= 27d567bc2d48d9f40e9ccc20ea370b15)
cupholder comments on Avoiding doomsday: a "proof" of the self-indication assumption - 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 (228)
Saw this come up in Recent Comments, taking the opportunity to simultaneously test the image markup and confirm Academian's Bayesian answer using boring old frequentist probability. Hope this isn't too wide... (Edit: yup, too wide. Here's a smaller-albeit-busier-looking version.)
Cupholder:
That is an excellent illustration ... of the many-worlds (or many-trials) case. Frequentist counting works fine for repeated situations.
The one-shot case requires Bayesian thinking, not frequentist. The answer I gave is the correct one, because observers do not gain any information about whether the coin was heads or tails. The number of observers that see each result is not the same, but the only observers that actually see any result afterwards are the ones in either heads-world or tails-world; you can't count them all as if they all exist.
It would probably be easier for you to understand an equivalent situation: instead of a coin flip, we will use the 1 millionth digit of pi in binary notation. There is only one actual answer, but assume we don't have the math skills and resources to calculate it, so we use Bayesian subjective probability.
Cupholder managed to find an analogous problem in which the Bayesian subjective probabilities mapped to the same values as frequentist probabilities, so that the frequentist approach really gives the same answer. Yes, it would be nice to just accept subjective probabilities so you don't have to do that, but the answer Cupholder gave is correct.
The analysis you label "Bayesian", on the other hand, is incorrect. After you notice that you have survived the killing you should update your probability that coin showed tails to
so you can then calculate
Or, as Academian suggested, you could have just updated to directly find
I disagree, but I am inclined to disagree by default: one of the themes that motivates me to post here is the idea that frequentist calculations are typically able to give precisely the same answer as Bayesian calculations.
I also see no trouble with wearing my frequentist hat when thinking about single coin flips: I can still reason that if I flipped a fair coin arbitrarily many times, the relative frequency of a head converges almost surely to one half, and that relative frequency represents my chance of getting a head on a single flip.
I believe that the observers who survive would. To clarify my thinking on this, I considered doing this experiment with a trillion doors, where one of the doors is again red, and all of the others blue. Let's say I survive this huge version of the experiment.
As a survivor, I know I was almost certainly behind a blue door to start with. Hence a tail would have implied my death with near certainty. Yet I'm not dead, so it is extremely unlikely that I got tails. That means I almost certainly got heads. I have gained information about the coin flip.
I think talking about 'observers' might be muddling the issue here. We could talk instead about creatures that don't understand the experiment, and the result would be the same. Say we have two Petri dishes, one dish containing a single bacterium, and the other containing a trillion. We randomly select one of the bacteria (representing me in the original door experiment) to stain with a dye. We flip a coin: if it's heads, we kill the lone bacterium, otherwise we put the trillion-bacteria dish into an autoclave and kill all of those bacteria. Given that the stained bacterium survives the process, it is far more likely that it was in the trillion-bacteria dish, so it is far more likely that the coin came up heads.
I don't think of the pi digit process as equivalent. Say I interpret 'pi's millionth bit is 0' as heads, and 'pi's millionth bit is 1' as tails. If I repeat the door experiment many times using pi's millionth bit, whoever is behind the red door must die, and whoever's behind the blue doors must survive. And that is going to be the case whether I 'have the math skills and resources to calculate' the bit or not. But it's not going to be the case if I flip fair coins, at least as flipping a fair coin is generally understood in this kind of context.
That would be like repeating the coin version of the experiment many times, using the exact same coin (in the exact same condition), flipping it in the exact same way, in the exact same environment. Even though you don't know all these factors of the initial conditions, or have the computational power to draw conclusions from it, the coin still lands the same way each time.
Since you are willing to suppose that these initial conditions are different in each trial, why not analogously suppose that in each trial of the digit of pi version of the experiment, that you compute a different digit of pi. or, more generally, that in each trial you compute a different logical fact that you were initially completely ignorant about.?
Yes, I think that would work - if I remember right, zeroes and ones are equally likely in pi's binary expansion, so it would successfully mimic flipping a coin with random initial conditions. (ETA: this is interesting. Apparently pi's not yet been shown to have this property. Still, it's plausible.)
This would also work, so long as your bag of facts is equally distributed between true facts and false facts.
That's probably why you don't understand the result; it is an anthropic selection effect. See my reply to Academician above.
That is not an analogous experiment. Typical survivors are not pre-selected individuals; they are post-selected, from the pool of survivors only. The analogous experiment would be to choose one of the surviving bacteria after the killing and then stain it. To stain it before the killing risks it not being a survivor, and that can't happen in the case of anthropic selection among survivors.
That's because you erroneously believe that your frequency interpretation works. The math problem has only one answer, which makes it a perfect analogy for the 1-shot case.
Okay.
I believe that situations A and B which you quote from Stuart_Armstrong's post involve pre-selection, not post-selection, so maybe that is why we disagree. I believe that because the descriptions of the two situations refer to 'you' - that is, me - which makes me construct a mental model of me being put into one of the 100 rooms at random. In that model my pre-selected consciousness is at issue, not that of a post-selected survivor.
By 'math problem' do you mean the question of whether pi's millionth bit is 0? If so, I disagree. The 1-shot case (which I think you are using to refer to situation B in Stuart_Armstrong's top-level post...?) describes a situation defined to have multiple possible outcomes, but there's only one outcome to the question 'what is pi's millionth bit?'
Presumably you heard the announcement.
This is post-selection, because pre-selection would have been "Either you are dead, or you hear that whoever was to be killed has been killed. What are your odds of being blue-doored now?"
There's only one outcome in the 1-shot case.
The fact that there are multiple "possible" outcomes is irrelevant - all that means is that, like in the math case, you don't have knowledge of which outcome it is.
The 'selection' I have in mind is the selection, at the beginning of the scenario, of the person designated by 'you' and 'your' in the scenario's description. The announcement, as I understand it, doesn't alter the selection in the sense that I think of it, nor does it generate a new selection: it just indicates that 'you' happened to survive.
I continue to have difficulty accepting that the millionth bit of pi is just as good a random bit source as a coin flip. I am picturing a mathematically inexperienced programmer writing a (pseudo)random bit-generating routine that calculated the millionth digit of pi and returned it. Could they justify their code by pointing out that they don't know what the millionth digit of pi is, and so they can treat it as a random bit?
Not seriously: http://www.xkcd.com/221/
Seriously: You have no reason to believe that the millionth bit of pi goes one way or the other, so you should assign equal probability to each.
However, just like the xkcd example would work better if the computer actually rolled the die for you every time rather than just returning '4', the 'millionth bit of pi' algorithm doesn't work well because it only generates a random bit once (amongst other practical problems).
In most pseudorandom generators, you can specify a 'seed' which will get you a fixed set of outputs; thus, you could every time restart the generator with the seed that will output '4' and get '4' out of it deterministically. This does not undermine its ability to be a random number generator. One common way to seed a random number generator is to simply feed it the current time, since that's as good as random.
Looking back, I'm not certain if I've answered the question.
I think so: I'm inferring from your comment that the principle of indifference is a rationale for treating a deterministic-but-unknown quantity as a random variable. Which I can't argue with, but it still clashes with my intuition that any casino using the millionth bit of pi as its PRNG should expect to lose a lot of money.
I agree with your point on arbitrary seeding, for whatever it's worth. Selecting an arbitrary bit of pi at random to use as a random bit amounts to a coin flip.
I'd be extremely impressed if a mathematically inexperienced programmer could pull of a program that calculated the millionth digit of pi!
I say yes (assuming they only plan on treating it as a random bit once!)
If 'you' were selected at the beginning, then you might not have survived.
Yeah, but the description of the situation asserts that 'you' happened to survive.
Adding that condition is post-selection.
Note that "If you (being asked before the killing) will survive, what color is your door likely to be?" is very different from "Given that you did already survive, ...?". A member of the population to which the first of these applies might not survive. This changes the result. It's the difference between pre-selection and post-selection.
FWIW, it's not that hard to calculate binary digits of pi:
http://oldweb.cecm.sfu.ca/projects/pihex/index.html
I think I'll go calculate the millionth, and get back to you.
EDIT: also turns out to be 0.