But in that case P(Awake|Heads) should, consistently with this interpretation, refer to your being awakened at all conditioned on the case where the coin landed heads. This is (1/3+2/3)/2 = 0.5 — Pierre-Normand
If you mean P(Awake) to refer to the probability of your being awakened at all (on at least one day) then P(Awake) is indeed 0.5. — Pierre-Normand
Then this goes back to what I said above. These are two different questions with, I believe, two different answers:
1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?
2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads? — Michael
1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?
2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads?
Thirders answer the second question, which I believe is the wrong answer to the first question. The experiment doesn't work by randomly selecting an interview from a set of interviews after repeating the experiment several times and then dropping Sleeping Beauty into it. — Michael
They describe completely different approaches to modelling the problem. That doesn't immediately tell us which SB ought to model the situation as, or whether they're internally coherent. — fdrake
One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating. — Pierre-Normand
I never buy betting arguments unless the random variables are set up! — fdrake
One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating. — Pierre-Normand
I would rather say that the experience works by ensuring that Sleeping Beauty finds herself being awoken in circumstances that she knows to be twice as likely to occur (because twice a frequent) as a result of a coin having landed heads than as a result of this coin having landed tails. — Pierre-Normand
There is a space of possible awakening/interview events A that are being characterised by the day in which they occur ((M)onday or (T)uesday) and by the state of a coin that has been flipped prior to them occurring ((H)eads or (T)ails). P(H) = P(T) = 0.5. — Pierre-Normand
Then you have to say the same about my extreme example. Even when she knows that the experiment is only being run once, Sleeping Beauty's credence that the coin landed heads 100 times in a row is greater than here credence that it didn't.
And I think that's an absurd conclusion, showing that your reasoning is false. — Michael
I'm not sure why you think this is absurd. Compare again my lottery study example. Suppose there are one billion people on the waiting list. If a coin lands heads 20 times in a row, then, 100 million people get pulled from the list. Else, one single person gets pulled from the list. I am then informed that I got pulled from the list (but not whether I am alone or one from 100 million). Is it absurd for me to believe that the coin landed heads 20 times in a row? My credence in this proposition should be roughly 99% — Pierre-Normand
In Sleeping Beauty's case, your intuition that her credence in the high probability of the sequence of heads ought to be absurd apparently stems from your unwillingness to contemplate the possibility of her being able to updating it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. — Pierre-Normand
There's actually two spaces. See here. — Michael
The difference is that the unconditional probability of being called up is very low, and so just being called up at all affects one's credence. In the Sleeping Beauty case (both the normal and my extreme version), she's guaranteed to be awoken either way. — Michael
She's certainly able to update it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. I'm saying that it's irrational of her to.
The only rational approach, upon waking, is to recognize that it landing heads 100 times in a row is so unlikely that it almost certainly didn't, and that this is her first and only interview. — Michael
Regarding betting, expected values, and probability:
Rather than one person repeat the experiment 2^100 times, the experiment is done on 2^100 people, with each person betting that the coin will land heads 100 times in a row. 2^100 - 1 people lose, and 1 person wins, with the winner's winnings exceeding the sum of the losers' losses. The expected value of betting that the coin will land heads 100 is greater than the cost, but the probability of winning is still 1/2 ^100
Even though I could win big, it is more rational to believe that I will lose. — Michael
We must rather imagine that in the unlikely event that the coin lands heads 100 times in a row, 2^101 persons get pulled from the waitlist (or else, only one person does). In that case, if you are one random person who has been pulled from the waitlist, it is twice as likely that your have been so pulled as a result of the coin having landed heads 100 times in a row than not. It is hence rational for you to make this bet and in the long run 2/3 of the people pulled from the waitlist who make that bet with be right. — Pierre-Normand
This isn't comparable to the Sleeping Beaty problem because being a participant isn't guaranteed. That makes all the difference. — Michael
I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all. — Pierre-Normand
I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all. — Pierre-Normand
As one of the 2^100 participants who has just been awoken, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, given that half the participant interviews (plus one) are with participants assigned the winning ticket. — Pierre-Normand
Finding out that today is Monday just removes the blue circle. — Michael
I agree with the idea that Sleeping Beauty's credence in H is updated to 1/2 after she learns that her current awakening is occurring on a Monday. The question is what was this credence updated from, before she learned that the current day was Monday. At that point she could not rule out that the current day was Tuesday. Does that not imply that her credence in H was lower than 1/2 before she learned that the day was Monday? Before she learned that, all three mutually exclulsive areas of the Venn diagram represent possible states for her to be in. She therefore should have non-zero credences for each one of them. — Pierre-Normand
The probability that the coin will land heads and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Tuesday is 1/2.
As the Venn diagram shows, there are two (overlapping) probability spaces, hence why the sum of each outcome's probability is greater than 1.
The double halfer approach does entail:
P(Heads & Monday) = P(Tails & Monday) = P(Tails and Tuesday) = 1/2 — Michael
From the external point of view of the experimenter, it makes sense that the tree probabilities add up to more than one since the three outcomes are non exclusive. The events (Tails & Monday) and (Tails and Tuesday) can (and indeed must) be realized jointly. The intersection zone in the Venn diagram highlights the fact that those two events aren't mutually exclusive. The perspective (and epistemic position) of Sleeping Beauty is different. When she is awoken, those three possible states are mutually exclusive since it is not possible for her to believe that, in her current state of awakening, the current day is both Monday and Tuesday. Her credences in the three mutually exclusive states that she can be in therefore must add up to 1. — Pierre-Normand
There are four people, each assigned a number (unknown to them) between 1 and 4. Two of them are to be put to sleep at random, determined by a single coin toss: if heads then 1 and 2 are put to sleep; if tails then 3 and 4 are put to sleep.
After the coin toss one of the two is put to sleep first. For each of the remaining three, what is the probability that they will be put to sleep? Is it 1/3, because there are three of them and only one is to be put to sleep, or is it 1/2, because that was the probability before the first was put to sleep?
So there are two ways for the participants to approach the problem:
1. I should reason as if I am randomly selected from the set of all participants
2. I should reason as if my interview is randomly selected from the set of all interviews — Michael
I think this is basically a Monty Hall problem. I would say that the probability that I will be put to sleep is 1/2, that the probability that the person to my left will be put to sleep is 1/2, that the probability that the person to my right will be put to sleep is 1/2, and that the probability that one of the other two will be put to sleep is 1/2. — Michael
In this modified scenario, imagine Monty picks one of the unchosen doors at random and reveals a goat by chance. — Pierre-Normand
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