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).
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).