The probability is 5/6 all the way through to the end. The two identical twins each know that they rolled six ifef the other didn't, but the pairing procedure gives them unequal measure; they do not exist to the same degree. Thus, the relation between die roll and measure acts as a piece of evidence that it is the other one who rolled six - but this is non-transferrable anthropic evidence, so they are prevented from using each others' evidence to reach agreement.
(This sort of thing is more sensible when you have a utility or preference function to embed the world-model into; this forces you to be clearer about what is being measured by probability.)
Probability theory doesn't automatically work on infinite sets. If you approach this problem as the well-defined limit of a finite problem, the answer is simple.
ETA: Say we take the limit as N goes to infinity, with everything else kept constant. We can end up with two countably infinite sets of twins, and apparently the same probabilities at each step, so we have a probability of 1/2 in the infinite case. Now, pretend that instead, we had X=N/3 (and the angel only tells us that, and nothing else). Then our probability of having a six is 1/3. As N approaches infinity, apparently our probability is still 1/3. But in this infinity land, we can still do the room pairing thing! There is a bijection between any two countably infinite sets. In fact, we could approach the infinite case with any ratio of sixes to non-sixes, as long as it's positive and less than one, and still end up with a bijection between the sixes and the non-sixes. Without a well-defined limit approaching the infinite case, we can produce any probability we want; limits need to be well-defined.
This is explained in Jaynes (2003). I don't have it with me, but if I recall it is in Chapter 15 or thereabouts on marginalization and other paradoxes.
Agreed with everything you say, but I don't think it addresses the main question. Suppose the angel does not say your (2), but the original (2): there is an actual countable infinity of copies of you. If I understand correctly, you are saying that probability theory breaks down under this information and cannot even address the question of how likely is your die to be 1/6. If this is so, isn't a serious problem for multiverse theories, as suggested in the next-to-last quoted paragraph?
(Argh, see edit. Of course I forgot to include the actual argument.)
Well it's certainly a hint that there is something we are confused about. If you are talking about quantum many-worlds, then we can at least prevent our heads from exploding by using the notion of measure to talk about the probability of a world, which even applies to the supposedly uncountably many worlds. (I'm out of my depth here but I think we would then be talking about physical quantum amplitudes in configuration space, rather than subjective probabilities... which mysteriously correspond.)
If you are talking about a single spatially infinite universe... then yeah I don't know how to deal with that. Although I'd at least note that it is a very strange epistemological state that I am in, if I think that I somehow came to have accurate, meaningful beliefs about infinitely many, spatially infinite objects. How did that information get to me?
When N=infinity, N/2=N/6, so Tsvi's version goes through with no changes :P
And as for how to resolve it - take limits. The end.
When N=infinity, N/2=N/6, so Tsvi's version goes through with no changes :P
I always wonder why people do things like say N/2=N/6 "Because it's infinite" then act surprise when it implies something weird. It is only very slightly more impressive than the trick from highschool algebra that makes 1=2 because someone divided by zero.
The main mathematical issue here is no uniform probability distribution on a countable set: if you try to assign each element the same probability, that probability can't be positive and it also can't be zero. In particular, there's no sense in which you can divide the number of twins who rolled a 5 by the total number of twins to get 5/6. The secondary mathematical issue in step 5 is that you're pairing up two infinite subsets of a countable measure space which aren't guaranteed to have the same measure (especially since, as just mentioned, there's no uniform measure here). This is, very roughly speaking, the same kind of monkeying around with infinities that gets you Banach-Tarski, and I would be very skeptical of it having any relevance to real-life decision making.
The sets don't have to be countable; if there are continuum-many of you indexed by the reals from 0 to 1, the angels could match the interval from 0 to 1/6 with the interval from 1/6 to 1. However, doing this does not preserve measure (as jimrandomh pointed out above), which is the real sleight-of-hand that makes this thought experiment akin to the one where everyone who rolled a six gets unwittingly duplicated up to five copies.
It's only if the sets are countable that we can probabilistically predict ahead of time that there is a pairing. To get the existence of a pairing, we need to know that the cardinality of those who rolled six is equal to the cardinality of those who didn't. It is a consequence of the Law of Large Numbers (or can be easily proved directly) that there are infinitely many sixes and infinitely many non-sixes. And any two infinite subsets of a countable set have the same cardinality. But in the uncountable case, while we can still conclude that there are there are infinitely many sixes and infinitely many non-sixes, I don't see how to get that the cardinality is the same. (In fact, events of the form "there are aleph_1 sixes" aren't going to be measurable in the usual product measure used to model independent events, I suspect.)
But of course if there are uncountably many rollers, then, assuming the Axiom of Countable Choice, we can choose a countably infinite subset and work with that.
The real interval [0,1) with Lebesgue measure is commonly used as a probability space; in this case, the measure of cases where one rolls a 6 has Lebesgue measure 1/6, we can without loss of generality say it's the interval [0,1/6), and we can linearly pair every point in this set with a point in the set [1/6,1) to get a nice measurable correspondence. It's just that this fails to preserve measure.
Also, welcome to Less Wrong! You might like to introduce yourself and your interests on a welcome thread (huh, looks like we need a new one).
The main mathematical issue here is no uniform probability distribution on a countable set:
Minor nitpick: You mean infinite set. Any finite set is, of course, both countable and has a uniform probability distribution, and your point is correct for all (measurable) infinite sets.
It depends on your conventions. Some people use "countable" and "countably infinite" to refer to what I refer to as "at most countable" and "countable."
Well, using infinite set would be better either way as this is a property of all infinite sets, not just countable (infinite) sets. It would also avoid any confusion due to this difference of convention.
That said, I didn't actually know about your convention*, so thank you for making me aware of it.
*probably because any easy way of saying "at most countable" in my first language would be confused with "very countable" or even "mostly countable", so I'm guessing none of my professors thought about/remembered this other convention when defining countable sets.
In particular, there's no sense in which you can divide the number of twins who rolled a 5 by the total number of twins to get 5/6.
I agree with this, but why is it relevant for the Bayesian sense of probability? In step 1, my credence on the die being a 6 is obviously 1/6, and this does not require taking any ratio of actual results, finite or infinite. Are you saying that if I learn that the multiverse hypothesis is correct, it stops being 1/6 and becomes ill-defined? Why would it?
The secondary mathematical issue in step 5 is that you're pairing up two infinite subsets of a countable measure space which aren't guaranteed to have the same measure (especially since, as just mentioned, there's no uniform measure here).
I agree that something "funny" happens in steps 4 and 5. The challenge is whether learning this changes your credence, and if so why and to what. I am not sure you are explaining this. (Maybe the answer should be obvious from your words and I'm just not seeing it).
If you want to assign probability 1/2 in step 5, you're implicitly doing it by using some symmetry of the problem (namely the symmetry that exchanges you with the twin sitting across from you). But the mathematical issue above means there's no reason to expect that this is actually a symmetry of the problem. If you agree that the number 5/6 doesn't come from looking at symmetries involving twins in step 2, there's no reason to get a second number by looking at symmetries involving twins in step 5.
Are you saying that if I learn that the multiverse hypothesis is correct, it stops being 1/6 and becomes ill-defined? Why would it?
No, I'm saying that it stays 1/6.
The challenge is whether learning this changes your credence, and if so why and to what.
Nothing changes. And a real Bayesian shouldn't believe the angel in the first place.
Something "funny" happens exactly in step 4. Specifically: Instead of pairing you with a random person, the angel is pairing you with a person selected by specific criteria "if you rolled 6, the other person didn't; and if you didn't, the other person did". Therefore, when meeting the other person, you should abandon the intuition that you both are a randomly selected pair.
The confusing part is that the situation is symetrical for both people. Yes, it is! But that is confusing only because we work with infinity here. You can rearrange an infinite set to increase or decrease a measure of its subset; and this is exactly what happens in step 4.
Step by thep: At the beginning, everyone's (that includes you) chance to have 6 was 1/6. So there was 1/6 of people with sixes, and 5/6 of people without sixes. Then the angel rearranged people, so the measure of people with sixes was increased to 1/2, and the measure of people without sixes was decreased to 1/2. Now at the end, your probability remains 1/6, so the other person's probability is 5/6. -- The confusing part is that the other person at the beginning also had chance 1/6 to have 6. But if they had 6 and you didn't, then there is a higher probability that the angel will assign them to you.
In other words, you should believe that their chance is 5/6 in the sense they if they didn't roll 6, they would be more likely to be assigned by the angel to someone who did; but instead they were assigned to you (and you with probability 5/6 are a person who didn't roll 6).
And the same holds for the other guy?
Yes, it does.
It is important to remember that first we rolled dice, and only later the angel decided that we two will be a pair. If both of us would not roll 6, the angel would simply decide to not bring us together. -- The paradox of infinity is that the angel will always have enough other people to bring, if we both happen to not roll 6.
Let's make it simple by removing the infinities.
There are exactly one million people, each of them rolls a die. An angel blindfolds them, and sorts them into two large groups: those who rolled a six, and those who didn't. The blindfolded people cannot compare the sizes of their groups. Then the angel asks you: "What is the probability of you having rolled six?" (You can assume that the angel speaks simultaneously with everyone, so there is nothing special about asking you this question, instead of asking it someone else.)
A logical answer would be: 1/6.
Then the angel says: "Now I choose a random person from the other group; let's call him Joe. What is the probability that Joe has rolled a six?"
A logical answer: 5/6.
The angel says: "However, Joe believes his chance of having rolled a six is 1/6, and he uses the same reasoning that you do. Is he wrong or what? Remember that I am having this dialogue simultaneously with each participant of the experiment."
A logical answer: Probability is in the mind. In my mind, Joe has a probability 5/6 of having rolled a six. In Joe's mind, he has a probability 1/6 of having rolled a six. These are two different questions.
Imagine that there are six people: me, Joe, Adam, Betty, Cecil, Daniel; and exactly one of us has rolled a 6. If it was me who rolled the six, you could have chosen any of {Joe, Adam, Betty, Cecil, Daniel} to be my twin. But if it was Joe who rolled the six, you didn't have another choice for my twin, only Joe. So the fact that Joe is my twin, is for me an evidence that Joe has rolled the six. And symetrically, if you chose me as a twin for Joe, that is for him an evidence that I have rolled the six.
Because we have this finite scenario, one person was chosen five times as a "twin", one person was chosen once, and four people we not chosen as "twins" at all. So we have only one of six people rolling 6, but we have five of six "twins" rolling 6. Therefore my probability of rolling 6 is 1/6 and my "twin"'s probability is 5/6.
...however in the infinite version, even if one subset is 5 times smaller than the other, the angel can choose a twin for everyone. But in some sense it means that if you rolled six, you are five times more likely to be selected as someone's pair, as you would be if the angel just ignored the numbers on dice and paired people randomly.
In my mind, Joe has a probability 5/6 of having rolled a six. In Joe's mind, he has a probability 1/6 of having rolled a six. These are two different questions.
While they are two different questions, you both have the same evidence and the same priors. They are symmetric questions, and should have identical answers.
So, I have 1/6 probability for the 6, he has 1/6 probability for the 6 and one of us has the 6 for sure?
Yes.
[ETA]: It's important to remember that the probabilities are relative; you have odds of 1/6, you should posit odds of 5/6 for your twin. Likewise, your twin has odds of 1/6, and your twin should posit odds of 5/6 for you.
Say, that the 6 is worth US$1. What is the fair price to sell it, for those two? 20 cents for sure?
In that case the buyer would profit 60 cents.
Are you selling the outcome? I/e, "If I rolled a 6, you can have the $1"?
Because yes, both players should choose to sell the option. And likewise, both should choose to buy if their twin offers. Or, alternatively, both players should rationally agree to exchange outcomes.
And precisely because of how the angel assigned the twins, people would be winning on average in this game.
With infinities, this is possible. Just like in the infinite world, pyramid schemes are possible. You can generate free money simply by having each person with number N send $100 to a person with number N/2.
Thanks for all the comments. For those who are drawn to not changing the probability all the way through the story--which, I agree, is the intuitively right answer--I recommend looking at this variant where sticking to your guns leads to incoherent probabilities.
Pairing the twins and putting them in a room doesn't change the proportion of people who rolled a six. It just finds a six-roller for each non-six-roller and sits them down across from each other.
That you get put into a room isn't evidence about your roll. Whatever you rolled, you still get put into a room, and you get the exact same experience. What you learn by getting paired up is information about the joint state of you and your twin's rolls - without the pairing, there are 36 possibilities, and with the pairing there are 10 (which one of you rolled the six, and five ways for the other roll to turn out).
This is a mess.
Maybe the probability did become 1/2 when you became face-to-face with your doppelganger? There's a small difference between a proof that something exists and giving an example. The existence of the list mentioned in 4 is a logical consequence of 1, 2, and 3. However, it seems like you do get new information when you get put in a room with your doppelganger: you know which pair you're in. You now know that one person in the room rolled a six, and one didn't. If the angels used some other pairing algorithm, then you'd learn something else.
Consider an exaggerated, finite case:
You have a lottery ticket. The drawing has taken place, but you haven't looked up the winning number yet. You then learn that a winning ticket was purchased at the same store that your ticket was bought from. That's good news, right?
That's good news, right?
Most players don't get this news. Most get the news that the winning ticket has been sold elsewhere.
Say, that there is a next step in this game. The angel reveals the number you diced.
Now, if you play it many times, do you think you will dice 6 in 50% or in about 17% of the cases?
Now, if you play it many times, do you think you will dice 6 in 50% or in about 17% of the cases?
There's a slight complication here in that, when you do this the second time, the angels can choose whether or not your twin rolled a six the first time or not. I don't know how much this matters.
Most players don't get this news. Most get the news that the winning ticket has been sold elsewhere.
Well, once you bring infinities into it, things get screwy; there's really no sense in which "more" people rolled "not-six" than rolled "six".
Now, if you play it many times, do you think you will dice 6 in 50% or in about 17% of the cases?
Considering all the other ways that infinities make things crazy, I think that 50% might actually be the right answer, but I can't prove this.
Considering all the other ways that infinities make things crazy, I think that 50% might actually be the right answer, but I can't prove this.
So, the angel pairing makes you dice 6 in one half of all the cases. Without this, only in the every day 17%?
Does this also holds for the 1000000 sides dice?
(I also think that infinities make things crazy. But just too crazy, inconsistent in fact.)
There's only inconsistency when you mix in the axiom of choice, which is implicitly done here.
There are only countably many people. The well-ordering of the integers is accepted in ZF. You only need the axiom of choice if you want to well-order a larger set.
This seems like a fun paradox related to Sleeping Beauty.
This is a standard "paradox" with infinities, all countable infinities are alike. Let's number all twins 1..infinity such that every 6th rolls 6. Now let's setup a pairing where twin n is matched with twin 6n, n not divisible by 6. Now all twins are matched in (rolled 6, did not roll 6) pairs. The act of picking one pair is what makes the probability become 1/2. And this happens in step 5.
Does this give you anthropic superpowers? Suppose that, instead of a die roll, you bought a lottery ticket, and the angels perform the same procedure, matching lottery winners and lottery losers. Did you just increase your chances of having won the lottery from one in several million to one in two? If you did this over and over, could you cheat the lottery?
Angels have the power, you don't. Compare this to a regular raffle where there is an actual human picking a winner from a hat, instead of an angel. If you bribe this person to pick your entry, you increase your odds of winning from 1:n to 100%.
Step 1, 2, and 3: 5/6 chance of not rolling six. Some infinities are bigger than others, by taking the limit you can show that there are 5 times as many people who didin't roll 6 than those who did.
Step 4 For every person who rolled a six and is on that list, there is one person who did not: So, 2/6=1/3 of all the people will be on that list, and of those on that list, 1/2 of them rolled a 6. Until you have more knowledge, there is no reason to suspect you are on that list.
Step 5: Assuming you can know that being transported to the room means you are on that list, then by being transported you just gained new information: You are on the list, which as said before, means your probability of rolling and not rolling a six is 1/2.
Step 4 For every person who rolled a six and is on that list, there is one person who did not: So, 2/6=1/3 of all the people will be on that list
This is wrong. The lists can be made (and in the problem, they are made) in such a way that each person will be in one of the lists. There is a one-one correspondence between people who rolled 6 and people who didn't, in the same way that, as Cantor showed, there is a one-one correspondence between even integers and all integers.
Yes. And there is also a way, to join every looser who does not have the 6 with the 999 winners in a bigger room.
Say, that you find yourself in one of those. How likely it is, that you are the sole looser among nearly 1000 winners?
If you want, you can make it even more extreme. You could have one room with one person, one with two, one with three, etc., each of which contain exactly one loser. Now it seems like the probability of having lost is infinitesimal.
I saw this conundrum at Alexander Pruss's blog and I thought LWers might enjoy discussing it: