The hundred-room problem
This thought-experiment has been on my mind for a couple of days, and no doubt it’s a special case of a more general problem identified somewhere by some philosopher that I haven’t heard of yet. It goes like this:
You are blindfolded, and then scanned, and ninety-nine atom-for-atom copies of you are made, each blindfolded, meaning a hundred in all. To each one is explained (and for the sake of the thought experiment, you can take this explanation as true (p is approx. 1)) that earlier, a fair-coin was flipped. If it came down heads, ninety-nine out of a hundred small rooms were painted red, and the remaining one was painted blue. If it came down tails, ninety-nine out of a hundred small rooms were painted blue, and the remaining one was painted red. Now, put yourself in the shoes of just one of these copies. When asked what the probability is that the coin came down tails, you of course answer “.5”. It is now explained to you that each of the hundred copies is to be inserted into one of the hundred rooms, and will then be allowed to remove their blindfolds. You feel yourself being moved, and then hear a voice telling you you can take your blindfold off. The room you are in is blue. The voice then asks you for your revised probability estimate that the coin came down tails.
It seems at first (or maybe at second, depending on how your mind works) that the answer ought to be .99 – ninety-nine out of the hundred copies will, if they follow the rule “if red, then heads, if blue then tails”, get the answer right.
However, it also seems like the answer ought to be .5, because you have no new information to update on. You already knew that at least one copy of you would, at this time, remove their blindfold and find themselves in a blue room. What have you discovered that should allow you to revise your probability of .5 to .99?
And the answer, of course, cannot be both .5 and .99. Something has to give.
Is there something basically quite obvious that I’m missing that will resolve this problem, or is it really the mean sonofabitch it appears to be? As it goes, I’m inclined to say the probability is .5 – I’m just not quite sure why. Thoughts?
If you look at all 100 copies, the problem disappears.
Before seeing rooms: All have probability estimate of 0.5 that coin was heads.
If the coin lands heads: All 100 give a probability of 0.5 that the coin landed heads.
If the coin lands tails: All 100 give a probability of 0.5 that the coin landed heads.
100 of 200 predictions with 0.5 confidence were correct. This is perfect calibration.
If the coin lands heads: 99 of the 100 will see a red room, and give a probability of 0.99 that the coin landed on heads. 1 of the 100 will see a blue room, and give a probability of 0.99 that the coin landed tails.
If the coin lands tails: 99 of the 100 will see a blue room, and give a probability of 0.99 that the coin landed on tails. 1 of the 100 will see a red room, and give a probability of 0.99 that the coin landed heads.
198 of 200 predictions with 0.99 confidence were correct. Once again, this is perfect calibration.
If you are in the 100 rooms situation, you should guess in such a way as to maximize the total accuracy of all the yous.
I knew there was an elegant solution to be had, my own reply became pointless. Thank you.
Briefly: a non-indexical construal of the evidence (“one of my copies is in a blue room”) gives the 0.5 answer, but if you strengthen the evidence to include the indexical information you have (“this particular copy is in a blue room”) you get the 0.99 answer (assuming something like Nick Bostrom’s SSA or Adam Elga’s restricted principle of indifference). Are you unaware of this, or do you reject the legitimacy of indexical evidence and don’t mention it?
I was glancingly aware of the Elga stuff (i.e. I knew about the Dr. Evil thought experiment, but hadn’t previously read the whole paper), and I’m still not sure how it impacts of this thought experiment. Need more time to process, but you seem fairly sure that it resolves the issue, so I’m leaning towards the “I missed something sort-of obvious” explanation. Will read the Bostrom stuff later.
Upvoted.
While I heartily recommend reading the Bostrom link (the linked section of chapter 3, and chapter 4 following it), here’s a bit more detail:
Let E := “(at least) one of the copies is in a blue room” Let E+ := “this particular copy is in a blue room”
Pr(E|Heads) = 1 and Pr(E|Tails) = 1, so the evidence E doesn’t push you away from the priors.
However, if you accept SSA or RPI, then while you are blindfolded your conditional probabilities for E+ should be Pr(E+|Heads) = 0.01 and Pr(E+|Tails) = 0.99. (Take, e.g. Elga’s point. Assuming that the copies are subjectively indistinguishable, Elga would say that you should split your conditional probability of being each particular copy evenly, so for each copy x you have Pr(I am copy x)=0.01. If Heads, only one copy is in a blue room, so the probability that you are that copy is 0.01. If Tails, 99 copies are in blue rooms, so the probability of being one of those copies is 0.99.) Plug these and your priors into Bayes’ theorem and you get Pr(Tails|E+) = 0.99.
As I understand it, though, from the Dr. Evil thought-experiment, the reason Dr. Evil ought to assign .5 probability to being the Dr. Evil in the battlestation and .5 to being the Dr. Evil-clone in the battlestation-clone is that his subjective state is exactly the same either way. But the copies in this thought-experiment have had it revealed to them which room their in. This copy is in a blue room—the probability that this particular copy is in a blue room is 1, the probability that this particular copy is in a red room is 0, and therefore P(E+|Heads)=1 and P(E+|Tails)=1 - no? Unless I’m wrong. Which there’s a good chance I am. I’m confused, is what I’m saying.
Note what I posted above (emphasis added):
The point is that there is a point in time at which the copies are subjectively indistinguishable. RPI then tells you to split your indexical credences equally among each copy. And this makes your anticipations (about what color you’ll see when the blindfold is lifted) depend on whether Heads or Tails is true, so when you see that you’re in a blue room you update towards Tails.
Sorry about that—didn’t catch it. I think this solution is probably right. I have a couple more minor objections scratching at the back of my mind, but I’m sure I’ll work them out.
Why did this comment get downvoted? (I up-voted it so that it is at 0.) But seriously, this comment is on point and gives helpful references. What gives?
Its on 6 karma now, but I suspect the downvote was due to the language being a little obtuse to those who don’t know the jargon. Indexical construction of evidence really means nothing to me (a PhD student in statistics), and I suspect it means nothing to others.
Sorry about that. I should have remembered that I was using what is mostly a philosophical term of art. The concept that JonathanLivengood explains is the one I’m referring to. Indexical statements/beliefs are also known as “centered worlds” (i.e. world-observer pairs as opposed to mere worlds), “attitudes de se”, or “self-locating beliefs”.
Ah! Yes, that makes sense. Philosophers use this language all the time, and I sometimes forget what is common speech and what is not. An index in this case is like a pointer. Words like “this,” “here,” “now,” “I,” are all indexicals—they indicate as if by gesturing (and sometimes, actually by gesturing).
That is, unless there is an ambiguity hidden in the question, where that ambiguity can be resolved in one of two ways yielding answers of 0.5 and 0.99. And there is. The word “probability” can be expanded out into a definition in terms of set measure. By bringing in cloning, you’ve introduced an ambiguity: it can either be measure over worlds, or measure over (world,observer) pairs.
When you add an observation (“you see X”), you are either excluding world-observer pairs in which the observer has not seen X, or else excluding worlds in which no observer ever sees X. When you say that the probability that it’s tails is 0.5, that’s because it’s tails in half of the worlds. When you say that the probability is 0.99, that’s because it’s tails, that’s because it’s tails for 0.99 of the world-observer pairs.
This strikes me as having similarities to the old chestnut of the mathematician’s children. If he tells you that he has at least one daughter, the probability that both his children are daughters is one-third (the options being (GG, GB, BG). But if he tells you that his eldest child is a daughter, the probability is one-half, because the options are (GB, GG). The two statements feel intuitively like the same kind of information, but they aren’t. Likewise with the difference between “one copy” and “this copy”.
Please use the more correct form of the problem which goes as follows:
Mathematician: I have two children
You: Is at least one of them a daughter?
Mathematician: Yes.
If he just offers the information “I have at least one daughter” the probability distributions all depends on WHY he offered that piece of information, and it may depend on all sort of weird criteria. The form I stated above removes that ambiguity.
Is this a joke?
Probability chestnuts make my teeth ache, and I frequently fall for them, but… this seems bogus to me.
In the first case, “I have at least one daughter”, why can’t I calculate the probability of two daughters as “I know one child is female, so I’m being asked for the probability that the other child is female, which is one-half”?
Introducing birth order to this reasoning seems entirely unjustified. Conversely, if it is justified, I don’t understand why “the options are (GG, BGX, BGY, GBX, GBY), therefore the probability of two girls is one-fifth” (where X and Y stand for any other factor I choose to take into consideration… for example, “taller than the median height for a member of that sex” and “shorter than...”, respectively) isn’t equally justified.
Put differently: if I’m going to consider birth order, it’s not clear why I should expect GG, BG, and GB to have equal likelihood.
Nope. You’ve fallen for this one.
You might also like the scenario where the additional piece of information is that the one child was born on Tuesday.
Never mind, I lost track of who was being responded to.
I wasn’t responding to you. I agree with your analysis.
OtherDave, it’s not just about birth order, the same difference in probability distributions would apply if we indexed the children according to height (e.g. ‘is your tallest child a daughter’) or according to foot-size or according to academic achievements or according to paycheck-size or whatever. As long as there’s indexing involved, “child A is daughter” gives us 1⁄2 possibility both are daughters, and “at least one child is daughter” gives us 1⁄3 possibility both are daughters. And more bizarre differentiations (see below) give us different results.
Because there’s no “other child” when the information he provides is “at least one”. There may be a single “other one”, or there may be no single “other one” if both of his children are girls.
But it may be better if you try to think of it in the sense of Bayesian evidence. Please consider the even more bizarre scenario below:
A. - I have two children
B—Is at least one of them a daughter born on a Tuesday?
A—Yes
Now the possibility he has two daughters is higher than 1⁄3 but less than 1⁄2.
Why? Because a person having two daughters has a greater chance of having a daughter born on a Tuesday, than a person that has only one daughter does. This means that having a daughter born on a Tuesday correlates MORE with people having two daughters than with people having only one daughter.
Likewise:
Do you have at least one daughter who’s class-president?
Yes
The more daughters someone has, the more likely that at least one of them is class-president. This means that it increases the probability he has many daughters. But if the question is
“Is your eldest daughter a class-president”?
Yes This doesn’t correlate with number of daughters, because we just care about one specific them, and the rest of them may equally well be zero or a dozen.
Now (to return to the simpler problem):
person having one daughter and one son—has 50% chance of having “eldest child be a daughter”, has 100% chance of having “at least one child be a daughter”
person having two daughters—has 100% chance of having “eldest child be a daughter”, has 100% chance of having “at least one child be a daughter”
I’m avoiding the math, but this mere fact ought suffice to show you that the worth of the respective evidence “at least one child is daughter ” and “eldest child is daughter” correlates different between the two hypotheses, so they support each hypothesis differently.
Not a joke at all. But it is a source of flamewars as virulent as Monty Hall and 0.999 repeating. :)
ArisKatsaris explained it well, giving the fully correct setup, so I won’t add anything more.
This is isomorphic to the Sleeping Beauty problem.
Depends on how you are scoring. If you weight the accuracy of the (identical) beliefs of the 99 copies that see the same color 99 times as much as the accuracy of the one other copy it’s .99. If you instead weight the accuracy the copies who see red and the copies who see blue equally, regardless of the number in both sets (perhaps because one color is randomly chosen to make a decision that affects all copies equally) it’s .5.
I am of the opinion that “You are placed in a red room” gives additional information beyond “Someone is placed in a red room”.
I’d like to add that, if you think that’s all the information, then you’d still conclude that there’s a 99% chance that it came down tails, on the basis that it’s 99 times more likely that someone ended up in a red room and every tiny detail that you didn’t feel the need to mention happened. That said, the probabilities don’t come out quite the same, and this does not dissolve the question.
If you don’t open your eyes and guess, you have a 0.5 chance of being right. The policy of ignoring the wall-color in front of you makes 50% of ’identical-you’s wrong about the coin toss.
Opening your eyes allows 99% of atomically ‘identical-you’s to guess the coin toss correctly, and gives the other 1% information that will lead them astray. The policy of taking the wall-colour as evidence makes 99% of ’identical-you’s correct about the coin toss.
This is as far as I am able to take the problem, to a point that I understand what’s going on. What aren’t I getting, if from here it doesn’t seem special?
Let’s consider writing a computer program to answer the question instead of doing it ourselves.
Program A: First it prints out 0.5 then it is given one of two inputs “red” or “blue”. If it is given “red” it prints out 0.01 and if it is given “blue” it prints out 0.99.
Program B: Prints out 0.5, ignores input then prints out 0.5 again.
Now we can run the red/blue room simulation many times and see which program makes better guesses.
I didn’t actually write the program, but with this point of view it is easy to see what happens: Strategy A works best in the majority of cases except for the one unlucky guy whose prediction is right off (due to being in a misleading difficult situation), Strategy B works badly in all cases except it’s better than A for that one guy.
So you may have found this a paradox in the sense that (for this one guy who was put into a confusing situation) being more rational can mean you make a much worse prediction!
I think it’s fairly simple, as I’ve encountered this problem before. The odds would be .5 in the terms that “either the coin landed heads or it didn’t”, with (assuming non-partial others) a 1% chance that you’re in the odd room. To you it might seem that it’s .5 because you’re either in a blue room, or you’re not. However, it’s also that you’re either in the odd room, or one of the 99 of the differently coloured ones.
One variant would be that there would be two seemingly identical buildings, one with 99 red rooms and one blue, one with 99 blue rooms and one red. You go into one of the buildings, and fall down a trapdoor into a blue room. The odds that you walked into the building with 99 blue rooms should be 99 %.
With a 99 percent certainty, you can guess what the result of the coinflip was.
This problem has a neat symmetry in that the players are copies of you; so all copies finding themselves in blue rooms will assign p(tails | blue)=x, and conversely all copies finding themselves in red rooms will assign p(tails | red)=1-x. This way (outside view), the problem reduces to finding what x gives the best sum log-score for all observers. Running the numbers gets x=99/100, but this problem has way too much symmetry to attribute this to any particular explanation. Basically, this is faul_sname’s reasoning with more math.
I’m going to try and slowly introduce indexicality into the setup, to argue that it’s really there all along and you ought to answer 0.99.
Do you have to be copied? The multiple coexisting copies are a bit of a distraction. Let’s say the setup is that you’re told you are going to be put in the first room and subsequently given a memory-erasing drug; then the second room and so on.
Now one more adjustment: the memory-erasing drug is imperfect; it merely makes your memory of each trial very fuzzy and uncertain. You can’t remember what you saw and deduced each time, but you remember that it happened. In particular, you know that you’re in the 65th room right now, although you have no idea what you saw in the previous 64.
Finally, instead of moving you from the first room to the 100th orderly, let’s say that a random room is chosen for you every time, and this is repeated a million times.
In this case, the answer 0.99 seems inevitable. The choice of the room for you was randomly independent in this particular trial, and brings you fresh legitimate evidence to bayes upon. Note that you can say that you would, with very high probability, see blue in one of the trials regardless of what the coin showed. This doesn’t faze you; if the coin were heads, it’s very unlikely that this particular 65th trial would be one to turn blue on you; and your answer reflects this unlikelyhood.
Now go back to orderly traversing the 100 rooms. This time you can say with certainty that you’d see blue in one of the trials no matter what. But this certainty is only very very slightly more than the very high probability of the previous paragraph; going from 0.99 to 0.5 on the strength of that change is hardly warranted. As before, if the coin were heads, it’s very unlikely that this particular 65th trial would be the one with the blue room; and your answer should reflect this unlikelyhood.
Now go back yet again: they improved the drug and you no longer remember how many trials you had before this one. Should this matter a lot to your response? When you knew this was the 65th trial, you knew nothing about your experience in the previous 64. And it was guaranteed that you’d eventually reach the 65th. It seems that the number of the trial isn’t giving you any real information, and your answer ought to remain unchanged without access to it.
I’m not a statistician, but I think that this dilemma might simply sound like other, much trickier probability issues.
One of the misleading aspects of it is this line: ‘When asked what the probability is that the coin came down tails, you of course answer “.5”.’ The problem is that this is posed in the past tense, but (without any other information whatsoever) the subject must treat it the same way as the question about a future situation: ‘What is the probability that a fair-coin would come down tails, all other things being equal?’
But the next time the question is asked, there’s a result to tie it to. It is an actual event with observable secondary effects, and you can hazard an educated guess based on the new information at hand.
Perhaps your thought-experiment vaguely references the counter-intuitive notion that if a coin is flipped 99 times and comes up tails every time, the chances of it coming up tails again is still just .5. But it’s more like if you had a jar or 100 marbles, containg either 99-white/1-black or 1-white/99-black and you reached in without looking. What is the probability that you managed to pick that 1 out of 100? It’s the same principle that is at work in the Monty Hall problem.
Your confusion confuses me.
If after removing the blindfold in the blue room you offer every copy of me the same pair of bets of “pay me X dollars now and I’ll give you 1 dollar if the coin was tails or pay me 1-X dollars and get 1 dollar if was heads” I bet tails when X < .99 and take heads when when .99 .99. If we’re both blindfolded and copied and both show up in a room I will offer you the bet “pay me $.49 and get $1 if the coin flip resulted in (tails when in a red room, heads when in a blue room) and the 100 copies of me will have $48 more than the copies of you (which we will split if we meet up later). If I clone you, someone else flips the coin and paints the rooms, then I open one room I offer you the same bet and make an expected (.99 x $0.49 - .01 x $0.51) = $0.48.
If all 100 copies of me have our sight networked and our perception is messed up such that we can’t see colours (but we can still tell at least one of us is in a blue room (say this problem is triggered by anyone on the network seeing the colour blue)); then we agree with the 50-50 odds on heads-tails. If all of my copies have bug-free networked sight I I’ll take bets whenever (X <> 1 given 99 blue rooms AND X <> 0 given 99 red rooms).
Is there another meaning of probability that I’m missing? Does this clarify the informational value of learning the colour of the room you wake up in?
I dunno, but this is gotta have something to do with “what do Born probabilities mean if the many-worlds hypothesis is true” issue.
SUDDENLY, ANSWERS! VROOOOOOOMMMM.