OK, well let’s start with Variation Alpha’. Consider that there are 20 equally likely possibilities, which we can label (x, y) where x belongs to {heads, tails} and y belongs to {1, …, 10}. Being in possibility (x, y) means “x is the result of the coin toss and y denotes the person we selected beforehand to be revived in the event of heads.”
Suppose that (like Patrick McGoohan) you are number 6. Then out of the 20 possibilities, there are 11 in which you are revived, namely (heads, 6) and (tails, 1) to (tails, 10). Therefore, applying Bayes’ theorem, given that you are revived, the probability of heads is 1⁄11.
OK. I have a quibble with your formalization but I get a similar result when working it out formally: if my background information consists of the Alpha procedure, then updating on being revived does give me 1⁄11.
The quibble is that I only know, algebrically, to condition on something that is a variable, so to work out the joint probability distribution at issue I had to introduce the variable z, with values {revived, not revived}. The triplet (H,3,NR) codes for “the coin comes up heads, person 3 gets picked to be revived in the event of heads, and I don’t get revived”. (Clearly this entails that I’m not person 3.)
The joint probability distribution P(x,y,z) factors out, per the product rule, into P(x)P(y)P(z|x,y) since x and y are independent.
Let’s use N=3 for the number of subjects involved, as I want to write out the full joint distribution (in case someone disagrees with that step) and N=10 makes it tedious. Arbitrarily I consider things from the perspective of Two.
(H,1,R)=0
(H,2,R)=1/6
(H,3,R)=0
(H,1,NR)=1/6
(H,2,NR)=0
(H,3,NR)=1/6
(T,1,R)=1/6
(T,2,R)=1/6
(T,3,R)=1/6
(T,1,NR)=0
(T,2,NR)=0
(T,3,NR)=0
This seems to check out: the marginal distribution for x is the expected 50⁄50, the marginal distribution for y is uniform, it all sums up to 1, it reproduces the setup as described. The conditional distribution P(x,y|z=R) is then:
(H,1)=0
(H,2)=1/4
(H,3)=0
(T,1)=1/4
(T,2)=1/4
(T,3)=1/4
Resulting in P(H|z=R)=1/4.
So I agree here that “I have been revived” is proper to update on, and yields 1/(N+1) credence for the coin having come up heads. (It wasn’t obvious to me to start out, and I still don’t rule out having made a mistake somewhere.)
I can see how this works out as equivalent to the variant I described, with z meaning “got the questionnaire” and y meaning “the label of the person picked to receive the questionnaire in the event of heads”. It shouldn’t matter, either, when we learn about the procedure.
Variations Beta and Gamma don’t seem to introduce anything that should matter, because nothing in the original formulation hinges crucially on particular differences in the memories of the N people involved.
I’m not quite sure what Delta means. My interpretation of Delta would be:
We give you a handout describing the procedure, and some time to absorb it, then put you to sleep. We flip a coin; if it comes up head we wake you, if tails we make an atom-level scan of you, and create and wake N-1 copies from the original scan on successive days, inserting the original on the y-th day.
The triplet (H,3,NR) codes for… um… “the coin came up heads, day 3 was picked to awaken the original me in the even of tails, I (someone other than the person to be awakened in the case of heads) was not revived”. Best I can do.
Something seems to have gone awry somewhere: Delta is not formally equivalent to the previous formulations.
Also, any interpretation of Delta has a big difference with Sleeping Beauty: it ends up with N distinct clones of me, whereas SB ends up with a single Beauty.
My description of Delta wasn’t great, to be fair. So I’ll clarify (and change it slightly) like this:
If (x, y) where x is in {H, T} and y is in {1,2,3} then:
If H then you are not cloned and wake up on day y. If T then a clone of you is created just before the beginning of day 1. Either you or the clone (doesn’t matter which) is woken for day 1 while the other is kept in storage. Then the one that was kept in storage is cloned just before the beginning of day 2. Etc.
The idea of moving from Gamma to (my new) Delta is “it shouldn’t matter whether the clones are created right away (and possibly never used) or ‘just in time’”.
Anyway, the following idea has occurred to me, for defending 1⁄3 as the answer to the original Sleeping Beauty problem: Imagine that there is a clock on the wall and that on any day when SB is woken, the time of day of her awakening is chosen randomly (from a uniform distribution). Then the information that SB gets on awakening is not simply “I was awakened at least once” but “I was awakened at least once at time x”...
...and I’ll leave you guys to do the calculation, but you get 1⁄3, not 1⁄2.
We still have the same problem: there is no value of z that corresponds to “I am a non-special member of the initial set of N people, and I happen to get unlucky and not be revived”. That makes Delta not equivalent to the other variants. It does very much matter whether “not revived” is subjectively possible!
It feels as if this might be the same point that neq1 made earlier in answer to one of the defenses of 1⁄3, so I’d urge you to press on with the formalization and calculation.
My take-away from the discussion (and the two occasions where I changed my mind so far) is that it confirms intuitions aren’t reliable and need to be backed by detailed formalization.
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.
If you are the person that was selected beforehand to be revived in the event of heads, then I agree with 1⁄11. Unfortunately, in variation beta we lose the ability to label someone ahead of time. This changes things.
No it doesn’t. Your clones are subjectively indistinguishable from you, but they’re all in different places at least. Perhaps they’re in rooms labelled 1-10, but not allowed to go outside and look at the number. So the experimenters can toss a D10 and randomly choose a subject without breaking the ‘clone condition’.
OK, well let’s start with Variation Alpha’. Consider that there are 20 equally likely possibilities, which we can label (x, y) where x belongs to {heads, tails} and y belongs to {1, …, 10}. Being in possibility (x, y) means “x is the result of the coin toss and y denotes the person we selected beforehand to be revived in the event of heads.”
Suppose that (like Patrick McGoohan) you are number 6. Then out of the 20 possibilities, there are 11 in which you are revived, namely (heads, 6) and (tails, 1) to (tails, 10). Therefore, applying Bayes’ theorem, given that you are revived, the probability of heads is 1⁄11.
OK. I have a quibble with your formalization but I get a similar result when working it out formally: if my background information consists of the Alpha procedure, then updating on being revived does give me 1⁄11.
The quibble is that I only know, algebrically, to condition on something that is a variable, so to work out the joint probability distribution at issue I had to introduce the variable z, with values {revived, not revived}. The triplet (H,3,NR) codes for “the coin comes up heads, person 3 gets picked to be revived in the event of heads, and I don’t get revived”. (Clearly this entails that I’m not person 3.)
The joint probability distribution P(x,y,z) factors out, per the product rule, into P(x)P(y)P(z|x,y) since x and y are independent.
Let’s use N=3 for the number of subjects involved, as I want to write out the full joint distribution (in case someone disagrees with that step) and N=10 makes it tedious. Arbitrarily I consider things from the perspective of Two.
(H,1,R)=0
(H,2,R)=1/6
(H,3,R)=0
(H,1,NR)=1/6
(H,2,NR)=0
(H,3,NR)=1/6
(T,1,R)=1/6
(T,2,R)=1/6
(T,3,R)=1/6
(T,1,NR)=0
(T,2,NR)=0
(T,3,NR)=0
This seems to check out: the marginal distribution for x is the expected 50⁄50, the marginal distribution for y is uniform, it all sums up to 1, it reproduces the setup as described. The conditional distribution P(x,y|z=R) is then:
(H,1)=0
(H,2)=1/4
(H,3)=0
(T,1)=1/4
(T,2)=1/4
(T,3)=1/4
Resulting in P(H|z=R)=1/4.
So I agree here that “I have been revived” is proper to update on, and yields 1/(N+1) credence for the coin having come up heads. (It wasn’t obvious to me to start out, and I still don’t rule out having made a mistake somewhere.)
I can see how this works out as equivalent to the variant I described, with z meaning “got the questionnaire” and y meaning “the label of the person picked to receive the questionnaire in the event of heads”. It shouldn’t matter, either, when we learn about the procedure.
Variations Beta and Gamma don’t seem to introduce anything that should matter, because nothing in the original formulation hinges crucially on particular differences in the memories of the N people involved.
I’m not quite sure what Delta means. My interpretation of Delta would be:
The triplet (H,3,NR) codes for… um… “the coin came up heads, day 3 was picked to awaken the original me in the even of tails, I (someone other than the person to be awakened in the case of heads) was not revived”. Best I can do.
Something seems to have gone awry somewhere: Delta is not formally equivalent to the previous formulations.
Also, any interpretation of Delta has a big difference with Sleeping Beauty: it ends up with N distinct clones of me, whereas SB ends up with a single Beauty.
My description of Delta wasn’t great, to be fair. So I’ll clarify (and change it slightly) like this:
If (x, y) where x is in {H, T} and y is in {1,2,3} then:
If H then you are not cloned and wake up on day y. If T then a clone of you is created just before the beginning of day 1. Either you or the clone (doesn’t matter which) is woken for day 1 while the other is kept in storage. Then the one that was kept in storage is cloned just before the beginning of day 2. Etc.
The idea of moving from Gamma to (my new) Delta is “it shouldn’t matter whether the clones are created right away (and possibly never used) or ‘just in time’”.
Anyway, the following idea has occurred to me, for defending 1⁄3 as the answer to the original Sleeping Beauty problem: Imagine that there is a clock on the wall and that on any day when SB is woken, the time of day of her awakening is chosen randomly (from a uniform distribution). Then the information that SB gets on awakening is not simply “I was awakened at least once” but “I was awakened at least once at time x”...
...and I’ll leave you guys to do the calculation, but you get 1⁄3, not 1⁄2.
We still have the same problem: there is no value of z that corresponds to “I am a non-special member of the initial set of N people, and I happen to get unlucky and not be revived”. That makes Delta not equivalent to the other variants. It does very much matter whether “not revived” is subjectively possible!
It feels as if this might be the same point that neq1 made earlier in answer to one of the defenses of 1⁄3, so I’d urge you to press on with the formalization and calculation.
My take-away from the discussion (and the two occasions where I changed my mind so far) is that it confirms intuitions aren’t reliable and need to be backed by detailed formalization.
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.
If you are the person that was selected beforehand to be revived in the event of heads, then I agree with 1⁄11. Unfortunately, in variation beta we lose the ability to label someone ahead of time. This changes things.
No it doesn’t. Your clones are subjectively indistinguishable from you, but they’re all in different places at least. Perhaps they’re in rooms labelled 1-10, but not allowed to go outside and look at the number. So the experimenters can toss a D10 and randomly choose a subject without breaking the ‘clone condition’.