= 27d567bc2d48d9f40e9ccc20ea370b15)
JGWeissman comments on Open Thread: July 2010 - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (653)
From Bob's perspective, he was more likely to be chosen as the one to talk to Al, if there are fewer scientist that observed exactly 25 big fish, which would happen if there are more big fish. So Bob should update on the evidence of being chosen.
This should be important to the finite case. The probability of being the first to see 25/100 is WAY higher (x 10^25 or so) if the lake is 3/4 full of big fish than if it is 1/4 full of big fish.
But in the infinite case the probability of being first is 0 either way...
There is a reason we consider infinities only as limits of sequences of finite quantities.
Suppose you tried to sum the log-odds evidence of the infinite scientist that the pond has more big fish. Well, some of them have positive evidence (summing to positive infinity), some have negative evidence (summing to negative infinity), and you can, by choosing the order of summation, get any result you want (up to some granularity) between negative and positive infinity.
You don't need anthropomorphic tricks to make things weird if you have actual infinities in the problem.
utilitymonster:
Maybe I'm misunderstanding your phrasing here, but it sounds fallacious. If there's a deck of cards and you're in a group of 52 people who are called out in random order and told to pick one card each from the deck, the probability of being the first person to draw an ace is exactly the same (1/52) regardless of whether it's a normal deck or a deck of 52 aces (or even a deck with 3 out of 4 aces replaced by other cards). This result doesn't even depend on whether the card is removed or returned into the deck after each person's drawing; the conclusion follows purely from symmetry. The only special case is when there are zero aces, in which the event becomes impossible, with p=0.
Similarly, if the order in which the scientists get their samples is shuffled randomly, and we ignore the improbable possibility that nobody sees 25/100, then purely by symmetry, the probability that Bob happens to be the first one to see 25/100 is the same regardless of the actual frequency of the 25/100 results: p = 1/N(scientists).
You're right, thanks.
I was considering an example with 10^100 scientists. I thought that since there would be a lot more scientists who got 25 big in the 1/4 scenario than in the 3/4 scenario (about 9.18 * 10^98 vs. 1.279 * 10^75), you'd be more likely to be first the 3/4 scenario. But this forgets about the probability of getting an improbable result.
In general, if there are N scientists, and the probability of getting some result is p, then we can expect Np scientists to get that result on average. If the order is shuffled as you suggest, then the probability of being the first to get that result is p * 1/(Np) = 1/N. So the probability of being the first to get the result is the same, regardless of the likelihood of the result (assuming someone will get the result).
EDIT: It occurs to me that I might have been thinking about the probability of being selected by Al conditional on getting 25/100. In that case, you're a lot more likely to be selected if the pond is 3/4 big than if it is 1/4 big, since WAY more people got similar results in the latter case. JGMWeissman was probably thinking the same.