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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 c

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, though. Being in one of cubicles 11+ /does/ eliminate the possibility of doom soon. So only passing the cubicle threshold gives any meaningful information about which of the two scenarios are true.

jerf 12y ago
That's an Aristotelian viewpoint, operating logic on the basis of True and False, in which case the argument presented is indeed a fallacy. However, when you operate on probabilities, Bayes' Rule does indeed allow you to drive the logic in a way that seems "backwards" from the Aristotelian point of view. This, by the way, has nothing to do with "Bayesian" vs. "Frequentism"... this is just the straight application of a simple and non-controversial thing call Bayes' Rule. Assuming you know the probabilities with the requisite precision you are perfectly justified in reasoning that way. It's also worth reading the end of the article carefully too... for instance, "doom" may be a bit leading, since "doom" could include a full Singularity after which we are no longer "humans" but nobody is upset about that, or other such scenarios that are still in some sense "doom" for The Human Race (imagine those glorious calligraphy letters for that) but aren't necessarily doom in a bad way. There's also just the casual observation that if society goes on for billions of years there are still simply some people who are born early. Sometimes people win the lottery, and we would not generally be too impressed if they reasoned backwards from there to the conclusion that winning the lottery was more likely than they would have initially thought. (That is not exactly a parallel, since a lottery is part of a much bigger net of information than the deliberately simplified version presented in the argument.) One of my objections would be that it is not clear that we can take a non-temporal line of logic as given in the beginning of the argument and directly apply it to a situation with time. And when I mean "not clear", I mean precisely that it is not clear to me, not that I'm saying it's impossible. In general I've observed that I'm not impressed with our attempts to apply our local probability rules at cosmological scales. Is Occam's Razor really applicable at cosmological scales? It's hard to say what's "simplest" when we don't even have any reasonable priors (or translate that to whatever math framework you may like, like "have no information about"). The orders of magnitude involved even with merely the visible universe, which we have no reason to believe is even the whole of reality, IMHO can easily swamp the simple probability of error in the argument too easily. (That is to say... there's an old observation in computer science that running a "probabilistic" algorithm with a probability of failure lower than the probability of hardware failure is indistinguishable from running a deterministic algorithm. It is not clear to me that we can overcome our probability of having something wrong in our argument.)
lmkg 12y ago
The reasoning in the article is correct. Cubicles 1-10 are occupied in both scenarios, but the observation of (self <= 10) is much less likely in the Doom Late scenario. This means that if you observe (self <= 10), the conditional probability of Doom Soon is much higher. They key point is that, if Doom Late occurred, then the probability of ending up in cubicle 1-10 is only 10%. So if the only observation that you have is that your cubicle is numbered 1-10, which is more likely: Doom Soon (50% absolute probability), or Doom Late (50% absolute) and you happened to end up in 1-10 (10% of the 50%)? Therefore, the conditional probability of Doom Soon is higher than 50% based on the observation of self <= 10.
- livingparadox 12y ago
  The premises in the argument can abstracted into the following: if doom_soon, then the probability of myself being in cubicles 1-10 is 100%; if doom_soon, then the probability of myself being in cubicles 11-100 is 0%; if doom_late, then the probability of myself being in cubicles 1-10 is 10%; if doom_late, then the probability of myself being in cubicles 11-100 is 90%; ??? Therefore if I am the cubicles 1-10, the probability of doom soon is 81%; In any logic class, this argument would quickly be pointed out as a fallacy. The statement of "if p, then q" says nothing about the truth of p, given q. Which is exactly what this argument seems to do, reason backwards from the conclusion when none of the premises are able to be logically inferred from the conclusion.
- livingparadox 12y ago
  I thought of a better way of explaining the invalidity of the argument. What would happen if, instead of inferring from our own presence in the cubicle, we take an external role. Furthermore, instead of taking the probability that a randomly selected person exists in of the two sets, we look at the probability that /someone/ exists in one of the two sets. If doom soon occurred, then the probability of cubicles 1-10 being occupied is 100%; If doom soon occurred, then the probability of cubicles 11-100 being occupied is 0%; If doom late occurred, then the probability of cubicles 1-10 being occupied is 100%; If doom late occurred, then the probability of cubicles 11-100 being occupied is 100%; So given those two probabilities, we have a 100% chance of cubicles 1-10 being occupied and a 50% chance of cubicles 11-100 being occupied. In either scenario, the occupancy status of cubicles 1-10 is assured. So checking in cubicles 1-10 gives us no meaningful information about which of the two scenarios occurred. Checking in cubicles 11-100 on the other hand /would/ affect our results. The occupancy of cubicles 11-100 on the other hand would tell us exactly which scenario occured.
rm445 12y ago
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 is 90.9% - we could go round in circles trying to choose words that everyone understands, but the numbers speak for themselves.
- livingparadox 12y ago
  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 <=10 individual to find themselves in (10 in doom soon, 10 in doom late). Of those 20 possibilities, 10 of those are in the doom soon scenario. The remaining 10 are in the doom late scenario. 10 doom soon to 10 doom late is a 50/50 probability. Edit: writing up a script as well, right now. Will add to link to it in my original comment when complete.
  - rm445 12y ago
    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.
livingparadox 12y ago
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/ 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

jerf 12y ago

That's an Aristotelian viewpoint, operating logic on the basis of True and False, in which case the argument presented is indeed a fallacy. However, when you operate on probabilities, Bayes' Rule does indeed allow you to drive the logic in a way that seems "backwards" from the Aristotelian point of view. This, by the way, has nothing to do with "Bayesian" vs. "Frequentism"... this is just the straight application of a simple and non-controversial thing call Bayes' Rule. Assuming you know the probabilities with the requisite precision you are perfectly justified in reasoning that way. It's also worth reading the end of the article carefully too... for instance, "doom" may be a bit leading, since "doom" could include a full Singularity after which we are no longer "humans" but nobody is upset about that, or other such scenarios that are still in some sense "doom" for The Human Race (imagine those glorious calligraphy letters for that) but aren't necessarily doom in a bad way. There's also just the casual observation that if society goes on for billions of years there are still simply some people who are born early. Sometimes people win the lottery, and we would not generally be too impressed if they reasoned backwards from there to the conclusion that winning the lottery was more likely than they would have initially thought. (That is not exactly a parallel, since a lottery is part of a much bigger net of information than the deliberately simplified version presented in the argument.) One of my objections would be that it is not clear that we can take a non-temporal line of logic as given in the beginning of the argument and directly apply it to a situation with time. And when I mean "not clear", I mean precisely that it is not clear to me, not that I'm saying it's impossible. In general I've observed that I'm not impressed with our attempts to apply our local probability rules at cosmological scales. Is Occam's Razor really applicable at cosmological scales? It's hard to say what's "simplest" when we don't even have any reasonable priors (or translate that to whatever math framework you may like, like "have no information about"). The orders of magnitude involved even with merely the visible universe, which we have no reason to believe is even the whole of reality, IMHO can easily swamp the simple probability of error in the argument too easily. (That is to say... there's an old observation in computer science that running a "probabilistic" algorithm with a probability of failure lower than the probability of hardware failure is indistinguishable from running a deterministic algorithm. It is not clear to me that we can overcome our probability of having something wrong in our argument.)

lmkg 12y ago

The reasoning in the article is correct. Cubicles 1-10 are occupied in both scenarios, but the observation of (self <= 10) is much less likely in the Doom Late scenario. This means that if you observe (self <= 10), the conditional probability of Doom Soon is much higher. They key point is that, if Doom Late occurred, then the probability of ending up in cubicle 1-10 is only 10%. So if the only observation that you have is that your cubicle is numbered 1-10, which is more likely: Doom Soon (50% absolute probability), or Doom Late (50% absolute) and you happened to end up in 1-10 (10% of the 50%)? Therefore, the conditional probability of Doom Soon is higher than 50% based on the observation of self <= 10.

livingparadox 12y ago
The premises in the argument can abstracted into the following: if doom_soon, then the probability of myself being in cubicles 1-10 is 100%; if doom_soon, then the probability of myself being in cubicles 11-100 is 0%; if doom_late, then the probability of myself being in cubicles 1-10 is 10%; if doom_late, then the probability of myself being in cubicles 11-100 is 90%; ??? Therefore if I am the cubicles 1-10, the probability of doom soon is 81%; In any logic class, this argument would quickly be pointed out as a fallacy. The statement of "if p, then q" says nothing about the truth of p, given q. Which is exactly what this argument seems to do, reason backwards from the conclusion when none of the premises are able to be logically inferred from the conclusion.
livingparadox 12y ago
I thought of a better way of explaining the invalidity of the argument. What would happen if, instead of inferring from our own presence in the cubicle, we take an external role. Furthermore, instead of taking the probability that a randomly selected person exists in of the two sets, we look at the probability that /someone/ exists in one of the two sets. If doom soon occurred, then the probability of cubicles 1-10 being occupied is 100%; If doom soon occurred, then the probability of cubicles 11-100 being occupied is 0%; If doom late occurred, then the probability of cubicles 1-10 being occupied is 100%; If doom late occurred, then the probability of cubicles 11-100 being occupied is 100%; So given those two probabilities, we have a 100% chance of cubicles 1-10 being occupied and a 50% chance of cubicles 11-100 being occupied. In either scenario, the occupancy status of cubicles 1-10 is assured. So checking in cubicles 1-10 gives us no meaningful information about which of the two scenarios occurred. Checking in cubicles 11-100 on the other hand /would/ affect our results. The occupancy of cubicles 11-100 on the other hand would tell us exactly which scenario occured.

rm445 12y ago

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 is 90.9% - we could go round in circles trying to choose words that everyone understands, but the numbers speak for themselves.

livingparadox 12y ago
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 <=10 individual to find themselves in (10 in doom soon, 10 in doom late). Of those 20 possibilities, 10 of those are in the doom soon scenario. The remaining 10 are in the doom late scenario. 10 doom soon to 10 doom late is a 50/50 probability. Edit: writing up a script as well, right now. Will add to link to it in my original comment when complete.
- rm445 12y ago
  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.

livingparadox 12y ago

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/ 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