Suppose that for exactly one day every week you sleep during the day and wake up in the evening, and for every other day you sleep at night and wake up in the morning.
Suppose that for a minute after waking, you can reason logically but cannot remember what day it is (and have no way of telling the time).
Then during that minute, surely your subjective probability of it being morning is 6⁄7.
OK now let’s change things up a little:
At the beginning of every week, a coin is flipped. If heads then rather than having 6 days diurnal and 1 day nocturnal, you just have 1 day nocturnal and six days in hibernation. If tails then you have 6 days diurnal and 1 day in hibernation.
Then surely the addition of the coin flip and the hibernation doesn’t change the fact that (for any given awakening) you have a 6⁄7 probability of waking in the morning.
Then surely the addition of the coin flip and the hibernation doesn’t change the fact that (for any given awakening) you have a 6⁄7 probability of waking in the morning.
Well, consider:
Scenario 1: You have a bag containing one red ball and an arbitrarily large number of green balls. You reach in and pull out one ball at random. What is the probability that the ball is red?
Scenario 2: You have a bag containing one red ball and another bag containing an arbitrarily large number of green balls. A fair coin is flipped; if heads, you are handed the bag with the red ball, and if tails you are handed the bag with the green balls (you can’t tell the difference between the bags). You reach in and pull out one ball at random. What is the probability that the ball is red?
In scenario 1, P(red) is vanishingly small. In scenario 2, P(red) is 1⁄2.
I think the coin flip does change things. In fact, I don’t see why it wouldn’t.
In case 1, you know you are somewhere along a path where you will wake up on one night and wake up on 6 mornings. You can’t determine where along that path you are, so you guess morning has probability 6⁄7
In case 2, there is a 50% chance you are somewhere on a path where you wake up once at night (and never in the morning), and a 50% chance you are somewhere on a path where you wake up on 6 mornings and 0 nights. So, probability it is morning is 1⁄2.
So even though, in the long run, 6⁄7 of your awakenings are in the morning, and you have (for that first minute) no information to help you work out which awakening this is, you still think that on any given awakening you ought to feel that it’s just as likely to be morning as evening?
Sure you can bite the bullet if you like, but quite frankly your intuitions are failing you if you can’t see why that sounds strange.
Suppose two weeks’ worth of coins are tossed ahead of time.
Then with probability 1⁄4, you will wake up twice in the evening.
With probability 1⁄2, you will wake up 6 times in the morning and once in the evening. And with probability 1⁄4 you will wake up 12 times in the morning.
Then by your logic, you ought to say that your probability of waking in the morning is (1/4)x(0/2) + (1/2)x(6/7) + (1/4)x(12/12) = 3⁄7 + 1⁄4 = 19⁄28, rather than 1⁄2 if the coins are tossed ‘just in time’.
How can whether the coins are tossed in advance or not change the subjective probability?
Then by your logic, you ought to say that your probability of waking in the morning is (1/4)x(0/2) + (1/2)x(6/7) + (1/4)x(12/12) = 3⁄7 + 1⁄4 = 19⁄28, rather than 1⁄2 if the coins are tossed ‘just in time’.
By neq1′s previous reasoning, there’s 50% chance of waking in the mornings and 50% chance of waking in the evening for any particular week. That is the case whether the coins are tossed in advance or not. The probability of a particular morning awakening would be 1⁄12.
I’m not sure where you got your (6/7) figure for neq1′s calculations.
I’m not sure where you got your (6/7) figure for neq1′s calculations.
neq1 admits that in my original scenario, before I introduced the coin and hibernations, you have a 6⁄7 probability of waking in the morning. The case where one of the two coins is heads and the other is tails is equivalent to this.
Your original question was about one week, not two (I thought).
At the beginning of every week, a coin is flipped. If heads then rather than having 6 days diurnal and 1 day nocturnal, you just have 1 day nocturnal and six days in hibernation. If tails then you have 6 days diurnal and 1 day in hibernation.
Are we just doing this twice? What happens between the weeks? Do they know the experiment has started over?
Could be, or could be a great number of weeks. Shouldn’t make any difference.
What happens between the weeks?
Nothing (except that, if necessary, the next week’s coin is tossed.)
Do they know the experiment has started over?
They know it will start over, and once the ‘minute of confusion’ has passed, they become aware of all that has happened up to now. But during the ‘minute of confusion’ they only know that ‘an experiment is in progress and it is week n for some n’ but don’t know which n.
Once you go more than 1 week it’s not the sleeping beauty problem anymore. Half the time she’s woken up once at night, 1⁄4 of the time she’s woken up 6 times in morn and once at night, 1⁄4 of the time she’s woken up 12 times in morn. This doesn’t have to do with when the coins are tossed. It’s just that, if you do it for 1 week you have the sleeping beauty problem; if you do it multiple weeks you don’t
(1) You got the numbers wrong. “Half the time” should say “1/4 of the time”, the first “1/4 of the time” should say “half the time”, and “once at night” should say “twice at night”.
(2) It’s all very well to state that the situation is different but you haven’t provided any reason why (i) a long sequence of (back-to-back) single week experiments should treated differently from a long sequence of two week experiments. Indeed, the two are the same in every respect except whether some of the coins are tossed in advance, or why (ii) a long sequence of back-to-back single week experiments should be treated differently from just one single week experiment.
(1) You’re right, I got the numbers wrong. Thanks.
(2) If she knows she is somewhere along a two week path, the probabilities are different than if she knows she is somewhere along a one week path. She’s conditioning on different information in the two cases.
Well, you do have to specify whether the subject knows in advance that the experiment is going on for two weeks, or if they’re separate experiments—it changes the subject’s knowledge about what’s going on. Though I’m not sure whether anyone thinks that makes much of a difference.
I’d be interested in your feedback on this and this, which is as close as I can get to a formalization of the original Sleeping Beauty problem, and (after all) does point to 1⁄3 as the answer.
You’re giving a very weak argument. AlephNeil is challenging how your math should work out here. Whether we’re talking about “the sleeping beauty problem” is not entirely relevant.
To an outside observer in the first scenario, the probability of a particular awakening being morning, picked at random, is 6⁄7.
To an outside observer in the second scenario, the probability of a particular awakening being morning is 1 in the tails case and 0 in the heads case, and each case is equally likely, so the probability of a particular awakening being morning is 1⁄2.
So adding the coin flip does change things about the scenario if you’re an outside observer, so I would not be surprised to find it changes things for the subject as well.
Maybe my previous reply didn’t really ‘defuse’ yours. I have to admit your objection was compelling—a good intuition pump if nothing else.
But anyway, moving from ‘intuition pumps’ to hopefully more rigorous arguments, I do have this up my sleeve.
(Edit) Looking back at the argument I linked to, I think I can reformulate it much more straightforwardly:
Consider the original Sleeping Beauty problem.
Suppose we fix a pair of symbols {alpha, beta} and say that with probability 1⁄2, alpha = “Monday” and beta = “Tuesday”, and with probability 1⁄2 alpha = “Tuesday” and beta = “Monday”. (These events are independent of the ‘coin toss’ described in the original problem.)
Sleeping beauty doesn’t know which symbol corresponds to which day. Whenever she is woken, she is shown the symbol corresponding to which day it is. Suppose she sees alpha—then she can reason as follows:
If the coin was heads then my probability of being woken on day alpha was 1⁄2. If the coin was tails then my probability of being woken on day alpha was 1. I know that I have been woken on day alpha (and this is my only new information). Therefore, by Bayes’ theorem, the probability that the coin was heads is 1⁄3.
(And then the final step in the argument is to say “of course it couldn’t possibly make any difference whether an ‘alpha or beta’ symbol was visible in the room.”)
Now, over the course of these debates I’ve gradually become more convinced that those arguing that the standard, intuitive notion of probability becomes ambiguous in cases like this are correct, so that the problem has no definitive solution. This makes me a little suspicious of the argument above—surely the 1/2-er should be able to write something equally “rigorous”.
For any such finite sequence of awakenings, there would (when viewed from the outside) be a 50% chance for a particular week of waking up in an evening, and a 50% chance for a particular week of waking up in the mornings—you can then assign a uniform distribution for particular weeks, getting a 1⁄6 probability of a particular morning in a tails week. If you pick an awakening randomly on that distribution, you have a 1⁄2 probability it’s an evening and a 1⁄12 probability it’s any particular morning (ETA: out of the week).
Another Rival Intution Pump
Suppose that for exactly one day every week you sleep during the day and wake up in the evening, and for every other day you sleep at night and wake up in the morning.
Suppose that for a minute after waking, you can reason logically but cannot remember what day it is (and have no way of telling the time).
Then during that minute, surely your subjective probability of it being morning is 6⁄7.
OK now let’s change things up a little:
At the beginning of every week, a coin is flipped. If heads then rather than having 6 days diurnal and 1 day nocturnal, you just have 1 day nocturnal and six days in hibernation. If tails then you have 6 days diurnal and 1 day in hibernation.
Then surely the addition of the coin flip and the hibernation doesn’t change the fact that (for any given awakening) you have a 6⁄7 probability of waking in the morning.
Well, consider:
Scenario 1: You have a bag containing one red ball and an arbitrarily large number of green balls. You reach in and pull out one ball at random. What is the probability that the ball is red?
Scenario 2: You have a bag containing one red ball and another bag containing an arbitrarily large number of green balls. A fair coin is flipped; if heads, you are handed the bag with the red ball, and if tails you are handed the bag with the green balls (you can’t tell the difference between the bags). You reach in and pull out one ball at random. What is the probability that the ball is red?
In scenario 1, P(red) is vanishingly small. In scenario 2, P(red) is 1⁄2.
The disanalogy is that you actually pull out all of the green balls, not just one.
Indeed—introducing amnesia and pulling out each of the green balls in turn might muddy this one up as well.
I think the coin flip does change things. In fact, I don’t see why it wouldn’t.
In case 1, you know you are somewhere along a path where you will wake up on one night and wake up on 6 mornings. You can’t determine where along that path you are, so you guess morning has probability 6⁄7
In case 2, there is a 50% chance you are somewhere on a path where you wake up once at night (and never in the morning), and a 50% chance you are somewhere on a path where you wake up on 6 mornings and 0 nights. So, probability it is morning is 1⁄2.
So even though, in the long run, 6⁄7 of your awakenings are in the morning, and you have (for that first minute) no information to help you work out which awakening this is, you still think that on any given awakening you ought to feel that it’s just as likely to be morning as evening?
Sure you can bite the bullet if you like, but quite frankly your intuitions are failing you if you can’t see why that sounds strange.
Not all awakenings are equally likely. 50% chance it’s one of the 6 morning awakening. 50% chance it’s the one night awakening.
Suppose two weeks’ worth of coins are tossed ahead of time.
Then with probability 1⁄4, you will wake up twice in the evening. With probability 1⁄2, you will wake up 6 times in the morning and once in the evening. And with probability 1⁄4 you will wake up 12 times in the morning.
Then by your logic, you ought to say that your probability of waking in the morning is (1/4)x(0/2) + (1/2)x(6/7) + (1/4)x(12/12) = 3⁄7 + 1⁄4 = 19⁄28, rather than 1⁄2 if the coins are tossed ‘just in time’.
How can whether the coins are tossed in advance or not change the subjective probability?
By neq1′s previous reasoning, there’s 50% chance of waking in the mornings and 50% chance of waking in the evening for any particular week. That is the case whether the coins are tossed in advance or not. The probability of a particular morning awakening would be 1⁄12.
I’m not sure where you got your (6/7) figure for neq1′s calculations.
neq1 admits that in my original scenario, before I introduced the coin and hibernations, you have a 6⁄7 probability of waking in the morning. The case where one of the two coins is heads and the other is tails is equivalent to this.
Sorry, I’m not following. What are you doing with these two weeks’ worth of coins?
In the situation I described previously, at the beginning of each week a coin is tossed.
What I’m doing is saying: Suppose week 1 AND week 2′s coin tosses both take place prior to the beginning of week 1.
Your original question was about one week, not two (I thought).
Are we just doing this twice? What happens between the weeks? Do they know the experiment has started over?
Could be, or could be a great number of weeks. Shouldn’t make any difference.
Nothing (except that, if necessary, the next week’s coin is tossed.)
They know it will start over, and once the ‘minute of confusion’ has passed, they become aware of all that has happened up to now. But during the ‘minute of confusion’ they only know that ‘an experiment is in progress and it is week n for some n’ but don’t know which n.
Once you go more than 1 week it’s not the sleeping beauty problem anymore. Half the time she’s woken up once at night, 1⁄4 of the time she’s woken up 6 times in morn and once at night, 1⁄4 of the time she’s woken up 12 times in morn. This doesn’t have to do with when the coins are tossed. It’s just that, if you do it for 1 week you have the sleeping beauty problem; if you do it multiple weeks you don’t
(1) You got the numbers wrong. “Half the time” should say “1/4 of the time”, the first “1/4 of the time” should say “half the time”, and “once at night” should say “twice at night”.
(2) It’s all very well to state that the situation is different but you haven’t provided any reason why (i) a long sequence of (back-to-back) single week experiments should treated differently from a long sequence of two week experiments. Indeed, the two are the same in every respect except whether some of the coins are tossed in advance, or why (ii) a long sequence of back-to-back single week experiments should be treated differently from just one single week experiment.
(1) You’re right, I got the numbers wrong. Thanks.
(2) If she knows she is somewhere along a two week path, the probabilities are different than if she knows she is somewhere along a one week path. She’s conditioning on different information in the two cases.
Well, you do have to specify whether the subject knows in advance that the experiment is going on for two weeks, or if they’re separate experiments—it changes the subject’s knowledge about what’s going on. Though I’m not sure whether anyone thinks that makes much of a difference.
I’d be interested in your feedback on this and this, which is as close as I can get to a formalization of the original Sleeping Beauty problem, and (after all) does point to 1⁄3 as the answer.
You’re giving a very weak argument. AlephNeil is challenging how your math should work out here. Whether we’re talking about “the sleeping beauty problem” is not entirely relevant.
To an outside observer in the first scenario, the probability of a particular awakening being morning, picked at random, is 6⁄7.
To an outside observer in the second scenario, the probability of a particular awakening being morning is 1 in the tails case and 0 in the heads case, and each case is equally likely, so the probability of a particular awakening being morning is 1⁄2.
So adding the coin flip does change things about the scenario if you’re an outside observer, so I would not be surprised to find it changes things for the subject as well.
Maybe my previous reply didn’t really ‘defuse’ yours. I have to admit your objection was compelling—a good intuition pump if nothing else.
But anyway, moving from ‘intuition pumps’ to hopefully more rigorous arguments, I do have this up my sleeve.
(Edit) Looking back at the argument I linked to, I think I can reformulate it much more straightforwardly:
Consider the original Sleeping Beauty problem.
Suppose we fix a pair of symbols {alpha, beta} and say that with probability 1⁄2, alpha = “Monday” and beta = “Tuesday”, and with probability 1⁄2 alpha = “Tuesday” and beta = “Monday”. (These events are independent of the ‘coin toss’ described in the original problem.)
Sleeping beauty doesn’t know which symbol corresponds to which day. Whenever she is woken, she is shown the symbol corresponding to which day it is. Suppose she sees alpha—then she can reason as follows:
If the coin was heads then my probability of being woken on day alpha was 1⁄2. If the coin was tails then my probability of being woken on day alpha was 1. I know that I have been woken on day alpha (and this is my only new information). Therefore, by Bayes’ theorem, the probability that the coin was heads is 1⁄3.
(And then the final step in the argument is to say “of course it couldn’t possibly make any difference whether an ‘alpha or beta’ symbol was visible in the room.”)
Now, over the course of these debates I’ve gradually become more convinced that those arguing that the standard, intuitive notion of probability becomes ambiguous in cases like this are correct, so that the problem has no definitive solution. This makes me a little suspicious of the argument above—surely the 1/2-er should be able to write something equally “rigorous”.
In my use of the words ‘every week’ I am implicitly—I take that back, I am explicitly supposing that every week the procedure is repeated.
So we would obtain an indefinitely long sequence of awakenings, of which 1⁄7 are in the evening and 6⁄7 in the morning.
For any such finite sequence of awakenings, there would (when viewed from the outside) be a 50% chance for a particular week of waking up in an evening, and a 50% chance for a particular week of waking up in the mornings—you can then assign a uniform distribution for particular weeks, getting a 1⁄6 probability of a particular morning in a tails week. If you pick an awakening randomly on that distribution, you have a 1⁄2 probability it’s an evening and a 1⁄12 probability it’s any particular morning (ETA: out of the week).