There is much obfuscation surrounding the SB problem, that appear in this thread. One is that when you ask about a probability in the present tense about an occurrence described in the past tense, you are asking about the conditional probability of that event given the present information. What this means is that there is only one question that is relevant: “Given SB’s knowledge when she is awake, what is the probability that the coin result was Heads?” Not “What was the probability, ignoring any information obtained since, that a coin flip result is Heads?” (For comparison, say I draw a card at random from a standard 52-card deck, and then tell you that it is a spade. The probability, that I drew the Ace of Spades, is 1⁄13. Not 1⁄52, the probability when I drew it.)
Then, and this will become important later, the question Adam Elga asked is equivalent to, but not identical to, the one he answered. In the one he asked, SB is woken once, or twice, based on a coin flip. In the one he answered, SB is always woken on Monday, and then again on Tuesday, if the coin result was Tails.
But in the one he asked, he added two details that presaged his solution. Neither detail has any impact on what was asked. The first is that he mentioned that the experiment occurs over two days. But days are irrelevant in what was asked. The only significance of these two days are in his solution. And he asked “when you are first awakened...”. His solution supposed that SB could be told that it is Monday, or that the coin landed Tails, and evaluated that probability after this information was revealed. The point of “when you are first awakened...” was simply to evaluate the probability before either revelation.
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My point is that these timing details are completely irrelevant, and are used to confuse the probability by establishing different kinds of conditions for each waking. A trivial solution can be obtained by making these conditions equivalent for each. Here is one way:
The experiment proceeds over three days, not two. Monday, Tuesday, and Wednesday.
Only one set of conditions produce a waking on each day.
Replace the coin with a six-sided die. Three faces will be green, representing “Heads,” and three will be red, representing “Tails.”
The words “Monday,” “Tuesday” and “Wednesday” are each written on two sides, once in each color.
The die is rolled on Sunday Night, and preserved throughout the experiment. If it [landed on/is currently showing] a green side, SB is woken only on the day indicated on that side. If red, SB is woken twice, on the two days not indicated on that side.
SB is asked for the probability that the die [landed on/is showing] a green side.
BY HALFER LOGIC, the probability that the die “landed on” each of the six sides is 1⁄6. But since she is awake, SB knows that there are only three sides that it could [have landed on/be showing] at the moment. They are still equally likely, and only one of the three is green.
“New information” does not mean learning something happened that was not guaranteed to happen. It means that the information eliminates at least one result that was possible without the information. My version does this transparently, by ruling out three die results.
In the classic version (the one in Elga’s solution), the information is that the combination of coin and day allows SB to be awake. Before that information was obtained, Heads+Tuesday is possible. By being awake, SB can eliminate one of the four equally-likely possibilities, leaving three. The obfuscation in that version happens because the two combinations that can occur on Monday are treated as equivalent to only one that can occur on Tuesday.
“BY HALFER LOGIC, the probability that the die “landed on” each of the six sides is 1⁄6. But since she is awake, SB knows that there are only three sides that it could [have landed on/be showing] at the moment. They are still equally likely, and only one of the three is green.”
Hello, I’m not sure what you’re trying to say. There are six possibilities here (M,W,T x R,G) and yes, I do think that each of them has a 1⁄6 chance. The chance of Red is equal to the chance of Green.
With all due respect, I don’t sense that you are trying to understand it. Your position seems to be firmly that SB has gained no information, so she can’t use conditional probability. Mine is that she does, and all I’m trying to do is make it hard to ignore why it is.
As you point out, there are six possible die rolls. But on any one day, only three remain possible. This is what is known, in probability theory, as a “condition.” It is what allows one to “update” a prior probability, which is what you calculated, to a “posterior” or “conditional” probability, which is what the Sleeping Beauty Problem asks for. This is true even if she gains no information. The condition will be the entire sample space, and the “updated” probability will be the same as the prior probability. So there is no reason to not explore what the condition is.
It’s easier if we number the days 0 (for “Monday”), 1 and 2. The set of possible die rolls—can we agree that this is the sample space? - is {G_0, G_1, G_2, R_0, R_1, R_2}. When she is awake, she does not know which day it is, but she does know that it is only one day. Call it day D. The condition is the event {G_D, R_MOD(D+1,3), R_MOD(D+2,3)}. The condition is the same regardless of what D is. Each outcome in this set has a prior probability of 1⁄6. The posterior (or conditional) probability is of G_D is (1/6)/[(1/6)+(1/6)+(1/6)]=1/3.
Let’s say she learns it’s Monday. Half the time, she’ll be awake on a random day. Half the time, she’ll be awake on 2⁄3 of days. P(M) = P(T) = P(W) = (1/2)(1/3) + (1/2)(2/3) = 1⁄2 (Note that this is distinct from the probability that a randomly chosen waking is M/T/W—more on that in a second)
What did I mean by “a randomly chosen waking is M/T/W”? An example would be the chance of Beauty receiving a knock on the door, if an outside experimenter flips a coin to decide whether to knock on her door on her first red waking or the second (or the only choice of green waking, if it was green).
The chance that she gets a knock on a Monday would be (1/2)(1/3) + (1/2)(2/3)(1/2) = 1⁄3.
The Thirder mistake is to treat different wakings as a random draw, rather than as sequential events that we know will both happen. (Also the same mistake made on the Doomsday Argument.) This is exactly analogous to the “Day 1 objection” to the Halfer position of the original problem, wherein Beauty learns that it’s Monday—similar to the scenario you’re describing with green and red sides of a dice, but simpler to reason about, so I think it would be better to discuss.
Meta-note for anyone else reading this: As far as I’ve seen, the Thirder logic results in credences of expected value that cannot be used without leading to losing bets (and depending on how it’s justified, can lead to more extreme obvious wrong conclusions such as the presumptuous philosopher). Where as I’ve never seen any contradiction or absurdity to arise from the Halver position. I used to be a Thirder myself at one point, and for a long time I was highly uncertain about what the true answer should be. But the more and more I’ve looked at objections to the Halfer position, the more and more I’ve found it to be resilient, and the more I believe there is a true sense in which the Halfer position is correct and the Thirder position is incorrect, when you’re talking about actual belief in the coin flip and not belief in something you know will be tallied more often with more wakings (e.g. guesses-over-repeated-experiments), and the that former is more useful thing to Beauty (i.e., more actionable; not able to be money-pumped) when she’s asked unspecifically, “What do you think the coin flipped?”
There are three die rolls where SB would be awake at thevmoment.
There are three where she would be asleep.
She is awake.
What is so difficult to see about this?
the Thirder logic results in credences of expected value that cannot be used without leading to losing bets
That depends entirely on how you evaluate the”bet.” If it is evaluated once, over the two days of the experiment, “1/2” produces balanced bets and “1/3“ produces unbalanced bets. If it is evaluated individually each time SB is asked the question, “1/2” produces unbalances bets and “1/3” produces balanced bets.
Please tell me why you think the question is not about the moment the question is asked, but instead includes another moment that may, or may not, happen.
That depends entirely on how you evaluate the”bet.”
I pick a random day to approach Beauty with an offer of a bet. If she guesses right, on Wednesday I’ll give her $1.5, otherwise she’ll give me $2. If she truly believes that the coin flipped Tails with a probability of 2⁄3, then she’ll take this bet.
What is so difficult to see about this?
You’re not making yourself clear. Yes, from the outside perspective of a randomly chosen day, there are different die rolls associated with whether Beauty is asleep or not on that day. How does that contradict the math I presented? Would you like to share some Bayesian equations of your own? Did you read my post about the Day 1 Objection, which I strongly suspect is very relevant?
Apologies if that comes off as rude, I’m just finding it difficult to respond usefully to this latest comment. (“What is so difficult to see about this?” isn’t a helpful question—I could just as easily pose the same question back to you!)
You say you pick a random day, but you do not specify the set from which you pick. Is it {Mon, Tue}, {H+Mon, T+Mon, T+Tue}, or {H+Mon, T+Mon, H+Tue, T+Tue}? In each case, it is possible that the bet will not be offered, and possible that SB is asked for a credence that is never “tested” with a bet. (And please, don’t say “Monday if Heads, and Monday or Tuesday if Tails.” That’s biasing the method to what the method is supposed to test.)
That is not how the problem works. You are manipulating the odds to make your answer look correct. This is why it does not represent the canonical version of the problem, where she is asked on each waking day, not a random one.
The bet must be offered on each waking day, and evaluated each time SB is asked fro a credence, to correctly represent her credence. If you do this, the experiment is a zero-sum game if and only if 2:1 odds are used. But I suspect you will disagree with me. Which is why I do not use betting arguments. Even though I know yours can’t be right based on the issues mentioned above, I know I won’t convince you with my betting scheme, either
You’re not making yourself clear.
I am. You are doing everything you can to avoid seeing it.
Would you like to share some Bayesian equations of your own?
I have described it well enough. But, since you insist:
In the canonical version, on any single day (and this includes Tuesday, after Heads), the sample space for that day includes four outcomes: {H+Mon, T+Mon, H+Tue, T+Tue}. Each has a prior probability—in Bayesian probability, this means a result of the experiment prior to evidence being considered—of 1⁄4. SB’s evidence is that it can’t be H+Tue, so the “new data” is the event {H+Mon, T+Mon, T+Tue}.
And again, the mistake you make in your equations is ignoring a valid outcome of the experiment because it goes unobserved by SB.
You have ignored how I point out that the need to consider it is demonstrated by letting SB observe it in a way that distinguishes it from the others: Wake her on Tuesday, after Heads, but take her shopping instead of asking about the coin. The above equation clearly gives her credence, in the situations where she is asked.
And the point is that the correct Bayesian solution does not depend, in any way, on what would happen on Tuesday, after Heads. Only that she makes he same observation for any outcome in {H+Mon, T+Mon, T+Tue}, and uses Bayesian probability based on that observation.
Did you read my post about the Day 1 Objection, which I strongly suspect is very relevant?
Did you note that the problem asks about her credence based on her current circumstance, and that there are circumstances that differ? Circumstances that can exist regardless of whether SB is awake.
So Pr(Awake) is not 1. “Awake” will not happen on Tuesday, after Heads, which is a possible result.
Please, please, please try to see that I am not saying she might not be awakened over the course of the experiment. I am saying that there are circumstances, consistent with how the question is posed on a single day that is made independent of the other day by amnesia, where she will not be wakened.
And the point of the six-sided die version was to isolate the problem to just the circumstance where she is awake. With no ambiguity about one or two wakings, and how to handle them. That when SB is asked the question, there are exactly three possible faces that could be showing, and one of them is green.
The bet must be offered on each waking day, and evaluated each time SB is asked fro a credence, to correctly represent her credence. If you do this, the experiment is a zero-sum game if and only if 2:1 odds are used. But I suspect you will disagree with me. Which is why I do not use betting arguments.
I actually think this might be useful to drill into further. Why couldn’t SB use her credence of the coin for the bet I described? Are you thinking that SB actually has two different credences about the coin, and which one she applies will depend on the exact question asked?
You have ignored how I point out that the need to consider it is demonstrated by letting SB observe it in a way that distinguishes it from the others: Wake her on Tuesday, after Heads, but take her shopping instead of asking about the coin.
Hoping for a clarification with this one: Do you mean that she goes shopping on Tuesday and is told on Tuesday? Or that a random day was chosen for her to go shopping, and she knew that it’d be a random day? I think I would agree with your presented math depending on exactly what she’s learning.
Okay I’m getting tired of fighting the editor, going to quote differently:
>> That when SB is asked the question, there are exactly three possible faces that could be showing, and one of them is green.
I’m not denying that those are the possibilities, but that they share equal probabilities.
>> Did you note that the problem asks about her credence based on her current circumstance, and that there are circumstances that differ? Circumstances that can exist regardless of whether SB is awake.
>> So Pr(Awake) is not 1. “Awake” will not happen on Tuesday, after Heads, which is a possible result.
Unfortunately I’m not following this part. P(Awake on Tues) is not 1, but that’s not what she’s observing, she’s observing being awake, which she already knew would happen.
Why couldn’t SB use her credence of the coin for the bet I described?
First, I’m still waiting for you to fully describe the random day. Is it chosen from a set of two, three, or four?
But to answer your question, she can. On the inside of the experiment, where there are three equivalently-placed choices. It’s a common justification for 1⁄3. You can’t, on the outside, where there is no way to make a random choice that is equivalent. It’s a different experiment than the one where she is asked about her credence.
Do you mean that she goes shopping on Tuesday and is told on Tuesday?
On Monday, and on Tuesday after Tails, she is asked about her credence. On Tuesday after Heads, she goes shopping. She doesn’t have to be told anything in this case, since her credences that it is Tuesday, and that the coin landed Heads, will both be 100%. This takes what some want to call “probability mass” away from her Sunday-Night credence in Heads. That was 1⁄2, and removing probability mass is what reduces it to 1⁄3.
And the point is that IT DOES NOT MATTER what would happen on H+Tue. That is still a day in the experiment, just as valid a result as T+Mon. If you do this, the proper odds for the bet if she is offered one are 2:1.
I’m not denying that those are the possibilities, but that they share equal probabilities.
Please justify how, without using the circular argument that it is the only way to get the answer you want.
But you gave me an idea, to alter the “shopping” version to the red/green die version.
Wake her on all three days. If the result was a green number matching the current day, or a red number not matching it, ask her for her credence in green. If it was a red number matching the current day, or a green one not matching it, ask her for her credence in red.
Before she is asked, there are six-equally likely faces (which contradicts your claim about them being unequal). If she is asked about green, it is reduced to three (still equally likely) faces, where one is green. There is absolutely no difference between these “ask about green” cases and the “awake” cases from before.
I’m not following this part.
And it is what I keep trying to describe. You, on the outside, see a two-day experiment where SB is wakened once, or twice. To SB, it is a one-day sample of that experiment, but limited to the three equally-likely situations where she will be awake. When the sun rose this morning (remember that it is a one-day sample), there were four possible combinations. Three of them, still equally likely, result in waking her. Her evidence is that it is one of these three, and not H+Tue.
Pr(Awake) vs. Pr(Awake on Tues) is awkward because you are confusing two-day and one-day probabilities. So I’ll use the shopping variant. Instead of “Awake” I’ll use “Ask” and “Shop.”
In the two-day experiment—that is, before the coin flip on Sunday and anticipating the next two days—Pr(Ask)=1 and Pr(Shop)=1/2. Note that these are not independent events, so these don’t add up to 1. When SB awakes, but before she learns what will happen today, Pr(Ask)=3/4 and Pr(Shop)=1/4. These are independent events that represent all possibilities, so they do add up to 1.
BOTH OF THESE ARE “PRIOR” PROBABILITIES, meaning they describe the possibilities with no evidence. What you are “not following,” is that the probability space for SB’s credence is the second one, not the first.
Your Bayesian analysis is wrong because it is based on the first. Pr(Awake)=1 means she will be awake at some time over two days. This is why you think you can choose one day at random from what I suspect you mean the two “days” Monday and Tuesday_If_Tails.
SB’s credence is about one day only. Its prior sample space (remember, no evidence) is {H+Mon,T+Mon, H+Tue,T+Tue} with a 1⁄4 probability for each. Her evidence is that H+Tue is eliminated, either because she is awake of because she didn’t go shopping.
Your Bayesian analysis “folds” H+Tue into H+Monday because that is the only way she is awake after Heads.
I’m hoping to find the time for a full response later, but right now:
>> First, I’m still waiting for you to fully describe the random day. Is it chosen from a set of two, three, or four?
It’s a random choice among the days she’ll be awoken. If she’s to awake just Mon, then she’ll be offered on Mon. If she’s to awake on Mon and Tues, then a coin is flipped.
You are right, I didn’t look closely at your betting argument, because (as I have said) betting arguments are invalid. They can be manipulated—as you do—to get the answer you want.
I pick a random day to approach Beauty with an offer of a bet. If she guesses right, on Wednesday I’ll give her $1.5, otherwise she’ll give me $2. If she truly believes that the coin flipped Tails with a probability of 2⁄3, then she’ll take this bet.
You are saying the odds (easier to work with for betting) you offer are 1.5:2 for Tails (representing probability 4⁄7), and so should be 2.67:2 (representing 3⁄7) for Heads. If Beauty thinks they are 1:2 and 4:2 (2/3 and 1⁄3), Beauty should indeed take this bet, on Tails.
But if this is what you mean, you did not do what claimed to do. You did not puck a random day, you choose a random timeline and inserted a bet into a waking moment in that timeline. She may as well have chosen the bet on Sunday.
By flipping the coin on Sunday and resolving the bet, once, on Wednesday, you have manipulated the wager. What happens in between—all the sleeping, waking, and amnesia—is completely irrelevant. So what is the point of what happens in between? Isn’t the entire point of the thought experiment to evaluate how, and yes, if, it matters? Yet you designed your question to eliminate any issue, not to resolve it.
Try this: actually choose a random day for the offer, but use her “thirder” odds. Note that now there is a chance that no bet will be offered, if the random day is Tuesday and the coin landed on Heads.
If the coin landed Tails, Beauty will be offered a wager. She will win $1 if she bet on Tails (at 1:2), and lose $2 if she bet on Heads. But if the coin landed Heads, she will lose an expected $1 if she chooses Tails when the bet is offered, and win an expected $2 (at 4:2) if she chooses Heads when the bet is offered.
In other words, if we remove the bias you built into your challenge, Beauty expects to wager $2 and a either choice wins $1 if correct.
What if you have Beauty put $2 into each of two identical envelopes? Each time she is awake, we give her one and she chooses either Tails (at 1:2 odds) or Heads (at 4:2). We again evaluate on Wednesday, but this time the number of bets evaluated can vary.
If the coin lands Tails, she is up $2 if she chooses Tails (two wins of $1 since we assume she is consistent over the two days) and down $2 if she chooses Heads. If it lands Heads, she is down $4 if she chooses Tails (two losses of $4) and up $4 if she chooses Heads. This is a balanced wager even though the balance is differs depending on the result.
But it is the differing balance that halfers object to. It is an invalid objection, but again it is why I try to avoid betting arguments altogether.
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It’s a random choice among the days she’ll be awoken. If she’s to awake just Mon, then she’ll be offered on Mon. If she’s to awake on Mon and Tues, then a coin is flipped.
Sorry, I somehow confused your similar response to another comment with this one. This models a very different experiment. One where Beauty is woken exactly once regardless of the coin. It happens on Monday if the coin landed Heads. Or on Monday or Tuesday, according to the second coin, if the first coin landed on Tails.
Certainly, since it is wrong.
There is much obfuscation surrounding the SB problem, that appear in this thread. One is that when you ask about a probability in the present tense about an occurrence described in the past tense, you are asking about the conditional probability of that event given the present information. What this means is that there is only one question that is relevant: “Given SB’s knowledge when she is awake, what is the probability that the coin result was Heads?” Not “What was the probability, ignoring any information obtained since, that a coin flip result is Heads?” (For comparison, say I draw a card at random from a standard 52-card deck, and then tell you that it is a spade. The probability, that I drew the Ace of Spades, is 1⁄13. Not 1⁄52, the probability when I drew it.)
Then, and this will become important later, the question Adam Elga asked is equivalent to, but not identical to, the one he answered. In the one he asked, SB is woken once, or twice, based on a coin flip. In the one he answered, SB is always woken on Monday, and then again on Tuesday, if the coin result was Tails.
But in the one he asked, he added two details that presaged his solution. Neither detail has any impact on what was asked. The first is that he mentioned that the experiment occurs over two days. But days are irrelevant in what was asked. The only significance of these two days are in his solution. And he asked “when you are first awakened...”. His solution supposed that SB could be told that it is Monday, or that the coin landed Tails, and evaluated that probability after this information was revealed. The point of “when you are first awakened...” was simply to evaluate the probability before either revelation.
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My point is that these timing details are completely irrelevant, and are used to confuse the probability by establishing different kinds of conditions for each waking. A trivial solution can be obtained by making these conditions equivalent for each. Here is one way:
The experiment proceeds over three days, not two. Monday, Tuesday, and Wednesday.
Only one set of conditions produce a waking on each day.
Replace the coin with a six-sided die. Three faces will be green, representing “Heads,” and three will be red, representing “Tails.”
The words “Monday,” “Tuesday” and “Wednesday” are each written on two sides, once in each color.
The die is rolled on Sunday Night, and preserved throughout the experiment. If it [landed on/is currently showing] a green side, SB is woken only on the day indicated on that side. If red, SB is woken twice, on the two days not indicated on that side.
SB is asked for the probability that the die [landed on/is showing] a green side.
BY HALFER LOGIC, the probability that the die “landed on” each of the six sides is 1⁄6. But since she is awake, SB knows that there are only three sides that it could [have landed on/be showing] at the moment. They are still equally likely, and only one of the three is green.
“New information” does not mean learning something happened that was not guaranteed to happen. It means that the information eliminates at least one result that was possible without the information. My version does this transparently, by ruling out three die results.
In the classic version (the one in Elga’s solution), the information is that the combination of coin and day allows SB to be awake. Before that information was obtained, Heads+Tuesday is possible. By being awake, SB can eliminate one of the four equally-likely possibilities, leaving three. The obfuscation in that version happens because the two combinations that can occur on Monday are treated as equivalent to only one that can occur on Tuesday.
“BY HALFER LOGIC, the probability that the die “landed on” each of the six sides is 1⁄6. But since she is awake, SB knows that there are only three sides that it could [have landed on/be showing] at the moment. They are still equally likely, and only one of the three is green.”
Hello, I’m not sure what you’re trying to say. There are six possibilities here (M,W,T x R,G) and yes, I do think that each of them has a 1⁄6 chance. The chance of Red is equal to the chance of Green.
With all due respect, I don’t sense that you are trying to understand it. Your position seems to be firmly that SB has gained no information, so she can’t use conditional probability. Mine is that she does, and all I’m trying to do is make it hard to ignore why it is.
As you point out, there are six possible die rolls. But on any one day, only three remain possible. This is what is known, in probability theory, as a “condition.” It is what allows one to “update” a prior probability, which is what you calculated, to a “posterior” or “conditional” probability, which is what the Sleeping Beauty Problem asks for. This is true even if she gains no information. The condition will be the entire sample space, and the “updated” probability will be the same as the prior probability. So there is no reason to not explore what the condition is.
It’s easier if we number the days 0 (for “Monday”), 1 and 2. The set of possible die rolls—can we agree that this is the sample space? - is {G_0, G_1, G_2, R_0, R_1, R_2}. When she is awake, she does not know which day it is, but she does know that it is only one day. Call it day D. The condition is the event {G_D, R_MOD(D+1,3), R_MOD(D+2,3)}. The condition is the same regardless of what D is. Each outcome in this set has a prior probability of 1⁄6. The posterior (or conditional) probability is of G_D is (1/6)/[(1/6)+(1/6)+(1/6)]=1/3.
What observation are you trying to update on?
P(G) = 1⁄2
P(waking) = 1
P(waking | G) = 1
P(G | waking) = P(waking | G)*P(G)/P(waking) = 1⁄2
Let’s say she learns it’s Monday. Half the time, she’ll be awake on a random day. Half the time, she’ll be awake on 2⁄3 of days.
P(M) = P(T) = P(W) = (1/2)(1/3) + (1/2)(2/3) = 1⁄2
(Note that this is distinct from the probability that a randomly chosen waking is M/T/W—more on that in a second)
P(M | G) = 1⁄3
P(G | M) = P(M | G)*P(G)/P(M) = (1/3)(1/2)/(1/2) = 1⁄3
Again, no updates are occurring.
What did I mean by “a randomly chosen waking is M/T/W”? An example would be the chance of Beauty receiving a knock on the door, if an outside experimenter flips a coin to decide whether to knock on her door on her first red waking or the second (or the only choice of green waking, if it was green).
The chance that she gets a knock on a Monday would be (1/2)(1/3) + (1/2)(2/3)(1/2) = 1⁄3.
The Thirder mistake is to treat different wakings as a random draw, rather than as sequential events that we know will both happen. (Also the same mistake made on the Doomsday Argument.) This is exactly analogous to the “Day 1 objection” to the Halfer position of the original problem, wherein Beauty learns that it’s Monday—similar to the scenario you’re describing with green and red sides of a dice, but simpler to reason about, so I think it would be better to discuss.
And I break down exactly why that objection fails here:
https://ramblingafter.substack.com/p/sleeping-beauty-meets-two-face
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Meta-note for anyone else reading this: As far as I’ve seen, the Thirder logic results in credences of expected value that cannot be used without leading to losing bets (and depending on how it’s justified, can lead to more extreme obvious wrong conclusions such as the presumptuous philosopher). Where as I’ve never seen any contradiction or absurdity to arise from the Halver position. I used to be a Thirder myself at one point, and for a long time I was highly uncertain about what the true answer should be. But the more and more I’ve looked at objections to the Halfer position, the more and more I’ve found it to be resilient, and the more I believe there is a true sense in which the Halfer position is correct and the Thirder position is incorrect, when you’re talking about actual belief in the coin flip and not belief in something you know will be tallied more often with more wakings (e.g. guesses-over-repeated-experiments), and the that former is more useful thing to Beauty (i.e., more actionable; not able to be money-pumped) when she’s asked unspecifically, “What do you think the coin flipped?”
There are three die rolls where SB would be awake at thevmoment.
There are three where she would be asleep.
She is awake.
What is so difficult to see about this?
That depends entirely on how you evaluate the”bet.” If it is evaluated once, over the two days of the experiment, “1/2” produces balanced bets and “1/3“ produces unbalanced bets. If it is evaluated individually each time SB is asked the question, “1/2” produces unbalances bets and “1/3” produces balanced bets.
Please tell me why you think the question is not about the moment the question is asked, but instead includes another moment that may, or may not, happen.
I pick a random day to approach Beauty with an offer of a bet. If she guesses right, on Wednesday I’ll give her $1.5, otherwise she’ll give me $2. If she truly believes that the coin flipped Tails with a probability of 2⁄3, then she’ll take this bet.
You’re not making yourself clear. Yes, from the outside perspective of a randomly chosen day, there are different die rolls associated with whether Beauty is asleep or not on that day. How does that contradict the math I presented? Would you like to share some Bayesian equations of your own? Did you read my post about the Day 1 Objection, which I strongly suspect is very relevant?
Apologies if that comes off as rude, I’m just finding it difficult to respond usefully to this latest comment. (“What is so difficult to see about this?” isn’t a helpful question—I could just as easily pose the same question back to you!)
You say you pick a random day, but you do not specify the set from which you pick. Is it {Mon, Tue}, {H+Mon, T+Mon, T+Tue}, or {H+Mon, T+Mon, H+Tue, T+Tue}? In each case, it is possible that the bet will not be offered, and possible that SB is asked for a credence that is never “tested” with a bet. (And please, don’t say “Monday if Heads, and Monday or Tuesday if Tails.” That’s biasing the method to what the method is supposed to test.)
That is not how the problem works. You are manipulating the odds to make your answer look correct. This is why it does not represent the canonical version of the problem, where she is asked on each waking day, not a random one.
The bet must be offered on each waking day, and evaluated each time SB is asked fro a credence, to correctly represent her credence. If you do this, the experiment is a zero-sum game if and only if 2:1 odds are used. But I suspect you will disagree with me. Which is why I do not use betting arguments. Even though I know yours can’t be right based on the issues mentioned above, I know I won’t convince you with my betting scheme, either
I am. You are doing everything you can to avoid seeing it.
I have described it well enough. But, since you insist:
In the canonical version, on any single day (and this includes Tuesday, after Heads), the sample space for that day includes four outcomes: {H+Mon, T+Mon, H+Tue, T+Tue}. Each has a prior probability—in Bayesian probability, this means a result of the experiment prior to evidence being considered—of 1⁄4. SB’s evidence is that it can’t be H+Tue, so the “new data” is the event {H+Mon, T+Mon, T+Tue}.
Bayesian probability says:
Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = Pr({H+Mon, T+Mon, T+Tue}|{H+Mon}) * Pr({H+Mon}) / Pr({H+Mon, T+Mon, T+Tue}}
Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = (1)*(1/4)/(3/4) Pr({H+Mon}|{H+Mon, T+Mon, T+Tue} = 1⁄3
And again, the mistake you make in your equations is ignoring a valid outcome of the experiment because it goes unobserved by SB.
You have ignored how I point out that the need to consider it is demonstrated by letting SB observe it in a way that distinguishes it from the others: Wake her on Tuesday, after Heads, but take her shopping instead of asking about the coin. The above equation clearly gives her credence, in the situations where she is asked.
And the point is that the correct Bayesian solution does not depend, in any way, on what would happen on Tuesday, after Heads. Only that she makes he same observation for any outcome in {H+Mon, T+Mon, T+Tue}, and uses Bayesian probability based on that observation.
Did you note that the problem asks about her credence based on her current circumstance, and that there are circumstances that differ? Circumstances that can exist regardless of whether SB is awake.
So Pr(Awake) is not 1. “Awake” will not happen on Tuesday, after Heads, which is a possible result.
Please, please, please try to see that I am not saying she might not be awakened over the course of the experiment. I am saying that there are circumstances, consistent with how the question is posed on a single day that is made independent of the other day by amnesia, where she will not be wakened.
And the point of the six-sided die version was to isolate the problem to just the circumstance where she is awake. With no ambiguity about one or two wakings, and how to handle them. That when SB is asked the question, there are exactly three possible faces that could be showing, and one of them is green.
I actually think this might be useful to drill into further. Why couldn’t SB use her credence of the coin for the bet I described? Are you thinking that SB actually has two different credences about the coin, and which one she applies will depend on the exact question asked?
Hoping for a clarification with this one: Do you mean that she goes shopping on Tuesday and is told on Tuesday? Or that a random day was chosen for her to go shopping, and she knew that it’d be a random day? I think I would agree with your presented math depending on exactly what she’s learning.
Okay I’m getting tired of fighting the editor, going to quote differently:
>> That when SB is asked the question, there are exactly three possible faces that could be showing, and one of them is green.
I’m not denying that those are the possibilities, but that they share equal probabilities.
>> Did you note that the problem asks about her credence based on her current circumstance, and that there are circumstances that differ? Circumstances that can exist regardless of whether SB is awake.
>> So Pr(Awake) is not 1. “Awake” will not happen on Tuesday, after Heads, which is a possible result.
Unfortunately I’m not following this part. P(Awake on Tues) is not 1, but that’s not what she’s observing, she’s observing being awake, which she already knew would happen.
First, I’m still waiting for you to fully describe the random day. Is it chosen from a set of two, three, or four?
But to answer your question, she can. On the inside of the experiment, where there are three equivalently-placed choices. It’s a common justification for 1⁄3. You can’t, on the outside, where there is no way to make a random choice that is equivalent. It’s a different experiment than the one where she is asked about her credence.
On Monday, and on Tuesday after Tails, she is asked about her credence. On Tuesday after Heads, she goes shopping. She doesn’t have to be told anything in this case, since her credences that it is Tuesday, and that the coin landed Heads, will both be 100%. This takes what some want to call “probability mass” away from her Sunday-Night credence in Heads. That was 1⁄2, and removing probability mass is what reduces it to 1⁄3.
And the point is that IT DOES NOT MATTER what would happen on H+Tue. That is still a day in the experiment, just as valid a result as T+Mon. If you do this, the proper odds for the bet if she is offered one are 2:1.
Please justify how, without using the circular argument that it is the only way to get the answer you want.
But you gave me an idea, to alter the “shopping” version to the red/green die version.
Wake her on all three days. If the result was a green number matching the current day, or a red number not matching it, ask her for her credence in green. If it was a red number matching the current day, or a green one not matching it, ask her for her credence in red.
Before she is asked, there are six-equally likely faces (which contradicts your claim about them being unequal). If she is asked about green, it is reduced to three (still equally likely) faces, where one is green. There is absolutely no difference between these “ask about green” cases and the “awake” cases from before.
And it is what I keep trying to describe. You, on the outside, see a two-day experiment where SB is wakened once, or twice. To SB, it is a one-day sample of that experiment, but limited to the three equally-likely situations where she will be awake. When the sun rose this morning (remember that it is a one-day sample), there were four possible combinations. Three of them, still equally likely, result in waking her. Her evidence is that it is one of these three, and not H+Tue.
Pr(Awake) vs. Pr(Awake on Tues) is awkward because you are confusing two-day and one-day probabilities. So I’ll use the shopping variant. Instead of “Awake” I’ll use “Ask” and “Shop.”
In the two-day experiment—that is, before the coin flip on Sunday and anticipating the next two days—Pr(Ask)=1 and Pr(Shop)=1/2. Note that these are not independent events, so these don’t add up to 1. When SB awakes, but before she learns what will happen today, Pr(Ask)=3/4 and Pr(Shop)=1/4. These are independent events that represent all possibilities, so they do add up to 1.
BOTH OF THESE ARE “PRIOR” PROBABILITIES, meaning they describe the possibilities with no evidence. What you are “not following,” is that the probability space for SB’s credence is the second one, not the first.
Your Bayesian analysis is wrong because it is based on the first. Pr(Awake)=1 means she will be awake at some time over two days. This is why you think you can choose one day at random from what I suspect you mean the two “days” Monday and Tuesday_If_Tails.
SB’s credence is about one day only. Its prior sample space (remember, no evidence) is {H+Mon,T+Mon, H+Tue,T+Tue} with a 1⁄4 probability for each. Her evidence is that H+Tue is eliminated, either because she is awake of because she didn’t go shopping.
Your Bayesian analysis “folds” H+Tue into H+Monday because that is the only way she is awake after Heads.
I’m hoping to find the time for a full response later, but right now:
>> First, I’m still waiting for you to fully describe the random day. Is it chosen from a set of two, three, or four?
It’s a random choice among the days she’ll be awoken. If she’s to awake just Mon, then she’ll be offered on Mon. If she’s to awake on Mon and Tues, then a coin is flipped.
You are right, I didn’t look closely at your betting argument, because (as I have said) betting arguments are invalid. They can be manipulated—as you do—to get the answer you want.
You are saying the odds (easier to work with for betting) you offer are 1.5:2 for Tails (representing probability 4⁄7), and so should be 2.67:2 (representing 3⁄7) for Heads. If Beauty thinks they are 1:2 and 4:2 (2/3 and 1⁄3), Beauty should indeed take this bet, on Tails.
But if this is what you mean, you did not do what claimed to do. You did not puck a random day, you choose a random timeline and inserted a bet into a waking moment in that timeline. She may as well have chosen the bet on Sunday.
By flipping the coin on Sunday and resolving the bet, once, on Wednesday, you have manipulated the wager. What happens in between—all the sleeping, waking, and amnesia—is completely irrelevant. So what is the point of what happens in between? Isn’t the entire point of the thought experiment to evaluate how, and yes, if, it matters? Yet you designed your question to eliminate any issue, not to resolve it.
Try this: actually choose a random day for the offer, but use her “thirder” odds. Note that now there is a chance that no bet will be offered, if the random day is Tuesday and the coin landed on Heads.
If the coin landed Tails, Beauty will be offered a wager. She will win $1 if she bet on Tails (at 1:2), and lose $2 if she bet on Heads. But if the coin landed Heads, she will lose an expected $1 if she chooses Tails when the bet is offered, and win an expected $2 (at 4:2) if she chooses Heads when the bet is offered.
In other words, if we remove the bias you built into your challenge, Beauty expects to wager $2 and a either choice wins $1 if correct.
What if you have Beauty put $2 into each of two identical envelopes? Each time she is awake, we give her one and she chooses either Tails (at 1:2 odds) or Heads (at 4:2). We again evaluate on Wednesday, but this time the number of bets evaluated can vary.
If the coin lands Tails, she is up $2 if she chooses Tails (two wins of $1 since we assume she is consistent over the two days) and down $2 if she chooses Heads. If it lands Heads, she is down $4 if she chooses Tails (two losses of $4) and up $4 if she chooses Heads. This is a balanced wager even though the balance is differs depending on the result.
But it is the differing balance that halfers object to. It is an invalid objection, but again it is why I try to avoid betting arguments altogether.
+++++
Sorry, I somehow confused your similar response to another comment with this one. This models a very different experiment. One where Beauty is woken exactly once regardless of the coin. It happens on Monday if the coin landed Heads. Or on Monday or Tuesday, according to the second coin, if the first coin landed on Tails.