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.
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.