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Academian comments on Avoiding doomsday: a "proof" of the self-indication assumption - Less Wrong
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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.
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.
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.
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.
Is that a "yes" or a "no" for the scenario as I posed it?
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:
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.)
That's wrong; behavioral tests (properly set up) can reveal what people really believe, bypassing talk of probabilities.
Under the strict conditions above and the other conditions I have outlined (long-time-after, no other observers in the multiverse besides the prisoners), then sure, I'd be a fool not to guess red.
But I wouldn't recommend it to others, because if there are more people, that would only happen in the blue case. This is a case in which the number of observers depends on the unknown, so maximizing expected average utility (which is appropriate for decision theory for a given observer) is not the same as maximizing expected total utility (appropriate for a class of observers).
More tellingly, once I find out the result (and obviously the result becomes known when I get paid or punished), if it is red, I would not be surprised. (Could be either, 50% chance.)
Not that I've answered your question, it's time for you to answer mine: What would you vote, given that the majority of votes determines what SB gets? If you really believe you are probably in a blue room, it seems to me that you should vote blue; and it seems obvious that would be irrational.
Then if you find out it was red, would you be surprised?
So in my scenario, groups of people like you end up with 99 survivors being tortured or 1 not, with equal odds (despite that their actions are independent and non-competitive), and groups of people like me end up with 99 survivors not tortured or 1 survivor tortured, with equal odds.
Let's say I'm not asserting that means I'm "right". But consider that your behavior may be more due to a ritual of cognition rather than systematized winning.
You might respond that "rationalists win" is itself a ritual of cognition to be abandoned. More specifically, maybe you disagree that "whatever rationality is, it should fare well-in-total, on average, in non-competitive thought experiments". I'm not sure what to do about that response.
In your scenario, I'd vote red, because when the (independent!) players do that, her expected payoff is higher. More precisely, if I model the others randomly, me voting red increases the probability that SB lands in world with a majority "red" vote, increasing her expectation.
This may seem strange because I am playing by an Updateless strategy. Yes, in my scenario I act 99% sure that I'm in a blue room, and in yours I guess red, even though they have same assumptions regarding my location. Weird eh?
What's happening here is that I'm planning ahead to do what wins, and planning isn't always intuitively consistent with updating. Check out The Absent Minded Driver for another example where planning typically outperforms naive updating. Here's another scenario, which involves interactive planning.
To be honest with you, I'm not sure how the "surprise" emotion is supposed to work in scenarios like this. It might even be useless. That's why I base my actions on instrumental reasoning rather than rituals of cognition like "don't act surprised".
By the way, you are certainly not the first to feel the weirdness of time inconsistency in optimal decisions. That's why there are so many posts working on decision theory here.