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A Primer on the Doomsday Argument

anthropic-principle.com

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Re: A Primer on the Doomsday Argument

#41

Post by rm445 »

livingparadox wrote: post #39
rm445 wrote: post #35

Earlier quoted context omitted.

It's probably not worth worrying about the fate of humanity till you've figured out the rooms example. Since you're on this site I'm sure you can trivially knock up a script to simulate what happens. Or just draw the probability tree. In any case the relevant question is,, out of those who find their room number is less than ten, what is the probability that the coin toss meant only ten rooms are occupied. The answer…

Okay, here's a simplified probability tree. Scenario 1: Doom soon 10 people have a room number less than or equal to 10. Scenario 2: Doom late 10 people have a room number less than or equal to 10. --- Given that each scenario has an equal probability in isolation, there are 20 possible positions for the the Of those 20 possibilities, 10 of those are in the doom soon scenario. The remaining 10 are in the doom late sc…

You're wrong. Not being mean, you've just arrived at the wrong numerical answer to the maths question.

(Let's leave aside mentions of doom and stick to the rooms: according to the article no-one has quite nailed down yet whether there are subtleties about unbounded future populations). Rather than try to draw a tree I'll write it out flat:

10 rooms (0.5) & you're in 1-10 (1.0): Probability 0.5

10 rooms (0.5) & you're in 11-100 (0.0): Probability 0.0

100 rooms (0.5) & you're in 1-10 (0.1): Probability 0.05

100 rooms (0.5) & you're in 11-100 (0.9): Probability 0.45

Work out any relative probabilities using the right-hand column. For instance, when you're in room 1-10, the chances of the coin having been heads is 0.5/0.55. Even if this bugs you, this approach of using the tree will let you get the right answers to tricky teasers about medical tests with false positives and false negatives.

Re: A Primer on the Doomsday Argument

#42

Post by typomatic »

titanomachy wrote: post #37
typomatic wrote: post #18

Let's go back to step II: Imagine that the coin that we're flipping isn't fair, and comes up heads with prior probability p. Further, suppose that the ratio between the small population and the large population is R (in the article this is R = 10 / 100 = .1). In this situation, upon discovering that you're inside the small population, the posterior probability of heads ends up as p / (R - p * (1 - R)) So what if the…

I played around with the model a bit and based on a couple assumptions it can be scaled to P(Doom Soon) = x/(1+x) where x = B/b * d where B is the number of humans born in the Doom Late scenario, b is the number of humans born so far, and d is the prior probability of Doom Soon. Assumptions: a) the number of humans born between now and Doom Soon is negligible and b) the Doom Late scenario has many more humans than Do…

As B -> \infty, also t -> \infty.

I'm reminded of the Fight Club quote: On a long enough timeline, the survival rate for everyone drops to zero. :)

Re: A Primer on the Doomsday Argument

#43

Post by saluk »

"Corresponding to the prior probability (50%) of the coin falling heads or tails, we now have some prior probability of Doom Soon or Doom Late."

Nice to just magically know this probability. I get that the actual number doesn't matter, but I worry about a predictor that doesn't care about any real probabilities based on observation.

Re: A Primer on the Doomsday Argument

#44

Post by abruzzi »

Ok, this may be mathematically naive, but...

In the cubicle example discovering that you are in cube 1-10 makes the likelihood of the 1-10 scenario much more likely than it was before you discovered you were in one of the first 10 cubes using a simple application of Bayes theorem.

With the 100 billion or 100 trillion people example, you are person 60 billion. Theoretically making 100 billion much more likely than if you didn't know where you fell. That probability approaches one if work off the assumption that doom(late) means hundreds of millennia of humans spreading across the galaxy at our current growth rate. the higher the Total possible humans in doom(late) the higher the probability that being in the first 100b indicated that there will only be 100b. (or similar numbers)

One difference I see between the two scenarios is time. The 100 cubes are not filled in sequence. In the 100billion/100trillion example, every single person that ever lived is in the 100billion, until they're not, then everyone is not in the 100 billion. I don't know that it affects the math, but it affects my thinking about the problem.

Re: A Primer on the Doomsday Argument

#45

Post by BoppreH »

scotty79 wrote: post #14

Imagine you have 100 ordered rooms. All initially are filled with gold and ponies. God tosses a coin and depending on the result of the flip God strips 10 or 100 rooms of ponies and gold. You are in one of the rooms, open your eyes and see that there is no gold and ponies around. You take a look at your room number and see 7. Then you know that best decision is to operate on assumption that there is 91% chance that o…

The Doomsday Argument works because the result (Doom Soon or Doom Late) affects the number of observers, and we can work backwards to update our beliefs on which one will be.

That is why replacing the words with ponies and gold won't work unless you are a pony or sentient gold.

Re: A Primer on the Doomsday Argument

#46

Post by bluecalm »

baddox wrote: post #28

This is a much better discussion of the Doomsday Argument, and the self-indication and self-sampling assumptions: http://www.scottaaronson.com/democritus/lec17.html The whole course is also very good, and will probably be interesting to many HN users.

Very nice read, from this lecture:

>>So you're in the room. Conditioned on that fact, how worried should you be? How likely is it that you're going to die?

    A: 1/36.
    Scott: OK. That would be one guess.
>>One answer is that the dice have a 1/36 chance of landing snake-eyes, so you should be only a "little bit" worried (considering). A second reflection you could make is to consider, of people who enter the room, what the fraction is of people who ever get out. Let's say that it ends at 1,000. Then, 110 people get out and 1,000 die. If it ends at 10,000, then 1,110 people get out and 10,000 die. In either case, about 8/9 of the people who ever go into the room will die.

But it's not really a problem. If you flip a coin, then 10 coins, then 100 coins and continue until you get all heads then vast majority of flipped coins will be heads while still any particular coin have 1/2 probability of landing heads. There just isn't any paradox or even anything surprising. Similarly here it's in my opinion obvious that chances of dying are 1/36 and they are the same for all people in the situation. The fact that most people die isn't a paradox at all.

Re: A Primer on the Doomsday Argument

#47

Post by livingparadox »

rm445 wrote: post #41
livingparadox wrote: post #39

Earlier quoted context omitted.

Okay, here's a simplified probability tree. Scenario 1: Doom soon 10 people have a room number less than or equal to 10. Scenario 2: Doom late 10 people have a room number less than or equal to 10. --- Given that each scenario has an equal probability in isolation, there are 20 possible positions for the the Of those 20 possibilities, 10 of those are in the doom soon scenario. The remaining 10 are in the doom late sc…

You're wrong. Not being mean, you've just arrived at the wrong numerical answer to the maths question. (Let's leave aside mentions of doom and stick to the rooms: according to the article no-one has quite nailed down yet whether there are subtleties about unbounded future populations). Rather than try to draw a tree I'll write it out flat: 10 rooms (0.5) & you're in 1-10 (1.0): Probability 0.5 10 rooms (0.5) & you're…

Just to confirm my understanding...

The situation you gave me was "out of those who find their room number is less than ten, what is the probability that the coin toss meant only ten rooms are occupied".

In my example, I composed my tree of solely those who with a room number less than or equal to 10, since in your question, we're only concerned about those who are in rooms 1-10.

In your example, it seems you've included those who are in rooms 11-100... If we are only concerned with rooms 1-10, what relevance does the probability of 11-100 play in this role?

More specifically, your seems to be what the odds are of /being in/ room 1-10, rather than the odds of each scenario occurring, assuming you are in room 10.

Re: A Primer on the Doomsday Argument

#48

Post by titanomachy »

typomatic wrote: post #42
titanomachy wrote: post #37

Earlier quoted context omitted.

I played around with the model a bit and based on a couple assumptions it can be scaled to P(Doom Soon) = x/(1+x) where x = B/b * d where B is the number of humans born in the Doom Late scenario, b is the number of humans born so far, and d is the prior probability of Doom Soon. Assumptions: a) the number of humans born between now and Doom Soon is negligible and b) the Doom Late scenario has many more humans than Do…

As B -> \infty, also t -> \infty. I'm reminded of the Fight Club quote: On a long enough timeline, the survival rate for everyone drops to zero. :)

Haha nice. Although, in this case we are looking ahead, basing our short-term survival on what a long-term survival would theoretically look like. So the more people there are in the hypothetical Doom Late, the more likely Doom Soon becomes. The more I think about this the more absurd it seems.

Re: A Primer on the Doomsday Argument

#49

Post by livingparadox »

livingparadox wrote: post #24

Perhaps I'm ignorant of some deeper concept... but it seems that this argument is deeply flawed on the basis that they are presuming that finding yourself in cubicles 1-10 somehow indicates that cubicles 11+ are probably vacant. In both doom soon and doom late, cubicles 1-10 are occupied. There's no aspect of doom late that would be made less likely as a result of cubicles 1-10 being occupied. The reverse works well,…

In response to rm445's challenge in a reply, I wrote up a script simulating the two scenarios, and in situations where a room in 1-10 occupied, added up the totals of which scenario the individual is in.

http://jsfiddle.net/3f49ry5u/

The script seems to support the conclusion that given the premises of the scenario, the odds for each scenario are about 50/50

Re: A Primer on the Doomsday Argument

#50

Post by livingparadox »

rm445 wrote: post #35
livingparadox wrote: post #24

Perhaps I'm ignorant of some deeper concept... but it seems that this argument is deeply flawed on the basis that they are presuming that finding yourself in cubicles 1-10 somehow indicates that cubicles 11+ are probably vacant. In both doom soon and doom late, cubicles 1-10 are occupied. There's no aspect of doom late that would be made less likely as a result of cubicles 1-10 being occupied. The reverse works well,…

It's probably not worth worrying about the fate of humanity till you've figured out the rooms example. Since you're on this site I'm sure you can trivially knock up a script to simulate what happens. Or just draw the probability tree. In any case the relevant question is,, out of those who find their room number is less than ten, what is the probability that the coin toss meant only ten rooms are occupied. The answer…

Here's a simulation to test this argument.

http://jsfiddle.net/3f49ry5u/

So... unless I did the simulation wrong, it supports my conclusion.

If I did do it wrong, please show me a corrected simulation which supports your position.

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