The Sleeping Beauty Problem is mis-characterized. It is not a single probability experiment, it is one sample from a series of experiments. Without the amnesia drug, it would be just one experiment since the sampling would be redundant. But with the drug, each sample becomes independent sample. The unique circumstances of the problem conspire to hide that.
The way to resolve this, is to expand the problem. And before you claim that I am changing something, I am not. I am extending the exact same mechanisms in order to make their effects more intuitive.
Keep the amnesia. Instead of a coin, use a six-sided die. Instead of two days, make it six as well. And add a second “waking” scenario. Wake Beauty in either the morning, or the afternoon, and put a clock on her wall so she knows which kind of waking it is. The experiment will be governed by a 6x6 calendar, where the rows indicate the die roll, the columns indicate the day, and each entry in the calendar is either an “A” (for AM waking), a “P” (for PM waking), or an “S” (for sleep). And each time Beauty is awake, she is asked to assign a probability to each of the die results in {1,2,3,4,5,6} based on the time.
For my first version, each row must contain each letter at least once, but the remaining 18 entries are random. Here’s an example:
APSSPP
AAPSSP
SAAPSP
AASAPS
ASASAP
PSPAPA
The Halfer argument is that Beauty gains no new evidence when she is awake, since she knew she would be awake during a morning, and an afternoon, at some time regardless of the die. Regardless of the clock, all of the probabilities should be 1⁄6. The Thirder Argument is that, if the clock says it is the afternoon, the probability that die rolled the number D is the number of times the letter “P” appears in row D, divided by the number of times it appears in the table; 12, in this example. So the probabilities are {3/12, 2⁄12, 2⁄12, 1⁄12, 1⁄12, 3⁄12}.
But what if we remove the restriction that each letter must appear in each row?
ASAPPS
SAASPP
ASAASA
SAAAAA
APPSPS
PSSPSP
The same Thirder argument can be applied to this. If the clock says it is morning, a Thirder Beauty would say the answer is {2/14, 2⁄14, 4⁄14, 5⁄14, 1⁄14, 0}. But the Halfer argument falls apart. Either the answer is that there is still a 1⁄6 chance that the die rolled a six, even though beauty cannot be awake in the morning if the die rolled a six. Or the answer is {1/5, 1⁄5, 1⁄5, 1⁄5, 1⁄5, 0) and Beauty has to admit that what she observes does constitute new evidence.
Finally, what if, when Beauty is awake, they tell her that the afternoon technician became sick and every “P” was replaced with an “S”, like this:
ASASSS
SAASSS
ASAASA
SAAAAA
ASSSSS
SSSSSS
This can’t change her second answer, regardless of whether she is a Halfer or Thirder. In other words, what Beauty bases her answers on, when she sees it is an AM waking, is whether each enrtry matches the evidence she has, or does not. Not what she would observe when it does not match.
Here is the crux of the Halfer’s anthropic solution. That an S in a cell not only removes that cell from Beauty’s sample space, it actually removes it from the experiment (i.e., from reality). And so the different numbers of “A”s in a row have to be collapsed into outcome, to make each row with an A have the same number—one—of “A”s.
Great example! I tried to get the same result with biased coin.
I think what Halfer misses is that probability of coin P is not the same that probability that the last toss of the coin was P. We can collect partial information about the last toss via different ways including observation selection effects. We can replace coin in your example with Sun exploding in the past and destroying earth with some chance. In that case, a priopi P of explosion is not the same that there was explosion but we survived (the example needs some polishing).
The Sleeping Beauty Problem is mis-characterized. It is not a single probability experiment, it is one sample from a series of experiments. Without the amnesia drug, it would be just one experiment since the sampling would be redundant. But with the drug, each sample becomes independent sample. The unique circumstances of the problem conspire to hide that.
The way to resolve this, is to expand the problem. And before you claim that I am changing something, I am not. I am extending the exact same mechanisms in order to make their effects more intuitive.
Keep the amnesia. Instead of a coin, use a six-sided die. Instead of two days, make it six as well. And add a second “waking” scenario. Wake Beauty in either the morning, or the afternoon, and put a clock on her wall so she knows which kind of waking it is. The experiment will be governed by a 6x6 calendar, where the rows indicate the die roll, the columns indicate the day, and each entry in the calendar is either an “A” (for AM waking), a “P” (for PM waking), or an “S” (for sleep). And each time Beauty is awake, she is asked to assign a probability to each of the die results in {1,2,3,4,5,6} based on the time.
For my first version, each row must contain each letter at least once, but the remaining 18 entries are random. Here’s an example:
A P S S P P
A A P S S P
S A A P S P
A A S A P S
A S A S A P
P S P A P A
The Halfer argument is that Beauty gains no new evidence when she is awake, since she knew she would be awake during a morning, and an afternoon, at some time regardless of the die. Regardless of the clock, all of the probabilities should be 1⁄6. The Thirder Argument is that, if the clock says it is the afternoon, the probability that die rolled the number D is the number of times the letter “P” appears in row D, divided by the number of times it appears in the table; 12, in this example. So the probabilities are {3/12, 2⁄12, 2⁄12, 1⁄12, 1⁄12, 3⁄12}.
But what if we remove the restriction that each letter must appear in each row?
A S A P P S
S A A S P P
A S A A S A
S A A A A A
A P P S P S
P S S P S P
The same Thirder argument can be applied to this. If the clock says it is morning, a Thirder Beauty would say the answer is {2/14, 2⁄14, 4⁄14, 5⁄14, 1⁄14, 0}. But the Halfer argument falls apart. Either the answer is that there is still a 1⁄6 chance that the die rolled a six, even though beauty cannot be awake in the morning if the die rolled a six. Or the answer is {1/5, 1⁄5, 1⁄5, 1⁄5, 1⁄5, 0) and Beauty has to admit that what she observes does constitute new evidence.
Finally, what if, when Beauty is awake, they tell her that the afternoon technician became sick and every “P” was replaced with an “S”, like this:
A S A S S S
S A A S S S
A S A A S A
S A A A A A
A S S S S S
S S S S S S
This can’t change her second answer, regardless of whether she is a Halfer or Thirder. In other words, what Beauty bases her answers on, when she sees it is an AM waking, is whether each enrtry matches the evidence she has, or does not. Not what she would observe when it does not match.
Here is the crux of the Halfer’s anthropic solution. That an S in a cell not only removes that cell from Beauty’s sample space, it actually removes it from the experiment (i.e., from reality). And so the different numbers of “A”s in a row have to be collapsed into outcome, to make each row with an A have the same number—one—of “A”s.
Great example! I tried to get the same result with biased coin.
I think what Halfer misses is that probability of coin P is not the same that probability that the last toss of the coin was P. We can collect partial information about the last toss via different ways including observation selection effects. We can replace coin in your example with Sun exploding in the past and destroying earth with some chance. In that case, a priopi P of explosion is not the same that there was explosion but we survived (the example needs some polishing).