I'll analyse that case if you can describe it very specifically. Like in the OP. — fdrake
Sleeping Beauty is put to sleep and a coin is tossed 100 times. If it lands heads every time then she is woken up, interviewed, and put back to sleep 2101 times, otherwise she is woken up, interviewed, and put back to sleep once. — Michael
If I understand right, if the coin is heads 100 times, she wakes up on Monday and is not woken up on Tuesday. If the coin is not heads 100 times, she wakes up on Monday and Tuesday? Then the experiment ends. — fdrake
Michael - you ruined my mind again god damnit. — fdrake
The simplest "experiment" is just to imagine yourself in Sleeping Beauty's shoes. — Michael
You're inviting us to imagine ourselves in Sleeping Beauty's shoes to support the halfer position. — Pierre-Normand
To make this scenario more directly analogous to the original problem, let's imagine that Sleeping Beauty, upon each awakening, can not only express her belief about the coin toss but also place a bet on it. In the long run, she would profit from taking the bet as a thirder, further reinforcing this position. — Pierre-Normand
How do you condition on such a thing? What values do you place into Bayes' theorem?
P(Heads|Questioned)=P(Questioned|Heads)∗P(Heads) / P(Questioned) — Michael
P(Selected|Heads) / P(Selected) is 2/1.5 = 4/3. — Pierre-Normand
Being able to bet twice if it lands tails, and so make more money, doesn’t make it more likely that it landed tails; it just means you get to bet twice.
You might as well just say: you can place a £1 bet on a coin toss. If you correctly guess heads you win £1; if you correctly guess tails you win £2.
Obviously it’s better to bet on tails, but not because tails is more probable. — Michael
It makes it twice as likely that individual bets are winning bets. Right? Likewise in Sleeping Beauty's problem, the fact that she is being awoken twice when the coin lands heads makes is more likely that a randomly selected awakening is the result of a coin having landed heads. — Pierre-Normand
After being woken up, which of these is the most rational consideration?
1. The coin almost certainly didn't land heads 100 times, and so this is most certainly my first and only interview, or
2. If this experiment was repeated 2^{100} times then the total number of interviews after the coin landed heads 100 times is greater than the total number of interviews after it didn't, and so if I was to pick an interview at random from that set then there is a greater probability that that interview would have followed the coin landing heads 100 times.
I think the first is the most (and only) rational consideration.
[Although] the second is true ... given that the experiment isn't conducted by picking an interview at random from that set and dropping Sleeping Beauty into it, it's also irrelevant.
Also this makes no sense. You can't have a probability of 2. — Michael
This is not a probability. It's a ratio of probabilities. The updated probability P(Heads|Awoken) is 2/3. The quoted ratio being larger than one just reflects the fact that Bayesian updating results in a probability increase in this case. — Pierre-Normand
I think you numbers there are wrong. See this. — Michael
I don't think it's rational for Sleeping Beauty to use the answer to the second question to answer the first question. I think it's only rational for Sleeping Beauty's credence that the coin landed heads 100 times in a row to be — Michael
The question which has been eating me is "What is the probability of the day being Tuesday?". I think it's necessary to be able to answer that question for the thirder position. But I've not found a way of doing it yet that makes much sense. Though I'm sure there is a way! — fdrake
P(Tuesday|Awoken) = (P(Awoken|Tuesday) / P(Awoken)) * P(Tuesday)
Sleeping Beauty is awoken with probability 3/4 on an average day (Monday or Tuesday). On Tuesdays, she is awoken with P = 1/2. Therefore, P(Awoken|Tuesday) / P(Awoken) = (1/2)/(3/4) = 2/3.
This (2/3) is the Bayesian updating factor. The unconditioned probability of her being awoken on Tuesday is 1/2. The updated probability therefore is P(Tuesday|Awoken) = (2/3)*(1/2) = 1/3, as expected. — Pierre-Normand
I don't think it makes sense to say P(Awake) = 3/4. P(Awake) is just the probability that she will be woken up, which is 1. — Michael
Let's say that there are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.
If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.
I am then using Bayes' theorem to deduce the probability of a random awakening having occurred on a Tuesday. — Pierre-Normand
My reasoning is that P(Awake) = 0.5 given that there are 6 possible outcomes and I will be awake if one of these is true:
1. Heads and I am 1
2. Tails and I am 2
3. Tails and I am 3 — Michael
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