There’s a question that’s held my fascination for months. At times, it’s had me spinning in circles, caught up in seemingly impossible contradictions. And it’s not the famous Sleeping Beauty problem… or at least, not exactly.
There is a certain Vincent Conitzer, of Duke University, who presents what he calls a “Devastating” example supposedly disproving the logic behind the “1/2″ answer to the original problem.
His argument is valid, but based on a faulty premise. He assumes that Halfers, to get their answer, always look at remaining possibilities and renormalize equally between them. (At no point does he justify this assumption.)
So Conitzer is making a mistake in his critique of the Halfer position. I would say that anyone who still believes in the “1/3” answer to the original SB problem is making a mistake. But I’m not any less error prone myself. Holy hells, have I made mistakes. Conitzer’s example is a variant of SB with an easier answer, because it’s a question where the correct answer and the intuitive answer are the same. Yet at one point I got so confused, I believed in an answer that really should be self-evidently wrong to anyone.
The question that’s fascinated me for months isn’t any particular object-level SB variant, but rather the meta-level question: What makes Sleeping Beauty so difficult?
I think the typical answer here is, in a word, “anthropics”.
Take, for instance, Eliezer Yudkowsky, who in Project Lawful writes (via a character he uses to serve as a mouthpiece for lecturing on rationality):
This however would get us into ‘anthropics’, and we are not getting into ‘anthropics’. That, by the way, is a general slogan of dath ilani classes on probability theory: We are not getting into ‘anthropics’. I’m not even going to translate the word. As long as you don’t make any copies of people, you can stay out of that kind of trouble. That trouble-free life should be our ambition for at least the next several weeks.
I’m not satisfied with that. “Anthropics” doesn’t get a free pass to be inexplicably mysterious. It should bend to the laws of math, like everything else.
This is a post with a two-part thesis:
The Equiprobability bias is really good at leading humans astray.
If you want the right answer, you don’t need the “SIA” or “SSA” or the like. Just use Bayes’ Theorem.
But before we dive into the specifics of Sleeping Beauty and its variants, first I’d like to start with a very different example with surprisingly similar underlying math...
Monty Hall
“Well, a remarkable thing about this problem, simple as it is, is that it has sparked just endless debate”
In the Monty Hall problem, when you first pick a door, the chance of it being the right one is 1⁄3. When Monty opens a door and then you choose to switch, you’ve now narrowed down a possibility space from 3 options to 2: It can either be behind door 1 or door 3. A 50⁄50 chance! Yay! That’s an improvement.
Well, actually, that’s wrong. It’s not 50⁄50. If you switch, your chances go up to ⅔.
According to Conitzer, the “Halfer rule” says that whenever possibilities are eliminated, you should renormalize between all remaining possibilities equally. Here, that gets the wrong answer of 2⁄3.
I should hope no one uses Conitzer’s inaptly-named “Halfer rule”. Let’s see instead how my proposed solution of relying on Bayes handles this problem:
Say that you guess door 1, Monty opens door 2 to reveal a zonk, and you decide to switch to door 3.
Before any choices have been made, our prior probabilities for each door hiding the sports car are as follows:
P(1) = P(2) = P(3) = 1⁄3
The prior probability it’s not behind door 2?
P(~2) = ⅔
We want to know the chance it’s behind door 3, given that it’s not behind door 2, AKA:
P(3 | ~2) = ?
We also know that if it’s behind door 3, it cannot be behind door 2:
Conitzer’s variant begins like this: Sleeping Beauty gets put to sleep, wakes up Monday morning, and is shown a coin flip toss. Then regardless of the coin flip’s result, her memory of Monday will be erased and the procedure repeated: She gets put to sleep, wakes up Tuesday, is shown a second coin toss, and has her memory again wiped.
Conitzer’s question: After waking and observing a coin flip to Heads, what should Beauty believe is the chance that the two coin flips will have had different results? I.e., what’s the chance that either today is Monday and the coins will be Heads-Tails (HT) OR today is Tuesday and the coins were Tails-Heads (TH)?
If you believe that, after observing Heads, you should update to P(coins are different) = 2⁄3, then consider that the situation is symmetrical with Tails. Upon observing Tails, you should also update to 2⁄3. But that means no matter what you observe, you will update to 2/3—and if you already know this update will happen, you should just go ahead and update now. But that would mean believing two fair coins have a 2⁄3 likelihood of flipping to different results, even before having observed anything.
(For a period of time, this is what I actually believed! The reason for my confusion wasn’t necessary to the main points of this post, but if you’re interested, I explain in the bonus section at the end.)
Again, the “Halfer rule” (of renormalizing equally among remaining possibilities) gets the wrong answer of ⅔. That logic goes like this: When you observe a Heads, you learn that Tails-Tails (TT) has not happened. Because you don’t know which day it is, however, you haven’t learned anything else. That means all of HH, HT and TH remain as possibilities…
…with equal likelihood. HT and TH are two of the three options, so the chance the coin flips are different must be two out of three.
These get the same results because they mean the same thing. Eliminate TT, and this is what happens.
This is a fun problem! It’s weird. Unless you already understand it perfectly, you should be surprised by this. If you feel confused, that’s a good thing! (And if you’re not confused, then that’s also a good thing, and kudos.)
We woke up, we observed a Heads, we logically eliminated TT from the possibilities, and of course could not eliminate any of HT, TH, or HH. So where’s the flaw in the above reasoning?
The Monty Hall solution
There’s a different Bayes calculation we can apply to the Monty Hall problem that will give us the correct answer:
Except this only works because I redefined the meaning of “~2”. Before, it referred to the initial chance that Door 2 would be hiding the car. Now, it means: Once you’ve chosen to open Door 1, what’s the chance that Monty will reveal a zonk by opening Door 2?
He has to choose either Door 2 or Door 3 to open, and they’re equally probable from our perspective, hence: P(~2) = 1⁄2. If the car is behind Door 3, then we know Door 2 must be opened, and thus: P(~2 | 3) = 1. Then the rest is just plug-and-play with Bayes.
Our earlier calculation wasn’t wrong. It just didn’t capture as much information about the situation. True and accurate statements can still be incomplete.
But…
…you might have noticed…
…this doesn’t actually resolve anything. If both mathematical calculations are accurate, how do we choose which one to use over the other?
The answer: You don’t have to choose one over the other. You can just use both! If you’ve got two pieces of evidence, A and B, then Bayes lets you update on A first, then followed by B. Or B followed by A. That’s true even if A happens to be a superset of the information with B. No matter how you slice it, the math will work out.
O2 implies ~2, so we get no change when considering ~2 as a subsequent, standalone update.
Sleeping Beauty works the same way
Earlier, I considered the event “TT” and described:
P(~TT) = 3⁄4
P(~TT | Different) = 1
Let’s instead consider the observation of not whether you will, on either Monday or Tuesday, eventually observe a Heads, but consider the event of observing a Heads right now, in this particular moment of observation, and let’s call that event “H”.
P(H) = 1⁄2
Because, you know… fair coin. But here’s a diagram anyway:
P(H | Different) = 1⁄2
Because if there’s one Heads and one Tails, but you don’t know which will be on which day, and you don’t know which day it is, then yeah… it’s still an even chance.
Like with the Monty Hall problem, we have two ways of looking at the situation and two similarly defined events, one which gets the right answer (in this case, “H”) and one that does not (“~TT”). It can seem like a paradox: Without already knowing the right answer, how do we know to use “H” instead of “~TT”? And as before, if you’re not sure about which event to update on, you can simply update on both.
Let’s confirm.
If we wake up and see Heads, then we know Tails-Tails isn’t happening. Therefore H → ~TT, and we should expect that if we update on H first, a subsequent update on ~TT should effect no change:
And in the other direction, if we update on ~TT first and then H?
Calculated previously:
P(different | ~TT) = ⅔
Also, if we know Tails-Tails didn’t happen, then we know the chance of experiencing a Heads waking is higher:
P(H | ~TT) = ⅔
Also, if we know the coins are different, then we already know ~TT. Since we already established that P(H | different) = 1⁄2, that means this is the same:
I love this kind of thing, math working out the way you expect it to.
So the general guidance I’d like to give, if you’re seeing Bayes spit out different answers and you’re not sure which to go with: Just go with both! Add as much information as you can assemble, then see where the math takes you.
Furthermore, I think there will always be a clue you can find in retrospect which will act to confirm your findings. For instance:
Noticing that H → ~TT, but ~TT → H doesn’t make sense
Noticing that the event “not Door 2” was being defined in different ways
Noticing that when Sleeping Beauty wakes up in her original problem, she has learned no new information (and therefore has nothing to update on)
From my bonus section: Noticing that the conditions of the bettor’s approach represents new information not yet considered
Why do people answer 1⁄3 to Sleeping Beauty?
“The Equiprobability Bias: [Possible] Events tend to be viewed as equally likely” — Joan B. Garfield
Roll two dice, and you can get any sum between 2 and 12. These sums are not all equally likely; there are more ways to sum to 7, for example, than to get snake eyes. Yet “people show a strong tendency to believe that 11 and 12 are equally likely” (Nicolas Gauvrit and Kinga Morsanyi).
A better version of this dice problem was made famous by the Grand Duke of Tuscany and Galileo, around the year 1620. The Grand Duke considered three dice rolls, then asked Galileo if the sum of “10” was any more likely than the sum of “9”.
As Florent Buisson explains here, both sums can be achieved in six ways, via the following permutations:
It’s easy to think that “9” and “10” must be equally probable, and yet: “10” is more likely than “9”, because though it has the same number of summing permutations, “10” has more combinations. (4+3+3 is three times as likely to result as 3+3+3, which can only result if all dice rolled a 3.)
I believe the Equiprobability Bias is why so many people get confused by the SB problem.
We saw the exact same dynamic at play with Monty Hall, thinking that the two doors that remain after one is eliminated must each be 1⁄2.
Are you surprised by Benford’s Law? If so, it counts towards the Equiprobability Bias as well.
It should make sense: When we first learn probability as kids, we’re trained on examples involving coins and dice and drawing cards from decks, all equiprobable events. It’s like the $1.10 ball-and-bat problem: a failure of over-generalizing simple rules and not working through the necessary steps.
With SB, we’re being presented three equal-seeming options. Then we over-generalize. I’m no exception to this—I used to be a Thirder myself, until I thought through the problem more. And the gods know I’ve made all sorts of reasoning and mathematical errors in my life.
But I also think there are some very smart people getting unnecessarily tripped up by SB due to a tendency to get tangled up into webs as they contend with the aforementioned topic of anthropics.
You don’t need the SIA, SSA, self-locality, or anthropics
There’s another problem I’ve yet to mention in this post, despite the fact that it is completely isomorphic to Conitzer’s SB variant.
And by “isomorphic” I mean the math is EXACTLY the same.
It’s called the “Boy or Girl paradox”, and I don’t think it should be any more or less difficult to understand than Conitzer’s problem.
Except it is.
I don’t think Bentham’s Bulldog and other Thirders would trip up on the Boy or Girl paradox in the same way. Where the Boy or Girl problem merely involves the meeting of kids or dogs in different contexts, Sleeping Beauty uses amnesia to create experientially identical “observer moments”. Thinking about observer moments, or anthropics, can lead you down twisty chains of thought.
For instance: In the Conitzer variant, some would argue that the difference between events “H” and “~TT” is one of “self-locality”: ~TT is a description of the world from a more removed, objective perspective. H is a description of you during a particular observer moment. This is the fundamental difference, they’d argue. But this sort of thinking gives rise to paradoxes, which then require all sorts of convoluted concepts like the “SIA” or “SSA” to resolve, and can lead to EXTREMELY absurd conclusions such as the presumptuous philosopher.
As I argue here and Ape in the Coat argues here, neither the SIA nor the SSA is correct because both rely on the Doomsday argument’s faulty anthropic assumptions.
We don’t need “self-locality” when we can simply recognize that “H” holds more informational content than “~TT”.
Except I’m not satisfied leaving it at that.
How is it I was once a Thirder myself, and yet I’ve never believed that the 1⁄2 Boy or Girl problem could have any other answer than 1/2? If the math is the same, where does the extra confusion come from?
I believe I’ve found my answer, and it’s the same answer I’ve already presented: The Equiprobability Bias.
With Boy or Girl, there are two kids or dogs with two possible genders each, so your brain naturally thinks of four options. You’re more likely to think of three options (and therefore probabilities of 1⁄3 and 2⁄3) after seeing a diagram that, say, eliminates “GG” as an option.
Whereas with SB, the wakings form a viscerally graphic and intuitive set of three.
That’s really it, I think: the seed from which all these other tangled roots have grown.
If you have any other SB variants you’ve struggled with, or that you’ve created, please share them! I bet that with Bayes, we can find their answers.
Let’s say that during Conitzer’s setup, after Beauty awakes and sees a Heads, she meets a bettor who explains:
I decided to pick a random “Heads” day to approach you.
(This means if TT happened, I would not have approached.)
Do you want to take the bet that both coins flipped Heads?
If you’re right, you’ll gain $3. Wrong and you’ll lose $2.
Previously, I showed that P(Different | observation) = 1⁄2. If that’s the case, then the expected payoff for accepting the bet would be (1/2)(+$3) + (1/2)(-$2), which is positive. You should accept the bet!
Right?
Well, actually…
The expected payoff is (1/3)(+$3) + (2/3)(-$2), a negative value, which makes this a bad bet.
It seems as if there’s a sort of logic that’s needed to get the right answer to Conitzer’s problem, which leads to 1⁄2, and there’s another sort of logic needed to get the right answer here, which leads to 2⁄3. We distinguished them with the descriptors of “Beauty will at some point see Heads” versus “Beauty is currently seeing Heads”—a more general statement versus a “self-locating” one.
That’s what a particular r/slatestarcodex redditor argued with me, and it tripped me up bad. We went in circles and circles on this, yet at no point did either one of us realize that those 2⁄3 values represent different things.
This becomes easy to see if we consider one last variant, wherein the bettor approaches Beauty but changes the conditions of his approach:
I chose a random day (Heads or Tails) to approach you.
Do you want to take the bet (same as before)?
In this case: Yes! Take the bet! The chance the coins are different or the same really is 1⁄2 and you stand to gain in expected value.
The key is this: “I decided to pick a random “Heads” day to approach you” is new information. “Beauty is currently seeing Heads” represented the most up-to-date information before the bet, with its answer of 1⁄2. Learning about the approach creates an update towards 2/3…
…and makes this yet another example of the guideline: When in doubt, just try Bayes again.
Just Use Bayes: Sleeping Beauty and Monty Hall
There’s a question that’s held my fascination for months. At times, it’s had me spinning in circles, caught up in seemingly impossible contradictions. And it’s not the famous Sleeping Beauty problem… or at least, not exactly.
There is a certain Vincent Conitzer, of Duke University, who presents what he calls a “Devastating” example supposedly disproving the logic behind the “1/2″ answer to the original problem.
His argument is valid, but based on a faulty premise. He assumes that Halfers, to get their answer, always look at remaining possibilities and renormalize equally between them. (At no point does he justify this assumption.)
So Conitzer is making a mistake in his critique of the Halfer position. I would say that anyone who still believes in the “1/3” answer to the original SB problem is making a mistake. But I’m not any less error prone myself. Holy hells, have I made mistakes. Conitzer’s example is a variant of SB with an easier answer, because it’s a question where the correct answer and the intuitive answer are the same. Yet at one point I got so confused, I believed in an answer that really should be self-evidently wrong to anyone.
The question that’s fascinated me for months isn’t any particular object-level SB variant, but rather the meta-level question: What makes Sleeping Beauty so difficult?
I think the typical answer here is, in a word, “anthropics”.
Take, for instance, Eliezer Yudkowsky, who in Project Lawful writes (via a character he uses to serve as a mouthpiece for lecturing on rationality):
I’m not satisfied with that. “Anthropics” doesn’t get a free pass to be inexplicably mysterious. It should bend to the laws of math, like everything else.
This is a post with a two-part thesis:
The Equiprobability bias is really good at leading humans astray.
If you want the right answer, you don’t need the “SIA” or “SSA” or the like. Just use Bayes’ Theorem.
But before we dive into the specifics of Sleeping Beauty and its variants, first I’d like to start with a very different example with surprisingly similar underlying math...
Monty Hall
“Well, a remarkable thing about this problem, simple as it is, is that it has sparked just endless debate”
In the Monty Hall problem, when you first pick a door, the chance of it being the right one is 1⁄3. When Monty opens a door and then you choose to switch, you’ve now narrowed down a possibility space from 3 options to 2: It can either be behind door 1 or door 3. A 50⁄50 chance! Yay! That’s an improvement.
Well, actually, that’s wrong. It’s not 50⁄50. If you switch, your chances go up to ⅔.
According to Conitzer, the “Halfer rule” says that whenever possibilities are eliminated, you should renormalize between all remaining possibilities equally. Here, that gets the wrong answer of 2⁄3.
I should hope no one uses Conitzer’s inaptly-named “Halfer rule”. Let’s see instead how my proposed solution of relying on Bayes handles this problem:
Say that you guess door 1, Monty opens door 2 to reveal a zonk, and you decide to switch to door 3.
Before any choices have been made, our prior probabilities for each door hiding the sports car are as follows:
The prior probability it’s not behind door 2?
We want to know the chance it’s behind door 3, given that it’s not behind door 2, AKA:
We also know that if it’s behind door 3, it cannot be behind door 2:
All that’s left is Bayes:
Somehow, we got the wrong answer of 1⁄2 again.
But that’s alright. We’ll come back to this.
The “devastating” problem
Conitzer’s variant begins like this: Sleeping Beauty gets put to sleep, wakes up Monday morning, and is shown a coin flip toss. Then regardless of the coin flip’s result, her memory of Monday will be erased and the procedure repeated: She gets put to sleep, wakes up Tuesday, is shown a second coin toss, and has her memory again wiped.
Conitzer’s question: After waking and observing a coin flip to Heads, what should Beauty believe is the chance that the two coin flips will have had different results? I.e., what’s the chance that either today is Monday and the coins will be Heads-Tails (HT) OR today is Tuesday and the coins were Tails-Heads (TH)?
If you believe that, after observing Heads, you should update to P(coins are different) = 2⁄3, then consider that the situation is symmetrical with Tails. Upon observing Tails, you should also update to 2⁄3. But that means no matter what you observe, you will update to 2/3—and if you already know this update will happen, you should just go ahead and update now. But that would mean believing two fair coins have a 2⁄3 likelihood of flipping to different results, even before having observed anything.
(For a period of time, this is what I actually believed! The reason for my confusion wasn’t necessary to the main points of this post, but if you’re interested, I explain in the bonus section at the end.)
Again, the “Halfer rule” (of renormalizing equally among remaining possibilities) gets the wrong answer of ⅔. That logic goes like this: When you observe a Heads, you learn that Tails-Tails (TT) has not happened. Because you don’t know which day it is, however, you haven’t learned anything else. That means all of HH, HT and TH remain as possibilities…
…with equal likelihood. HT and TH are two of the three options, so the chance the coin flips are different must be two out of three.
What about Bayes?
These get the same results because they mean the same thing. Eliminate TT, and this is what happens.
This is a fun problem! It’s weird. Unless you already understand it perfectly, you should be surprised by this. If you feel confused, that’s a good thing! (And if you’re not confused, then that’s also a good thing, and kudos.)
We woke up, we observed a Heads, we logically eliminated TT from the possibilities, and of course could not eliminate any of HT, TH, or HH. So where’s the flaw in the above reasoning?
The Monty Hall solution
There’s a different Bayes calculation we can apply to the Monty Hall problem that will give us the correct answer:
Except this only works because I redefined the meaning of “~2”. Before, it referred to the initial chance that Door 2 would be hiding the car. Now, it means: Once you’ve chosen to open Door 1, what’s the chance that Monty will reveal a zonk by opening Door 2?
He has to choose either Door 2 or Door 3 to open, and they’re equally probable from our perspective, hence: P(~2) = 1⁄2. If the car is behind Door 3, then we know Door 2 must be opened, and thus: P(~2 | 3) = 1. Then the rest is just plug-and-play with Bayes.
Our earlier calculation wasn’t wrong. It just didn’t capture as much information about the situation. True and accurate statements can still be incomplete.
But…
…you might have noticed…
…this doesn’t actually resolve anything. If both mathematical calculations are accurate, how do we choose which one to use over the other?
The answer: You don’t have to choose one over the other. You can just use both! If you’ve got two pieces of evidence, A and B, then Bayes lets you update on A first, then followed by B. Or B followed by A. That’s true even if A happens to be a superset of the information with B. No matter how you slice it, the math will work out.
A penguin or a cow? It’s both! Source: This wonderful visual anagrams gallery
The math
Let “2” mean the event that Door 2 is hiding the car.
Let “O2” mean the event that Monty will Open 2.
Our initial calculation showed:
Now we want to update on O2, in addition to just ~2:
If door 2 doesn’t have the car but door 3 does, Monty will be forced to open door 2:
If you only know that the car’s not behind door 2, then there’s a 1⁄2 chance it’s behind door 3 and 1⁄2 it’s behind door 1:
Then applying Bayes:
Now let’s try going in the opposite direction, starting with the update we calculated using event O2, and seeing what happens when add in “2”:
This one’s easy:
O2 implies ~2, so we get no change when considering ~2 as a subsequent, standalone update.
Sleeping Beauty works the same way
Earlier, I considered the event “TT” and described:
Let’s instead consider the observation of not whether you will, on either Monday or Tuesday, eventually observe a Heads, but consider the event of observing a Heads right now, in this particular moment of observation, and let’s call that event “H”.
Because, you know… fair coin. But here’s a diagram anyway:
Because if there’s one Heads and one Tails, but you don’t know which will be on which day, and you don’t know which day it is, then yeah… it’s still an even chance.
And now the calculation:
Like with the Monty Hall problem, we have two ways of looking at the situation and two similarly defined events, one which gets the right answer (in this case, “H”) and one that does not (“~TT”). It can seem like a paradox: Without already knowing the right answer, how do we know to use “H” instead of “~TT”? And as before, if you’re not sure about which event to update on, you can simply update on both.
Let’s confirm.
If we wake up and see Heads, then we know Tails-Tails isn’t happening. Therefore H → ~TT, and we should expect that if we update on H first, a subsequent update on ~TT should effect no change:
And in the other direction, if we update on ~TT first and then H?
Calculated previously:
Also, if we know Tails-Tails didn’t happen, then we know the chance of experiencing a Heads waking is higher:
Also, if we know the coins are different, then we already know ~TT. Since we already established that P(H | different) = 1⁄2, that means this is the same:
Putting it all together:
I love this kind of thing, math working out the way you expect it to.
So the general guidance I’d like to give, if you’re seeing Bayes spit out different answers and you’re not sure which to go with: Just go with both! Add as much information as you can assemble, then see where the math takes you.
Furthermore, I think there will always be a clue you can find in retrospect which will act to confirm your findings. For instance:
Noticing that H → ~TT, but ~TT → H doesn’t make sense
Noticing that the event “not Door 2” was being defined in different ways
Noticing that when Sleeping Beauty wakes up in her original problem, she has learned no new information (and therefore has nothing to update on)
From my bonus section: Noticing that the conditions of the bettor’s approach represents new information not yet considered
Why do people answer 1⁄3 to Sleeping Beauty?
“The Equiprobability Bias: [Possible] Events tend to be viewed as equally likely” — Joan B. Garfield
Roll two dice, and you can get any sum between 2 and 12. These sums are not all equally likely; there are more ways to sum to 7, for example, than to get snake eyes. Yet “people show a strong tendency to believe that 11 and 12 are equally likely” (Nicolas Gauvrit and Kinga Morsanyi).
A better version of this dice problem was made famous by the Grand Duke of Tuscany and Galileo, around the year 1620. The Grand Duke considered three dice rolls, then asked Galileo if the sum of “10” was any more likely than the sum of “9”.
As Florent Buisson explains here, both sums can be achieved in six ways, via the following permutations:
9 = 6 + 2 + 1 = 5 + 2 + 2 = 5 + 3 + 1 = 4 + 3 + 2 = 4 + 4 + 1 = 3 + 3 + 3
10 = 6 + 2 + 2 = 6 + 3 + 1 = 5 + 3 + 2 = 5 + 4 + 1 = 4 + 4 + 2 = 4 + 3 + 3
It’s easy to think that “9” and “10” must be equally probable, and yet: “10” is more likely than “9”, because though it has the same number of summing permutations, “10” has more combinations. (4+3+3 is three times as likely to result as 3+3+3, which can only result if all dice rolled a 3.)
I believe the Equiprobability Bias is why so many people get confused by the SB problem.
We saw the exact same dynamic at play with Monty Hall, thinking that the two doors that remain after one is eliminated must each be 1⁄2.
The same thing happens again with Bertrand’s box paradox.
Are you surprised by Benford’s Law? If so, it counts towards the Equiprobability Bias as well.
It should make sense: When we first learn probability as kids, we’re trained on examples involving coins and dice and drawing cards from decks, all equiprobable events. It’s like the $1.10 ball-and-bat problem: a failure of over-generalizing simple rules and not working through the necessary steps.
With SB, we’re being presented three equal-seeming options. Then we over-generalize. I’m no exception to this—I used to be a Thirder myself, until I thought through the problem more. And the gods know I’ve made all sorts of reasoning and mathematical errors in my life.
But I also think there are some very smart people getting unnecessarily tripped up by SB due to a tendency to get tangled up into webs as they contend with the aforementioned topic of anthropics.
You don’t need the SIA, SSA, self-locality, or anthropics
There’s another problem I’ve yet to mention in this post, despite the fact that it is completely isomorphic to Conitzer’s SB variant.
And by “isomorphic” I mean the math is EXACTLY the same.
It’s called the “Boy or Girl paradox”, and I don’t think it should be any more or less difficult to understand than Conitzer’s problem.
Except it is.
I don’t think Bentham’s Bulldog and other Thirders would trip up on the Boy or Girl paradox in the same way. Where the Boy or Girl problem merely involves the meeting of kids or dogs in different contexts, Sleeping Beauty uses amnesia to create experientially identical “observer moments”. Thinking about observer moments, or anthropics, can lead you down twisty chains of thought.
For instance: In the Conitzer variant, some would argue that the difference between events “H” and “~TT” is one of “self-locality”: ~TT is a description of the world from a more removed, objective perspective. H is a description of you during a particular observer moment. This is the fundamental difference, they’d argue. But this sort of thinking gives rise to paradoxes, which then require all sorts of convoluted concepts like the “SIA” or “SSA” to resolve, and can lead to EXTREMELY absurd conclusions such as the presumptuous philosopher.
As I argue here and Ape in the Coat argues here, neither the SIA nor the SSA is correct because both rely on the Doomsday argument’s faulty anthropic assumptions.
We don’t need “self-locality” when we can simply recognize that “H” holds more informational content than “~TT”.
Except I’m not satisfied leaving it at that.
How is it I was once a Thirder myself, and yet I’ve never believed that the 1⁄2 Boy or Girl problem could have any other answer than 1/2? If the math is the same, where does the extra confusion come from?
I believe I’ve found my answer, and it’s the same answer I’ve already presented: The Equiprobability Bias.
With Boy or Girl, there are two kids or dogs with two possible genders each, so your brain naturally thinks of four options. You’re more likely to think of three options (and therefore probabilities of 1⁄3 and 2⁄3) after seeing a diagram that, say, eliminates “GG” as an option.
Whereas with SB, the wakings form a viscerally graphic and intuitive set of three.
That’s really it, I think: the seed from which all these other tangled roots have grown.
If you have any other SB variants you’ve struggled with, or that you’ve created, please share them! I bet that with Bayes, we can find their answers.
A sleeping beauty by the talented Ngan Pham
BONUS: My added confusion
Let’s say that during Conitzer’s setup, after Beauty awakes and sees a Heads, she meets a bettor who explains:
Previously, I showed that P(Different | observation) = 1⁄2. If that’s the case, then the expected payoff for accepting the bet would be (1/2)(+$3) + (1/2)(-$2), which is positive. You should accept the bet!
Right?
Well, actually…
The expected payoff is (1/3)(+$3) + (2/3)(-$2), a negative value, which makes this a bad bet.
It seems as if there’s a sort of logic that’s needed to get the right answer to Conitzer’s problem, which leads to 1⁄2, and there’s another sort of logic needed to get the right answer here, which leads to 2⁄3. We distinguished them with the descriptors of “Beauty will at some point see Heads” versus “Beauty is currently seeing Heads”—a more general statement versus a “self-locating” one.
That’s what a particular r/slatestarcodex redditor argued with me, and it tripped me up bad. We went in circles and circles on this, yet at no point did either one of us realize that those 2⁄3 values represent different things.
This becomes easy to see if we consider one last variant, wherein the bettor approaches Beauty but changes the conditions of his approach:
In this case: Yes! Take the bet! The chance the coins are different or the same really is 1⁄2 and you stand to gain in expected value.
The key is this: “I decided to pick a random “Heads” day to approach you” is new information. “Beauty is currently seeing Heads” represented the most up-to-date information before the bet, with its answer of 1⁄2. Learning about the approach creates an update towards 2/3…
…and makes this yet another example of the guideline: When in doubt, just try Bayes again.