Mallah comments on Avoiding doomsday: a "proof" of the self-indication assumption - Less Wrong

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Comment author: Mallah 30 March 2010 03:34:33AM -1 points [-]

A - A hundred people are created in a hundred rooms. Room 1 has a red door (on the outside), the outsides of all other doors are blue. You wake up in a room, fully aware of these facts; what probability should you put on being inside a room with a blue door?

Here, the probability is certainly 99%.

Sure.

B - same as before, but an hour after you wake up, it is announced that a coin will be flipped, and if it comes up heads, the guy behind the red door will be killed, and if it comes up tails, everyone behind a blue door will be killed. A few minutes later, it is announced that whoever was to be killed has been killed. What are your odds of being blue-doored now?

There should be no difference from A; since your odds of dying are exactly fifty-fifty whether you are blue-doored or red-doored, your probability estimate should not change upon being updated.

Wrong. Your epistemic situation is no longer the same after the announcement.

In a single-run (one-small-world) scenario, the coin has a 50% to come up tails or heads. (In a MWI or large universe with similar situations, it would come up both, which changes the results. The MWI predictions match yours but don't back the SIA). Here I assume the single-run case.

The prior for the coin result is 0.5 for heads, 0.5 for tails.

Before the killing, P(red|heads) = P(red|tails) = 0.01 and P(blue|heads) = P(blue|tails) = 0.99. So far we agree.

P(red|before) = 0.5 (0.01) + 0.5 (0.01) = 0.01

Afterwards, P'(red|heads) = 0, P'(red|tails) = 1, P'(blue|heads) = 1, P'(blue|tails) = 0.

P(red|after) = 0.5 (0) + 0.5 (1) = 0.5

So after the killing, you should expect either color door to be 50% likely.

This, of course, is exactly what the SIA denies. The SIA is obviously false.

So why does the result seem counterintuitive? Because in practice, and certainly when we evolved and were trained, single-shot situations didn't occur.

So let's look at the MWI case. Heads and tails both occur, but each with 50% of the original measure.

Before the killing, we again have P(heads) =P(tails) = 0.5

and P(red|heads) = P(red|tails) = 0.01 and P(blue|heads) = P(blue|tails) = 0.99.

Afterwards, P'(red|heads) = 0, P'(red|tails) = 1, P'(blue|heads) = 1, P'(blue|tails) = 0.

Huh? Didn't I say it was different? It sure is, because afterwards, we no longer have P(heads) = P(tails) = 0.5. On the contrary, most of the conscious measure (# of people) now resides behind the blue doors. We now have for the effective probabilities P(heads) = 0.99, P(tails) = 0.01.

P(red|after) = 0.99 (0) + 0.01 (1) = 0.01

Comment author: Academian 06 April 2010 09:07:08PM * 0 points [-]

P(red|after) = 0.5 (0) + 0.5 (1) = 0.5

So after the killing, you should expect either color door to be 50% likely.

No; you need to apply Bayes theorem here. Intuitively, before the killing you are 99% sure you're behind a blue door, and if you survive you should take it as evidence that "yay!" the coin in fact did not land tails (killing blue). Mathematically, you just have to remember to use your old posteriors as your new priors:

P(red|survival) = P(red)·P(survival|red)/P(survival) = 0.01·(0.5)/(0.5) = 0.01

So SIA + Bayesian updating happens to agree with the "quantum measure" heuristic in this case.

However, I am with Nick Bodstrom in rejecting SIA in favor of his "Observation Equation" derived from "SSSA", precisely because that is what maximizes the total wealth of your reference class (at least when you are not choosing whether to exist or create dupcicates).

Comment author: Mallah comment score below threshold [+] (40 children)

Comment author: Academian 07 April 2010 11:49:45AM * 0 points [-]

P(red|before) =0.01 is not equal to P(red).

I omitted the "|before" for brevity, as is customary in Bayes' theorem.

Cupholder's excellent diagram should help make the situation clear. Here is a written explanation to accompany:

R = "you are in a red room"

K = "at some time, everyone in a red/blue room is killed according as a coin lands heads/tails"

H = "the killing has happened"

A = "you are alive"

P(R) means your subjective probability that you are in a red room, before knowing K or H. Once you know all three, by Bayes' theorem:

P(R|KHA) = P(R)·P(KHA|R)/P(KHA) = 0.01·(0.5)/(0.5) = 0.01

0.01 is not equal to P(red). P(red) would be the probability of being in a red room given no information about whether the killing has occured; i.e. no information about what time it is.

I'd denote that by P(R|KA) -- with no information about H -- and you can check that it indeed equals 0.01. Again, Cupholder's diagram is an easy way to see this intuitively. If you want a verbal/mathematical explanation, first note from the diagram that the probability of being alive in a red room before killings happen is also 0.01:

P(R|K~HA) = #(possible living observers in red rooms before killings)/#(possible living observers before killings) = 0.01

So we have P(R|KHA)=P(R|K~HA)=0.01, and therefore by the usual independence trick,

So even when you know about a killing, but not whether it has happened, you still believe you are in a red room with probability 0.01.

Comment author: Mallah 07 April 2010 04:10:36PM * 0 points [-]

I omitted the "|before" for brevity, as is customary in Bayes' theorem.

That is not correct. The prior that is customary in using Bayes' theorem is the one which applies in the absence of additional information, not before an event that changes the numbers of observers.

For example, suppose we know that x=1,2,or 3. Our prior assigns 1/3 probability to each, so P(1) = 1/3. Then we find out "x is odd", so we update, getting P(1|odd) = 1/2. That is the standard use of Bayes' theorem, in which only our information changes.

OTOH, suppose that before time T there are 99 red door observers and 1 blue door one, and after time T, there is 1 red door are 99 blue door ones. Suppose also that there is the same amount of lifetime before and after T. If we don't know what time it is, clearly P(red) = 1/2. That's what P(red) means. If we know that it's before T, then update on that info, we get P(red|before)=0.99.

Note the distinction: "before an event" is not the same thing as "in the absence of information". In practice, often it is equivalent because we only learn info about the outcome after the event and because the number of observers stays constant. That makes it easy for people to get confused in cases where that no longer applies.

Now, suppose we ask a different question. Like in the case we were considering, the coin will be flipped and red or blue door observers will be killed; and it's a one-shot deal. But now, there will be a time delay after the coin has been flipped but before any observers are killed. Suppose we know that we are such observers after the flip but before the killing.

During this time, what is P(red|after flip & before killing)? In this case, all 100 observers are still alive, so there are 99 blue door ones and 1 red door one, so it is 0.01. That case presents no problems for your intuition, because it doesn't involve changes in the #'s of observers. It's what you get with just an info update.

Then the killing occurs. Either 1 red observer is killed, or 99 blue observers are killed. Either outcome is equally likely.

In the actual resulting world, there is only one kind of observer left, so we can't do an observer count to find the probabilities like we could in the many-worlds case (and as cupholder's diagram would suggest). Whichever kind of observer is left, you can only be that kind, so you learn nothing about what the coin result was.

Actually, if we consider that you could have been an observer-moment either before or after the killing, finding yourself to be after it does increase your subjective probability that fewer observers were killed. However, this effect goes away if the amount of time before the killing was very short compared to the time afterwards, since you'd probably find yourself afterwards in either case; and the case we're really interested in, the SIA, is the limit when the time before goes to 0.

See here

Comment author: Academian 08 April 2010 11:36:14PM * 0 points [-]

Given that others seem to be using it to get the right answer, consider that you may rightfully believe SIA is wrong because you have a different interpretation of it, which happens to be wrong.

I am using an interpretation that works -- that is, maximizes the total utility of equivalent possible observers -- given objectively-equally-likely hypothetical worlds (otherwise it is indeed problematic).

The prior that is customary in using Bayes' theorem is the one which applies in the absence of additional information, not before an event that changes the numbers of observers.

That's correct, and not an issue. In case it appears an issue, the beliefs in the update yielding P(R)=0.01 can be restated non-indexically (with no reference to "you" or "now" or "before"):

R = "person X is/was/will be in a red room"

K = "at some time, everyone in a red/blue room is killed according as a coin lands heads/tails

S = "person X survives/survived/will survive said killing"

Anthropic reasoning just says "reason as if you are X", and you get the right answer:

1) P(R|KS) = P(R|K)·P(S|RK)/P(S|K) = 0.01·(0.5)/(0.5) = 0.01

If you still think this is wrong, and you want to be prudent about the truth, try finding which term in the equation (1) is incorrect and which possible-observer count makes it so. In your analysis, be sure you only use SIA once to declare equal likelihood of possible-observers, (it's easiest at the beginning), and be explicit when you use it. Then use evidence to constrain which of those equally-likely folk you might actually be, and you'll find that 1% of them are in red rooms, so SIA gives the right answer in this problem.

Cupholder's diagram, ignoring its frequentist interpretation if you like, is a good aid to count these equally-likely folk.

In the actual resulting world, there is only one kind of observer left, so we can't do an observer count to find the probabilities like we could in the many-worlds case (and as cupholder's diagram would suggest). Whichever kind of observer is left, you can only be that kind, so you learn nothing about what the coin result was.

SIA doesn't ask you to count observers in the "actual world". It applies to objectively-equally-likely hypothetical worlds:

http://en.wikipedia.org/wiki/Self-Indication_Assumption

"SIA: Given the fact that you exist, you should (other things equal) favor hypotheses according to which many observers exist over hypotheses on which few observers exist."

Quantitatively, to work properly it say to consider any two observer moments in objectively-equally-likely hypothetical worlds as equally likely. Cupholder's diagram represents objectively-equally-likely hypothetical worlds in which to count observers, so it's perfect.

Some warnings:

make sure SIA isn't the only information you use... you have to constrain the set of observers you're in (your "reference class"), using any evidence like "the killing has happened".
don't count observers before and after the killing as equally likely -- they're not in objectively-equally-likely hypothetical worlds. Each world-moment before the killing is twice as objectively-likely as the world-moments after it.

Comment author: Mallah 13 April 2010 03:31:42AM * 0 points [-]

Given that others seem to be using it to get the right answer, consider that you may rightfully believe SIA is wrong because you have a different interpretation of it, which happens to be wrong.

Huh? I haven't been using the SIA, I have been attacking it by deriving the right answer from general considerations (that is, P(tails) = 1/2 for the 1-shot case in the long-time-after limit) and noting that the SIA is inconsistent with it. The result of the SIA is well known - in this case, 0.01; I don't think anyone disputes that.

P(R|KS) = P(R|K)·P(S|RK)/P(S|K) = 0.01·(0.5)/(0.5) = 0.01

If you still think this is wrong, and you want to be prudent about the truth, try finding which term in the equation (1) is incorrect and which possible-observer count makes it so.

Dead men make no observations. The equation you gave is fine for before the killing (for guessing what color you will be if you survive), not for after (when the set of observers is no longer the same).

So, if you are after the killing, you can only be one of the living observers. This is an anthropic selection effect. If you want to simulate it using an outside 'observer' (who we will have to assume is not in the reference class; perhaps an unconscious computer), the equivalent would be interviewing the survivors.

The computer will interview all of the survivors. So in the 1-shot case, there is a 50% chance it asks the red door survivor, and a 50% chance it talks to the 99 blue door ones. They all get an interview because all survivors make observations and we want to make it an equivalent situation. So if you get interviewed, there is a 50% chance that you are the red door one, and a 50% chance you are one of the blue door ones.

Note that if the computer were to interview just one survivor at random in either case, then being interviewed would be strong evidence of being the red one, because if the 99 blue ones are the survivors you'd just have a 1 in 99 chance of being picked. P(red) > P(blue). This modified case shows the power of selection.

Of course, we can consider intermediate cases in which N of the blue survivors would be interviewed; then P(blue) approaches 50% as N approaches 99.

The analogous modified MWI case would be for it to interview both the red survivor and one of the blue ones; of course, each survivor has half the original measure. In this case, being interviewed would provide no evidence of being the red one, because now you'd have a 1% chance of being the red one and the same chance of being the blue interviewee. The MWI version (or equivalently, many runs of the experiment, which may be anywhere in the multiverse) negates the selection effect.

If you are having trouble following my explanations, maybe you'd prefer to see what Nick Bostrom has to say. This paper talks about the equivalent Sleeping Beauty problem. The main interesting part is near the end where he talks about his own take on it. He correctly deduces that the probability for the 1-shot case is 1/2, and for the many-shot case it approaches 1/3 (for the SB problem). I disagree with his 'hybrid model' but it is pretty easy to ignore that part for now.

Also of interest is this paper which correctly discusses the difference between single-world and MWI interpretations of QM in terms of anthropic selection effects.

Comment author: Academian 16 April 2010 10:16:42AM * 1 point [-]

I have been attacking it by deriving the right answer from general considerations (that is, P(tails) = 1/2 for the 1-shot case

Let me instead ask a simple question: would you actually bet like you're in a red room?

Suppose you were told the killing had happened (as in the right column of Cupholder's diagram, and required to guess the color of your room, with the following payoffs:

Guess red correctly -> you earn $1.50
Guess blue correctly -> you earn $1.00
Guess incorrectly -> you are terribly beaten.

Would you guess red? Knowing that under independent repeated or parallel instances of this scenario (although merely hypothetical if you are concerned with the "number of shots"),

"guess heads" mentality typically leads to large numbers of people (99%) being terribly beaten
"guess blue" mentality leads to large numbers of people (99%) earning $1 and not being beaten
this not an interactive scenario like the Prisoner's dilemma, which is interactive in a way that renders a sharp distinction between group rationality and individual rationality,

would you still guess "red"? Not me. I would take my survival as evidence that blue rooms were not killed, and guess blue.

If you would guess "blue" for "other reasons", then we would exhibit the same behavior, and I have nothing more to discuss. At least in this case, our semantically different ways of managing possibilities are resulting in the same decision, which is what I consider important. You may disagree about this importance, but I apologize that I'm not up for another comment thread of this length.

If you would really guess "red", then I have little more to say than to reconsider your actions, and to again excuse me from this lengthy discussion.

Comment author: Mallah 16 April 2010 04:06:44PM * 0 points [-]

The way you set up the decision is not a fair test of belief, because the stakes are more like $1.50 to $99.

To fix that, we need to make 2 changes:

1) Let us give any reward/punishment to a third party we care about, e.g. SB.

2) The total reward/punishment she gets won't depend on the number of people who make the decision. Instead, we will poll all of the survivors from all trials and pool the results (or we can pick 1 survivor at random, but let's do it the first way).

The majority decides what guess to use, on the principle of one man, one vote. That is surely what we want from our theory - for the majority of observers to guess optimally.

Under these rules, if I know it's the 1-shot case, I should guess red, since the chance is 50% and the payoff to SB is larger. Surely you see that SB would prefer us to guess red in this case.

OTOH if I know it's the multi-shot case, the majority will be probably be blue, so I should guess blue.

In practice, of course, it will be the multi-shot case. The universe (and even the population of Earth) is large; besides, I believe in the MWI of QM.

The practical significance of the distinction has nothing to do with casino-style gambling. It is more that 1) it shows that the MWI can give different predictions from a single-world theory, and 2) it disproves the SIA.

Comment author: Academian 16 April 2010 06:08:41PM * 0 points [-]

Is that a "yes" or a "no" for the scenario as I posed it?

The way you set up the decision is not a fair test of belief.

I agree. It is only possible to fairly "test" beliefs when a related objective probability is agreed upon, which for us is clearly a problem. So my question remains unanswered, to see if we disagree behaviorally:

the stakes are more like $1.50 to $99.

That's not my intention. To clarify, assume that:

the other prisoners' decisions are totally independent of yours (perhaps they are irrational), so that you can in no sense effect 99 real other people to guess blue and achieve a $99 payoff with only one beating, and
the payoffs/beatings are really to the prisoners, not someone else,

Then, as I said, in that scenario I would guess that I'm in a blue room.

Would you really guess "red", or do we agree?

(My "reasons" for blue would be to note that I started out overwhelmingly (99%) likely to be in a blue room, and that my surviving the subsequent coin toss is evidence that it did not land tails and kill blue-roomed prisoners, or equivalently, that counterfactual-typically, people guessing red would result in a great deal of torture. But please forget why; I just want to know what you would do.)

continue this thread »

Comment author: cupholder 14 April 2010 08:17:10AM * 0 points [-]

Dead men make no observations. The equation you gave is fine for before the killing (for guessing what color you will be if you survive), not for after (when the set of observers is no longer the same).

Under a frequentist interpretation it is not possible for the equation to work pre-killing and yet not work post-killing: if one's estimate of P(R|KS) = 0.01 is correct, that implies one has correctly estimated the relative frequency of having been red-doored given that one survives the killing. That estimate of the relative frequency cannot then change after the killing, because that is precisely the situation for which the relative frequency was declared correct!

The computer will interview all of the survivors. So in the 1-shot case, there is a 50% chance it asks the red door survivor, and a 50% chance it talks to the 99 blue door ones. They all get an interview because all survivors make observations and we want to make it an equivalent situation. So if you get interviewed, there is a 50% chance that you are the red door one, and a 50% chance you are one of the blue door ones.

I don't agree, because in my judgment the greater number of people initially behind blue doors skews the probability in favor of 'you' being behind a blue door.

If you are having trouble following my explanations, maybe you'd prefer to see what Nick Bostrom has to say. This paper talks about the equivalent Sleeping Beauty problem. The main interesting part is near the end where he talks about his own take on it. He correctly deduces that the probability for the 1-shot case is 1/2, and for the many-shot case it approaches 1/3 (for the SB problem).

Reading Bostrom's explanation of the SB problem, and interpreting 'what should her credence be that the coin will fall heads?' as a question asking the relative frequency of the coin coming up heads, it seems to me that the answer is 1/2 however many times Sleeping Beauty's later woken up: the fair coin will always be tossed after she awakes on Monday, and a fair coin's probability of coming up heads is 1/2.

Comment author: Mallah 14 April 2010 04:00:22PM * 0 points [-]

Under a frequentist interpretation

In the 1-shot case, the whole concept of a frequentist interpretation makes no sense. Frequentist thinking invokes the many-shot case.

Reading Bostrom's explanation of the SB problem, and interpreting 'what should her credence be that the coin will fall heads?' as a question asking the relative frequency of the coin coming up heads, it seems to me that the answer is 1/2 however many times Sleeping Beauty's later woken up: the fair coin will always be tossed after she awakes on Monday, and a fair coin's probability of coming up heads is 1/2.

I am surprised you think so because you seem stuck in many-shot thinking, which gives 1/3.

Maybe you are asking the wrong question. The question is, given that she wakes up on Monday or Tuesday and doesn't know which, what is her creedence that the coin actually fell heads? Obviously in the many-shot case, she will be woken up twice as often during experiments where it fell tails, so in 2/3 or her wakeups the coin will be tails.

In the 1-shot case that is not true, either she wakes up once (heads) or twice (tails) with 50% chance of either.

Consider the 2-shot case. Then we have 4 possibilities:

coins , days , fraction of actual wakeups where it's heads
HH , M M , 1
HT , M M T , 1/3
TH , M T M , 1/3
TT , M T M T , 0

Now P(heads) = (1 + 1/3 + 1/3 + 0) / 4 = 5/12 = 0.417

Obviously as the number of trials increases, P(heads) will approach 1/3.

This is assuming that she is the only observer and that the experiments are her whole life, BTW.

Comment author: JGWeissman 14 April 2010 11:05:19PM * 1 point [-]

Now P(heads) = (1 + 1/3 + 1/3 + 0) / 4 = 5/12 = 0.417

This should be a weighted average, reflecting how many coin flips are observed in the four cases:

P(heads) = (2*1 + 3*1/3 + 3*1/3 + 4*0)/(2+3+3+4) = (2+1+1+0)/12 = 4/12 = 1/3

continue this thread »

Comment author: cupholder 14 April 2010 10:39:41PM * 0 points [-]

In the 1-shot case, the whole concept of a frequentist interpretation makes no sense. Frequentist thinking invokes the many-shot case.

Maybe I misunderstand what the frequentist interpretation involves, but I don't think the 2nd sentence implies the 1st. If I remember rightly, a frequentist interpretation of probability as long-run frequency in the case of Bernoulli trials (e.g. coin flips) can be justified with the strong law of large numbers. So one can do that mathematically without actually flipping a coin arbitrarily many times, from a definition of a single Bernoulli trial.

Maybe you are asking the wrong question.

My initial interpretation of the question seems to differ from the intended one, if that's what you mean.

The question is, given that she wakes up on Monday or Tuesday and doesn't know which, what is her creedence that the coin actually fell heads?

This subtly differs from Bostrom's description, which says 'When she awakes on Monday', rather than 'Monday or Tuesday.' I think your description probably better expresses what Bostrom is getting at, based on a quick skim of the rest of Bostrom's paper, and also because your more complex description makes both of the answers Bostrom mentions (1/2 and 1/3) defensible: depending on how I interpret you, I can extract either answer from the one-shot case, because the interpretation affects how I set up the relative frequency.

If I count how many times on average the coin comes up heads per time it is flipped, I must get the answer 1/2, because the coin is fair.

If I count how many times on average the coin comes up heads per time SB awakes, the answer is 1/3. Each time I redo the 'experiment,' SB has a 50% chance of waking up twice with the coin tails, and a 50% chance of waking up once with the coin heads. So on average she wakes up 0.5×2 + 0.5×1 = 1.5 times, and 0.5×1 = 0.5 of those 1.5 times correspond to heads: hence 0.5/1.5 = 1/3.

I'm guessing that the Bayesian analog of these two possible thought processes would be something like

SB asking herself, 'if I were the coin, what would I think my chance of coming up heads was whenever I'm awake?'
SB asking herself, 'from my point of view, what is the coin about to be/was the coin yesterday whenever I wake up?'

but I may be wrong. At any rate, I haven't thought of a rationale for your 2-shot calculation. Repeating the experiment twice shouldn't change the relative frequencies - they're relative! So the 2-shot case should still have 1/2 or 1/3 as the only justifiable credences.

(Edited to fix markup/multiplication signs.)

continue this thread »

Comment author: cupholder 07 April 2010 04:40:37AM * 0 points [-]

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.)

Comment author: Mallah 07 April 2010 01:35:55PM -2 points [-]

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.

Comment author: JGWeissman 08 April 2010 04:57:32AM * 1 point [-]

The one-shot case requires Bayesian thinking, not frequentist.

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

p(tails|survival) = p(tails) * p(survival|tails) / p(survival) = .5 * .01 / .5 = .01

so you can then calculate

"P(red|after)" = p(heads|survival) * "p(red|heads)" + p(tails|survival) * "p(red|tails)" = .99 * 0 + .01 * 1
= .01

Or, as Academian suggested, you could have just updated to directly find

p(red|survival) = p(red) * p(survival|red) / p(survival)

Comment author: cupholder 08 April 2010 03:48:42AM 0 points [-]

The one-shot case requires Bayesian thinking, not frequentist.

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.

The answer I gave is the correct one, because observers do not gain any information about whether the coin was heads or tails.

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.

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.

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.

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.

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.

Comment author: JGWeissman 08 April 2010 04:32:30AM 1 point [-]

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.

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.?

Comment author: cupholder 08 April 2010 05:41:09AM * 0 points [-]

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.

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.)

or, more generally, that in each trial you compute a different logical fact that you were initially completely ignorant about.?

This would also work, so long as your bag of facts is equally distributed between true facts and false facts.

Comment author: Mallah 13 April 2010 04:04:50AM * 0 points [-]

I think talking about 'observers' might be muddling the issue here.

That's probably why you don't understand the result; it is an anthropic selection effect. See my reply to Academician above.

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.

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.

I don't think of the pi digit process as equivalent.

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.

Comment author: cupholder 14 April 2010 05:56:41AM 0 points [-]

See my reply to Academician above.

Okay.

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.

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.

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.

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?'

Comment author: Mallah 14 April 2010 03:28:20PM 1 point [-]

A few minutes later, it is announced that whoever was to be killed has been killed. What are your odds of being blue-doored now?

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?"

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?'

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.

Comment author: cupholder 14 April 2010 08:28:23PM 0 points [-]

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?"

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.

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.

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?

continue this thread »

Comment author: wnoise 07 April 2010 09:36:31PM * 0 points [-]

FWIW, it's not that hard to calculate binary digits of pi:

http://oldweb.cecm.sfu.ca/projects/pihex/index.html

The Quadrillionth Bit of Pi is '0'! The Forty Trillionth Bit of Pi is '0'! The Five Trillionth Bit of Pi is '0'!

I think I'll go calculate the millionth, and get back to you.

EDIT: also turns out to be 0.

Mallah comments on Avoiding doomsday: a "proof" of the self-indication assumption - Less Wrong

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