4/5 accuracy, and you can get 2/3 by ignoring all previous guesses and looking only at your own draw: Yes, rewarding correct contrarians at least 20% more than correct majoritarians would provide enough incentive to break the information cascade. Only until you've accumulated enough extra information to make the majoritarian answer confident enough to overcome the difference between rewards, of course, but it would still equilibrate at a higher accuracy.
The math is pretty simple: as soon as the line has a red/blue discrepancy of more than one ball, ignore your ball and vote with the line.
Why not just do the math ?
Primarily because I'm at work and secondarily because I'm lazy.
Downvoted for laziness.