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DanArmak comments on Avoiding doomsday: a "proof" of the self-indication assumption - Less Wrong
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Final edit: I now understand that the argument in the article is correct (and p=.99 in all scenarios). The formulation of the scenarios caused me some kind of cognitive dissonance but now I no longer see a problem with the correct reading of the argument. Please ignore my comments below. (Should I delete in such cases?)
I don't understand what precisely is wrong with the following intuitive argument, which contradicts the p=.99 result of SIA:
In scenarios E and F, I first wake up after the other people are killed (or not created) based on the coin flip. No-one ever wakes up and is killed later. So I am in a blue room if and only if the coin came up heads (and no observer was created in the red room). Therefore P(blue)=P(heads)=0.5, and P(red)=P(tails)=0.5.
Edit: I'm having problems wrapping my head around this logic... Which prevents me from understanding all the LW discussion in recent months about decision theories, since it often considers such scenarios. Could someone give me a pointer please?
Before the coin is flipped and I am placed in a room, clearly I should predict P(heads)=0.5. Afterwards, to shift to P(heads)=0.99 would require updating on the evidence that I am alive. How exactly can I do this if I can't ever update on the evidence that I am dead? (This is the scenario where no-one is ever killed.)
I feel like I need to go back and spell out formally what constitutes legal Bayesian evidence. Is this written out somewhere in a way that permits SIA (my own existence as evidence)? I'm used to considering only evidence to which there could possibly be alternative evidence that I did not in fact observe. Please excuse a rookie as these must be well understood issues.
There's nothing wrong with this argument. In E and F (and also in D in fact), the probability is indeed 50%.
How would you go about betting on that?
If I were actually in situation A, B, or C, I would expect a 99% chance of a blue door, and in D, E, or F, a 50%, and I would actually bet with this expectation.
There is really no practical way to implement this, however, because of the assumption that random events turn out in a certain way, e.g. it is assumed that there is only a 50% chance that I will survive, yet I always do, in order for the case to be the one under consideration.
Omega runs 10,000 trials of scenario F, and puts you in touch with 100 random people still in their room who believe there is a %50 chance they have red doors, and will happily take 10 to 1 bets that they are.
You take these bets, collect $1 each from 98 of them, and pay out $10 each to 2.
Were their bets rational?
You assume that the 100 people have been chosen randomly from all the people in the 10,000 trials. This is not valid. The appropriate way for these bets to take place is to choose one random person from one trial, then another random person from another trial, and so on. In this way about 50 of the hundred persons will be behind red doors.
The reason for this is that if I know that this setup has taken place 10,000 times, my estimate of the probability that I am behind a blue door will not be the same as if the setup has happened only once. The probability will slowly drift toward 99% as the number of trials increases. In order to prevent this drift, you have to select the persons as stated above.
If you find yourself in such a room, why does your blue door estimate go up with the number of trials you know about? Your coin was still 50-50.
How much does it go up for each additional trial? ie what are your odds if omega tells you you're in one of two trials of F?
The reason is that "I" could be anyone out of the full set of two trials. So: there is a 25% chance there both trials ended with red-doored survivors; a 25% chance that both trials ended with blue-doored survivors; and a 50% chance that one ended with a red door, one with a blue.
If both were red, I have a red door (100% chance). If both were blue, I have a blue door (100% chance). But if there was one red and one blue, then there are a total of 100 people, 99 blue and one red, and I could be any of them. So in this case there is a 99% chance I am behind a blue door.
Putting these things together, if I calculate correctly, the total probability here (in the case of two trials) is that I have a 25.5% chance of being behind a red door, and a 74.5% chance of being behind a blue door. In a similar way you can show that as you add more trials, your probability will get ever closer to 99% of being behind a blue door.
You could only be in one trial or the other.
What if Omega says you're in the second trial, not the first?
Or trial 3854 of 10,000?
"I could be any of them" in the sense that all the factors that influence my estimate of the probability, will influence the estimate of the probability made by all the others. Omega may tell me I am in the second trial, but he could equally tell someone else (or me) that he is in the first trial. There are still 100 persons, 99 behind blue doors and 1 behind red, and in every way which is relevant, I could be any of them. Thinking that the number of my trial makes a difference would be like thinking that if Omega tells me I have brown eyes and someone else has blue, that should change my estimate.
Likewise with trial 3854 out of 10,000. Naturally each person is in one of the trials, but the persons trial number does not make a significant contribution to his estimate. So I stand by the previous comments.
Thanks! I think this comment is the best so far for demonstrating the confusion (well, I was confused :-) about the different possible meanings of the phrase "you are an observer chosen from such and such set". Perhaps a more precise and unambiguous phrasing could be used.
Replace death with the light in the room being shut off.
That's not applicable to scenarios E and F which is where I have a problem. The observers there never wake up or are never created (depending on the coin toss), I can't replace that with a conscious observer and the light going off.
Whereas in scenarios A through D, you don't need SIA to reach the (correct) p=.99 conclusion, you don't even need the existence of observers other than yourself. Just reformulate as: I was moved to a room at random; the inhabitants of some rooms, if any, were killed based on a coin flip; etc.
Do it anyway. Take a scenario in which the light is shut off while you are sleeping, or never turned on. What does waking up with the lights on (or off) tell you about the color of the door?
Even in A thru D, the dead can't update.
The state of the lights tells me nothing about the color of the door. Whatever color room I happen to be in, the coin toss will turn my lights on or off with 50% probability.
I don't see what you intend me to learn from this example...
That dead or alive you are still most likely behind a blue door. You can use the lights being on as evidence just as well as your being alive.
That in B through D you are already updating based on your continued existence.
Beforehand you would expect a 50% chance of dying. Later, If you are alive, then the coin probably came up heads. In E and F, You wake up, You know the coin flip is in your past, You know that most 'survivors' of situations like this come out of blue doors.
If you play Russian roulette and survive, you can have a much greater than 5/6 confidence that the chamber wasn't loaded.
You can be very certain that you have great grandparents, given only your existence and basic knowledge about the world.
In E-F this is not correct. Your words "dead or alive" simply don't apply: the dead observers never were alive (and conscious) in these scenarios. They were created and then destroyed without waking up. There is no possible sense in which "I" could be one of them; I am by definition alive now or at least were alive at some point in the past. Even under the assumptions of the SIA, a universe with potential observers that never actually materialize isn't the same as one with actual observers.
I still think that in E-F, I'm equally likely to be behind a blue or a red door.
Correct. The crucial difference is that in B-D I could have died but didn't. In other Everett branches where the coin toss went the other way I did die. So I can talk about the probability of the branch where I survive, and update on the fact that I did survive.
But in E-F I could never have died! There is no branch of possibility where any conscious observer has died in E-F. That's why no observer can update on being alive there; they are all alive with p=1.
Yes, because in our world there are people who fail to have grandchildren, and so there are potential grandchildren who don't actually come to exist.
But in the world of scenarios E and F there is no one who fails to exist and to leave a "descendant" that is himself five minutes later...
I now understand that the argument in the article is correct (and p=.99 in all scenarios). The formulation of the scenarios caused me some kind of cognitive dissonance but now I no longer see a problem with the correct reading of the argument. Please ignore my comments below. (Should I delete in such cases?)
I wouldn't delete, if nothing else it serves as a good example of working through the dissonance.
edit It would also be helpful if you explained from your own perspective why you changed your mind.
Second James's preference and note that I find it useful as a reader to see an edit note of some sort in comments that are no longer supported.
I now understand that the argument in the article is correct (and p=.99 in all scenarios). The formulation of the scenarios caused me some kind of cognitive dissonance but now I no longer see a problem with the correct reading of the argument.