There are several, valid solutions that do not always introduce the details that are misinterpreted as ambiguities. The two-coin version is one, which says the answer is 1⁄3.
Here’s another, that I think also proves the answer is 1⁄3, but I’m sure halfers will disagree with that. But it does prove that 1⁄2 can’t be right.
Instead of leaving SB asleep on Tuesday, after Heads, we wake her but do not interview her. We do something entirely different, like take her on a $5000 shopping spree on Rodeo Drive. (She can get maybe one nice dress.)
This way, when she is first wakened—which can only mean before she learns if it is for an interview or a shopping spree, since she can’t know about any prior/subsequent waking—she is certain that the probability of Heads and Tails are each 50%. But when she is interviewed, she knows that something that only happens after a Heads has been “eliminated.” So the probability of Heads must be reduced, and the probability of Tails must be increased. I think that she must add “Heads and it is Tuesday” to the sample space Elga used, and each observation has a probability of 25%. Which makes the conditional probability of Heads, given that she is interviewed, 1⁄3.
BUT IT DOES NOT MATTER WHAT HAPPENS ON “HEADS AND IT IS TUESDAY.” The “ambiguity” is created by ignoring that “HEADS and it is Tuesday” happens even if SB sleeps through it.
OR, we could use four volunteers but only one coin. Let each on sleep through a different combination of “COIN and it is DAY.” Ask each for the probability that the coin landed on the side where she might sleep through a day. On each day, three will be wakened. For two of them, the coin landed on the side that means waking twice. For one, it is the side for waking once.
All three will be brought into a room where they can discuss the answer, but not share their combination. Each has the same information that defines the correct answer. Each must give the same answer, but only one matches the condition. That answer is 1⁄3.
+++++
Yes, these all are just different ways of presenting the same information: that, in the popular version, Tuesday after Heads still happens, but cannot be observed. This is what is wrong in all the debates; they treat it as if Tuesday after Heads does not happen.
In my card example, the “prior” sample space includes all 52 cards. The probability distribution is 1⁄52 for each card. When I say that one is an Ace, it does not mean that it was impossible for a Seven of Clubs to have been drawn, it means that an observation was made, and in that observation the card wasn’t the Seven of Clubs.
In the popular version of the SB problem, there are four possible states that can occur. Three can be observed; and because of the amnesia drug, they are all independent to SB as an observer. Regardless of whether she can know the day, it is part of the observation. Since she is awake, (Heads, Tuesday) is eliminated—as an observation, not from the experiment as a whole—and the updated probability—for this observation, not the experiment as a whole—is 1⁄3.
Now, you can use this problem to evaluate epistemic probability. It isn’t really am epistemic problem, but I supposed you can apply it. The answer is 1⁄3, and the correct epistemic solution is the one that says so.
You keep repeating the same points and they are all based on faulty assumptions. Which you would have already seen if you properly evaluated my example with balls in a box. Let me explicitly do it for you:
Two coins are tossed. Then if it’s not Heads Heads one ball is put into a box. Then the second coin is turned to the other side and again if it’s not Heads Heads a ball is put into a box. After this procedure is done, you are given a random ball from the box. What is the probability that the first coin is Heads after you’ve got the ball?
The correct answer here is unambiguosly 1⁄2, which we can check by running the experiment multiple times. On every iteration you get only one ball and on 1⁄2 of them the first coin is Heads. Getting a ball is not evidence in favor of anything because you get it regardless of the outcome of the coin toss.
But if we reason about this problem the same way you try to reason about Sleeping Beauty we inevitably arrive to the conclusion that it has to be 1⁄3. After all, there are four equiprobable possible states {HH, TT, HT, TH}. The ball you’ve just got couldn’t be put in the box on HH, so we have to update to three equiprobable states {HT, TH, TT} and the only one of them where the first coin is Heads is HT. P(HT) = 1⁄3.
This show that such reasoning method can’t generally produce correct answers. So when you applied it to Two-Coin-Toss version of Sleeping Beauty you didn’t actually show that 1⁄3 is the correct answer.
I keep repeating, because you keep misunderstanding how my example is very different than yours.
In yours, there is one “sampling” of the balls (that is, a check on the outcome and a query about it). This one sampling is done only after two opportunities to put a ball into box have occurred. The probability you ask about depends on what happened in both. Amnesia is irrelevant. She is asked just once.
In mine, there are two “samplings.” The probability in each is completely independent of the other. Amnesia is important to maintain the independence.
SPECIFICALLY: SB’s belief is based entirely one what happens in these steps:
Two coins are randomly arranged so that each of the four combinations {HH, HT, TH, TT} has a 25% chance to be the outcome.
If the random combination is HH, one option happens, that does not involve asking for a probability. Otherwise, another option happens, and it does involve asking for a probability.
SB has full knowledge of these three steps, and knows that the second option was chosen. She can assign a probability based ENTIRELY on these three steps.
This happens twice. What you seem to ignore, is that the method used to arrange the coins is different in the first pass through these three steps, and the second. In the first, it is flipping the coins. In the second, it is a modification of this flips. But BECAUSE OF AMNESIA, this modification does no, in any way, affect SB’s assessment that sample space is {HH, HT, TH, TT}, or that each has a 25% chance to be the outcome.
Her answer is unambiguously 1⁄3 anytime she is asked.
you keep misunderstanding how my example is very different than yours.
I understand that they are different, that’s the whole point. They are different in such a way that we can agree that the answer to my problem is clearly 1⁄2, while we can’t agree to the answer to your problem.
But none of their differences actually affect the mathematical argument you have constructed, so the way you arrive to an answer 1⁄3 in your problem, would arrive to the same answer in mine.
Amnesia is irrelevant
What amnesia does in Sleeping Beauty is ensuring that the Beauty can’t order the outcomes. So when she is awaken she doesn’t know whether it’s the first awakening or the second. She is unable to observe the event “I’ve been awaken twice in this experiment”. The similar effect is achieved by the fact that she is given a random ball from the box. She doesn’t know whether it’s the first ball or the second. And she can’t directly observe whether there are two balls in the box or only one.
In mine, there are two “samplings.”
Which is completely irrelevant to your mathematical argument about four equiprobable states because you’ve constructed it in such a manner, that the same probabilities are assigned to all of them regardless of whether the Beauty is awake or not. All your argument is based on “There are four equiprobable states and one of them is incompatible with the observations”, it is not dependent on the number of observations.
Now, there is a different argument that you could’ve constructed that would take advantage of two awakening. You could’ve said that when the first coin is Tails there are twice as many awakenings as when it’s Heads and claim that we should interpret it as P(Awake|T1)=2P(Awake|H1) but it’s very much not the argument you were talking about. In a couple of days, in my next post I’m explicitly exploring both of them.
The probability in each is completely independent of the other.
This is wrong. And it’s very easy to check. You may simulate your experiment, a large number of times, writing down the states of the coins on every awakening and notice that there is a clear way to predict the next token beter than chance:
if i-th token == TH and i-1-th token != TT then
i+1-th token = TT
elseif i-th token == TT and i-1-th token != TH:
i+1-th token = TH
Which is absolutely not the case in a situation where your argument actually works:
Two coins are tossed on Heads Heads event doesn’t happen, on every other outcome it does. Event has happened, what is the probability that the first coin came Heads?
What you seem to ignore, is that the method used to arrange the coins is different in the first pass through these three steps
Ignore? On the contrary. This is the exact reason for why your argument doesn’t work. You treat correlated events as independant. That’s what I’m trying to explain to you the whole time and why I brought up the problem with balls being put in the box, because there this kind of mistake is more obvious there.
But BECAUSE OF AMNESIA, this modification does no, in any way, affect SB’s assessment that sample space is {HH, HT, TH, TT}, or that each has a 25% chance to be the outcome
I suppose this is our crux.
I see two possible avenues for disagreement: about the territory and about the map
First is, whether having an amnesia actually modify the statistical properties of the experiment you are participating in. Do we agree that it’s not the case?
Second, statement about the map, which, correct me if I’m wrong, you actually hold, is that the Beauty should reason about her awakenings as independent because this represents her new state of knowledge due to amnesia?
This would be a correct statement if the Beauty was made to forgot that the events are correlated, if on the awakenings she had different information about which experiment she is participating in, than before she was put to sleep.
But in our case, the Beauty remembers the design of the experiment, she is aware that the states of the coins are not independent between two passes and if she reasons as if they are—she makes a mistake.
Her answer is unambiguously 1⁄3 anytime she is asked.
Her answer is 1⁄3 to the question “What is the probability that the coin is Heads in a random awakening throughout multiple iterations of such experiment”.
Her answer is 1⁄2 to the question “What is the probability that the coin is Heads in this particular experiment”.
But I don’t think that ambiguity is really the problem here.
Why couldn’t the ball I’ve just got been put into the box on HH? On HH, after we turn the second coin we get HT which is not HH, so a ball is put into the box, no?
Well, sure but then it would mean that the ball wasn’t put into the box on HH, it was put into the box on HT.
If this explanation still feels confusing as if something unlawful is going on—it’s because it is. It’s the exact kind of slight of hand that JeffJo uses to show that 1⁄3 is the correct answer to Two-Coin-Toss version of Sleeping Beauty. If you are able to spot the mistake here you should be able to spot it in his reasoning as well.
I see, I haven’t yet read that one. But yes, we should be clear what we denote with HH/HT/TT/TH, the coins before, or after the turning of second coin.
There are several, valid solutions that do not always introduce the details that are misinterpreted as ambiguities. The two-coin version is one, which says the answer is 1⁄3.
Here’s another, that I think also proves the answer is 1⁄3, but I’m sure halfers will disagree with that. But it does prove that 1⁄2 can’t be right.
Instead of leaving SB asleep on Tuesday, after Heads, we wake her but do not interview her. We do something entirely different, like take her on a $5000 shopping spree on Rodeo Drive. (She can get maybe one nice dress.)
This way, when she is first wakened—which can only mean before she learns if it is for an interview or a shopping spree, since she can’t know about any prior/subsequent waking—she is certain that the probability of Heads and Tails are each 50%. But when she is interviewed, she knows that something that only happens after a Heads has been “eliminated.” So the probability of Heads must be reduced, and the probability of Tails must be increased. I think that she must add “Heads and it is Tuesday” to the sample space Elga used, and each observation has a probability of 25%. Which makes the conditional probability of Heads, given that she is interviewed, 1⁄3.
BUT IT DOES NOT MATTER WHAT HAPPENS ON “HEADS AND IT IS TUESDAY.” The “ambiguity” is created by ignoring that “HEADS and it is Tuesday” happens even if SB sleeps through it.
OR, we could use four volunteers but only one coin. Let each on sleep through a different combination of “COIN and it is DAY.” Ask each for the probability that the coin landed on the side where she might sleep through a day. On each day, three will be wakened. For two of them, the coin landed on the side that means waking twice. For one, it is the side for waking once.
All three will be brought into a room where they can discuss the answer, but not share their combination. Each has the same information that defines the correct answer. Each must give the same answer, but only one matches the condition. That answer is 1⁄3.
+++++
Yes, these all are just different ways of presenting the same information: that, in the popular version, Tuesday after Heads still happens, but cannot be observed. This is what is wrong in all the debates; they treat it as if Tuesday after Heads does not happen.
In my card example, the “prior” sample space includes all 52 cards. The probability distribution is 1⁄52 for each card. When I say that one is an Ace, it does not mean that it was impossible for a Seven of Clubs to have been drawn, it means that an observation was made, and in that observation the card wasn’t the Seven of Clubs.
In the popular version of the SB problem, there are four possible states that can occur. Three can be observed; and because of the amnesia drug, they are all independent to SB as an observer. Regardless of whether she can know the day, it is part of the observation. Since she is awake, (Heads, Tuesday) is eliminated—as an observation, not from the experiment as a whole—and the updated probability—for this observation, not the experiment as a whole—is 1⁄3.
Now, you can use this problem to evaluate epistemic probability. It isn’t really am epistemic problem, but I supposed you can apply it. The answer is 1⁄3, and the correct epistemic solution is the one that says so.
You keep repeating the same points and they are all based on faulty assumptions. Which you would have already seen if you properly evaluated my example with balls in a box. Let me explicitly do it for you:
Two coins are tossed. Then if it’s not Heads Heads one ball is put into a box. Then the second coin is turned to the other side and again if it’s not Heads Heads a ball is put into a box. After this procedure is done, you are given a random ball from the box. What is the probability that the first coin is Heads after you’ve got the ball?
The correct answer here is unambiguosly 1⁄2, which we can check by running the experiment multiple times. On every iteration you get only one ball and on 1⁄2 of them the first coin is Heads. Getting a ball is not evidence in favor of anything because you get it regardless of the outcome of the coin toss.
But if we reason about this problem the same way you try to reason about Sleeping Beauty we inevitably arrive to the conclusion that it has to be 1⁄3. After all, there are four equiprobable possible states {HH, TT, HT, TH}. The ball you’ve just got couldn’t be put in the box on HH, so we have to update to three equiprobable states {HT, TH, TT} and the only one of them where the first coin is Heads is HT. P(HT) = 1⁄3.
This show that such reasoning method can’t generally produce correct answers. So when you applied it to Two-Coin-Toss version of Sleeping Beauty you didn’t actually show that 1⁄3 is the correct answer.
I keep repeating, because you keep misunderstanding how my example is very different than yours.
In yours, there is one “sampling” of the balls (that is, a check on the outcome and a query about it). This one sampling is done only after two opportunities to put a ball into box have occurred. The probability you ask about depends on what happened in both. Amnesia is irrelevant. She is asked just once.
In mine, there are two “samplings.” The probability in each is completely independent of the other. Amnesia is important to maintain the independence.
SPECIFICALLY: SB’s belief is based entirely one what happens in these steps:
Two coins are randomly arranged so that each of the four combinations {HH, HT, TH, TT} has a 25% chance to be the outcome.
If the random combination is HH, one option happens, that does not involve asking for a probability. Otherwise, another option happens, and it does involve asking for a probability.
SB has full knowledge of these three steps, and knows that the second option was chosen. She can assign a probability based ENTIRELY on these three steps.
This happens twice. What you seem to ignore, is that the method used to arrange the coins is different in the first pass through these three steps, and the second. In the first, it is flipping the coins. In the second, it is a modification of this flips. But BECAUSE OF AMNESIA, this modification does no, in any way, affect SB’s assessment that sample space is {HH, HT, TH, TT}, or that each has a 25% chance to be the outcome.
Her answer is unambiguously 1⁄3 anytime she is asked.
I understand that they are different, that’s the whole point. They are different in such a way that we can agree that the answer to my problem is clearly 1⁄2, while we can’t agree to the answer to your problem.
But none of their differences actually affect the mathematical argument you have constructed, so the way you arrive to an answer 1⁄3 in your problem, would arrive to the same answer in mine.
What amnesia does in Sleeping Beauty is ensuring that the Beauty can’t order the outcomes. So when she is awaken she doesn’t know whether it’s the first awakening or the second. She is unable to observe the event “I’ve been awaken twice in this experiment”. The similar effect is achieved by the fact that she is given a random ball from the box. She doesn’t know whether it’s the first ball or the second. And she can’t directly observe whether there are two balls in the box or only one.
Which is completely irrelevant to your mathematical argument about four equiprobable states because you’ve constructed it in such a manner, that the same probabilities are assigned to all of them regardless of whether the Beauty is awake or not. All your argument is based on “There are four equiprobable states and one of them is incompatible with the observations”, it is not dependent on the number of observations.
Now, there is a different argument that you could’ve constructed that would take advantage of two awakening. You could’ve said that when the first coin is Tails there are twice as many awakenings as when it’s Heads and claim that we should interpret it as P(Awake|T1)=2P(Awake|H1) but it’s very much not the argument you were talking about. In a couple of days, in my next post I’m explicitly exploring both of them.
This is wrong. And it’s very easy to check. You may simulate your experiment, a large number of times, writing down the states of the coins on every awakening and notice that there is a clear way to predict the next token beter than chance:
Which is absolutely not the case in a situation where your argument actually works:
Two coins are tossed on Heads Heads event doesn’t happen, on every other outcome it does. Event has happened, what is the probability that the first coin came Heads?
Ignore? On the contrary. This is the exact reason for why your argument doesn’t work. You treat correlated events as independant. That’s what I’m trying to explain to you the whole time and why I brought up the problem with balls being put in the box, because there this kind of mistake is more obvious there.
I suppose this is our crux.
I see two possible avenues for disagreement: about the territory and about the map
First is, whether having an amnesia actually modify the statistical properties of the experiment you are participating in. Do we agree that it’s not the case?
Second, statement about the map, which, correct me if I’m wrong, you actually hold, is that the Beauty should reason about her awakenings as independent because this represents her new state of knowledge due to amnesia?
This would be a correct statement if the Beauty was made to forgot that the events are correlated, if on the awakenings she had different information about which experiment she is participating in, than before she was put to sleep.
But in our case, the Beauty remembers the design of the experiment, she is aware that the states of the coins are not independent between two passes and if she reasons as if they are—she makes a mistake.
Her answer is 1⁄3 to the question “What is the probability that the coin is Heads in a random awakening throughout multiple iterations of such experiment”.
Her answer is 1⁄2 to the question “What is the probability that the coin is Heads in this particular experiment”.
But I don’t think that ambiguity is really the problem here.
Why couldn’t the ball I’ve just got been put into the box on HH? On HH, after we turn the second coin we get HT which is not HH, so a ball is put into the box, no?
Well, sure but then it would mean that the ball wasn’t put into the box on HH, it was put into the box on HT.
If this explanation still feels confusing as if something unlawful is going on—it’s because it is. It’s the exact kind of slight of hand that JeffJo uses to show that 1⁄3 is the correct answer to Two-Coin-Toss version of Sleeping Beauty. If you are able to spot the mistake here you should be able to spot it in his reasoning as well.
I see, I haven’t yet read that one. But yes, we should be clear what we denote with HH/HT/TT/TH, the coins before, or after the turning of second coin.