Marana said:
Sleeping beauty can consider herself to be randomly selected from all such sleeping beauties. That would give a probability of 1/2. The question is whether, from that starting point of 1/2, there is any way to update to 1/3. That's what I haven't seen so far.
I went through the numbers, and it's 2/3. If you do the experiment over and over, with different starting times, then of the "active" Sleeping Beauties (the ones in day 1 or day 2 of the experiment), 1/3 are on day 1 after a coin flip of heads, 1/3 are on day 2 after a coin flip of heads, and 1/3 are on day 1 after a coin flip of tails. If the actual Sleeping Beauty thinks of herself as a random choice among those, then she would come up with 2/3 heads.
It's sort of similar to the situation where there is some country where people have a 50% chance of producing one offspring, and 50% chance of producing two offspring. If you take a random adult and ask the probability that it will have two offspring, the answer is 50%. If you take a random child and ask what is the probability that their parent had two children, it's 2/3.
Equivalently, she can consider her experiment (the coin toss followed by either M or MT) to be randomly selected from all such experiments, and use the relative frequency of 1/2. This is justified because the experiment is an experiment. Again, the question is whether she can change from that 1/2.
Yeah, in the repeated sleeping beauty experiment, there are two different relative frequencies to compute:
- Let's call a coin flip "active" if it's being used in a sleeping beauty experiment, and it was flipped less than two days ago (regardless of the result).
- Let's call a sleeping beauty "active" if she is experiencing day 1 or day 2.
Then:
- 2/3 of active sleeping beauties have an associated coin toss result of heads
- 1/2 of all active coin flips are heads
So, if you point to a random coin flip result (recorded on a piece of paper) and ask a Sleeping Beauty what are the odds that it is heads, she should answer 1/2. But if you tell her that it's her result, she should say 2/3.
Relative frequency can be used, but not haphazardly. It needs to be justified... and we all agree on that! Because clearly using relative frequency of heads awakenings is the wrong way to compute the subjective probability of a heads coin toss without the drug. That is a misuse. And I'm arguing that, even with the drug, it is still a misuse to use relative frequency of heads awakenings as the subjective probability of a heads coin toss. But it is a proper use to use relative frequency with experiments, since they are experiments.
To call it a misuse, you need to say what reason is there not to. What harm comes from it?
To me, the best example of a counter-intuitive result coming from the thirder position is to change it to a lottery. A person has a one in a million chance of winning. But you can make it subjectively 50/50 by waking the winner a million days in a row. That's strange.
It is at issue in the most popular argument for 1/3. The argument goes: after conditioning on "it is monday", we must get 1/2.
That is, P(monday and heads)/P(monday) = P(monday and tails)/P(monday) = 1/2. Therefore P(monday and heads) = P(monday and tails).
I assume we all agree that you wouldn't use regular conditioning on "it is now 2:00" and "it is now 2:01" in my example.
That's not the same thing. You can eliminate that problem by making statements about connections between events. "The first time I looked at the clock, it was 2:00." "The second time I looked at the clock, it was 2:01". No contradiction.
I argue that the same holds here. You can't use regular conditioning on "it is monday", and the argument for 1/3 doesn't work.
I think that's barking up the wrong tree. Instead of Monday, think about Tuesday. You ask Sleeping Beauty what the probability of getting heads was. Now you tell her that today is Tuesday. Then she knows that she got heads. Of course, telling her "Today is Tuesday" gave her new information, and it allowed her to update her likelihood estimates. So if you have a theory of subjective probability that can't accommodate such an update, then I would say something is wrong with that theory.
This isn't really a new thing: none of us condition on the flow of time and immediately throw out every probability computation. It's unspoken, but we avoid it intuitively. Until the sleeping beauty problem, a bizarre setup that makes us want to use it.
I don't think that there is anything wrong with conditioning on the flow of time. If it's possible to forget what day it is, then knowing what day it is is additional information, and it can change your subjective likelihood estimates. It's a language problem for how to deal with words like "now" in a consistent way, not a problem with the applicability of probability.
Rather than using conditioning, what we usually do (correctly) is to know that, by itself, the passage of time doesn't change things that don't change with time.
But that just doesn't seem true. If today is Tuesday, she knows that the coin toss result was heads. So knowing it's Tuesday changes the odds from whatever to 1.
Now, if you did really want to condition on the time, you could treat the time as if it was randomly selected. That's fine. Then you would be conditioning on "it is time X and time X was randomly selected". Using this method you would get P(heads) = 1/2 and P(heads|monday) = 2/3.
How do you get that? That's truly nonsensical, for the following reason (pointed out by
@PeroK): If the memory wipe happens on the morning of the awakening, right before sleeping beauty wakes up, then there is no need to even toss the coin until Tuesday morning. So on Monday, the coin hasn't even been tossed (under this variant). How could the knowledge that today is Monday tell you about a coin that has not yet been tossed?
It makes a lot more sense to say that knowing it's Monday doesn't tell you anything about the likelihood of a coin toss in the future. So P(H | Monday) = 1/2. Knowing that it's Tuesday tells you exactly what the coin toss result was: P(H | Tuesday) = 1. If you don't know whether it's a Monday or a Tuesday, then the probability of heads would be: P(H) = P(H|Monday) P(Monday) + P(H|Tuesday) P(Tuesday) > 1/2
But thirders are trying to have it both ways. They do not want to treat time as randomly selected or static, yet they want to condition on it.
If you want to randomly select a time, as well as randomly select heads or tails, then there are 4 possibilities, equally likely:
- Heads and Monday
- Heads and Tuesday
- Tails and Monday
- Tails and Tuesday
Then you rule out 4 on the basis that the rules say no memory wipe in that case. (So if Sleeping Beauty doesn't know whether it's Monday or Tuesday, then she can't be in situation 4). Eliminating situation 4 still leaves the other three equally likely.