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).
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.
[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.
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.
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.
Fair. The issue I identified as secondary is in fact primary.
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).
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.
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?
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.
No, I’m saying that it stays 1⁄6.
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?
Or he must employ a different logic here?
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.
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.