The calculation is a little bit awkward because seemingly one has to condition on an event of zero probability (which entails division by zero). But we can proceed as follows:
Suppose the number of moments in a day is finite but ‘very large’, call it N.
Let’s list all of the possible outcomes:
If x = heads then SB is woken on Monday, and there are now N equally likely possibilities for when this will be.
If x = tails then SB is woken on Monday and again on Tuesday. There are N^2 equally likely possibilities for the two waking times.
Suppose SB wakes at time t0. Then she can reason thusly: If the coin toss was heads, then the probability of me seeing a clock show t0 was 1/N. Or if the coin toss was tails: Out of the N^2 possibilities, there are N where I see t0 on Monday and N where I see t0 on Tuesday, but I’ve double counted the case where I see t0 on both Monday and Tuesday, so in fact there are 2N-1 equally likely ways this could have happened. Note that (2N-1)/N^2 is roughly equal to 2/N.
So let H be the event “coin is heads” and let T0 be the event “SB sees clock pointing to t0″.
We have: P(T0 | H) = 1/N and P(T0 | ~H) = about 2/N
So the posterior probabilities for H and ~H must be (about) 1⁄3 and 2⁄3 respectively.
The posterior probabilities converge to 1⁄3, 2⁄3 as N goes to infinity.
(Note: The reason for the discrepancy (i.e. the fact that P(H | T0) is not exactly1⁄3) is that SB’s reasoning about ‘double-counting’ the instance when she is woken at t0 both times is actually invalid, and this possibility ought to be double counted. But the entire dispute centers around showing why it has to work this way in the case N = 1, so I think I’m entitled to pretend that the anti-double-counting argument is valid in order to show the contrary.)
I can reformulate the argument above 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”.
Sorry, I meant to say I’d urge you to press on with the formalization and calculation in your interpretation of the Delta case.
I’ll punt on the wall-clock idea. I’m not planning to spend any time working out the formalization for anything that involves large numbers of values for any given variable—my skills aren’t up to doing that confidently, and we seem to have enough to go on with formulations of the problem that only involve smaller sets.
OK but intuitively it can’t make any difference whether SB is woken at a fixed or a random time of day, and it can’t make any difference whether there is a clock on the wall.
So the solution to the ‘random-waking, clock on wall variation’ must be the same as the solution of the original SB problem.
The calculation is a little bit awkward because seemingly one has to condition on an event of zero probability (which entails division by zero). But we can proceed as follows:
Suppose the number of moments in a day is finite but ‘very large’, call it N.
Let’s list all of the possible outcomes:
If x = heads then SB is woken on Monday, and there are now N equally likely possibilities for when this will be.
If x = tails then SB is woken on Monday and again on Tuesday. There are N^2 equally likely possibilities for the two waking times.
Suppose SB wakes at time t0. Then she can reason thusly: If the coin toss was heads, then the probability of me seeing a clock show t0 was 1/N. Or if the coin toss was tails: Out of the N^2 possibilities, there are N where I see t0 on Monday and N where I see t0 on Tuesday, but I’ve double counted the case where I see t0 on both Monday and Tuesday, so in fact there are 2N-1 equally likely ways this could have happened. Note that (2N-1)/N^2 is roughly equal to 2/N.
So let H be the event “coin is heads” and let T0 be the event “SB sees clock pointing to t0″.
We have: P(T0 | H) = 1/N and P(T0 | ~H) = about 2/N
From Bayes’ theorem: P(H | T0) / P(~H | T0) = (P(H)/P(~H)) (P(T0|H) / P(T0|~H)) = (1/2)/(1/2) (1/N)/(2/N) = 1 * 1⁄2 = 1⁄2 (roughly)
So the posterior probabilities for H and ~H must be (about) 1⁄3 and 2⁄3 respectively.
The posterior probabilities converge to 1⁄3, 2⁄3 as N goes to infinity.
(Note: The reason for the discrepancy (i.e. the fact that P(H | T0) is not exactly 1⁄3) is that SB’s reasoning about ‘double-counting’ the instance when she is woken at t0 both times is actually invalid, and this possibility ought to be double counted. But the entire dispute centers around showing why it has to work this way in the case N = 1, so I think I’m entitled to pretend that the anti-double-counting argument is valid in order to show the contrary.)
I can reformulate the argument above 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”.
Sorry, I meant to say I’d urge you to press on with the formalization and calculation in your interpretation of the Delta case.
I’ll punt on the wall-clock idea. I’m not planning to spend any time working out the formalization for anything that involves large numbers of values for any given variable—my skills aren’t up to doing that confidently, and we seem to have enough to go on with formulations of the problem that only involve smaller sets.
OK but intuitively it can’t make any difference whether SB is woken at a fixed or a random time of day, and it can’t make any difference whether there is a clock on the wall.
So the solution to the ‘random-waking, clock on wall variation’ must be the same as the solution of the original SB problem.
See this for a crisp, simple formalization which appears to show where the ambiguity between 1⁄2 and 1⁄3 comes from.