How do you condition on such a thing? What values do you place into Bayes' theorem? — Michael
I don't think your numbers are accurate there. — Michael
But in this case we're not asking about an outsider's analysis of frequency, but Sleeping Beauty's when it is known that just a single experiment is being run. — Michael
And how do you expect to apply Bayes's rule without any base rate information? SB can reason as I have described to determine what those base rates would be were the experiment repeated a number of times, and set her subjective probabilities accordingly. — Srap Tasmaner
The probability that I will be questioned if the coin lands heads is 1. The probability that I will be questioned is 1. — Michael
What you want is the odds that this interview is a heads-type interview. — Srap Tasmaner
What you want is the odds that this interview is a heads-type interview. — Srap Tasmaner
Yes, that's the left hand side of the theorem: P(Heads|Questioned). — Michael
The trouble is on the RHS, the probability that I am being interviewed given that my coin was heads. — Srap Tasmaner
In our version, the base rate of heads interviews is 1 in 3. Make it 1 in 1000. (That is, 999 awakenings on tails, not 2.) Isn't it obvious that if I'm a subject in such an experiment, I know it's far more likely I'm being asked for my credence because my coin came up tails? If I'm one of 1200 subjects, I know there are 600,000 interviews, only 600 of which were for heads, while 599,400 were for tails. Equally likely that this interview is for heads as for tails? Not by a long shot. — Srap Tasmaner
import numpy as np import matplotlib.pyplot as plt def experiment(trials): headsProbability = 0.0 for heads in range(trials + 1): headArrangements = np.math.factorial(trials)/np.math.factorial(trials-heads)/np.math.factorial(heads) headsRightAnswer = heads / (heads + (trials - heads)*2.0) headsProbability += headArrangements / np.power(2.0, trials) * headsRightAnswer return headsProbability trials = [i+1 for i in range(50)] results = [experiment(t) for t in trials] plt.plot(trials, results, '-') plt.ylabel("heads probability") plt.xlabel("trials") plt.show()
((This is, I don't know, maybe the third time I've argued with Michael about something and then concluded he was right all along.)) — Srap Tasmaner
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.
If Michael is woken then what is his credence that the coin landed heads?
…
Michael's credence before the experiment is P(1) = 1/3, so if woken he ought to continue to have a credence of P(1) = 1/3 since he gains no new relevant evidence if he wakes up during the experiment.
And given that if woken the coin landed heads iff he is 1, he ought to have a credence of P(Heads) = 1/3. — Michael
Four volunteers will be assigned a random number but each will undergo an experiment that is functionally equivalent to the popular version of the problem. The same sleep and amnesia drugs will be used, and each will be awoken at least once, but maybe twice, based on the same fair coin toss. Only their schedules and the question they are asked will differ, but end up being equivalent to the popular problem. On Monday and Tuesday:
#1 Will be awoken unless it is Tuesday, after Heads.
#2 Will be awoken unless it is Tuesday, after Tails.
#3 Will be awoken unless it is Monday, after Heads.
#4 Will be awoken unless it is Monday, after Tails.
Each will be asked for their credence that this is the only time they will be awoken. For #1 and #3, that means credence in Heads. For #2 and #4, it is credence in Tails. For all four, the answer has to be the same as the correct answer to the popular version of the Sleeping Beauty Problem.
On each day, we can bring the three awake volunteers together to discuss their answers. Of these three, exactly one will not be, or was not, awakened on the other day of the experiment. But none of the three can have more, or less, credence that she is that one instead of one of the others.
So with three awake volunteers, one of whom will be awakened only once, the answer is 1/3.
Why was the scientific american wasting time on this? — sime
mean(rbinom(10000,1,0.5))
It's just right, look at the code!
All coding something up does is let you check a calculation for it being correct, not whether it's the appropriate calculation to do. — fdrake
No, the question is what is the probability SB experiences an awakening with the coin being heads — hypericin
For a life-span can be divided into countless parts, each completely independent of the others, so that from my existing at one time it doesn’t follow that I exist at later times, unless some cause keeps me in existence – one might say that it creates me afresh at each moment.
The probability that I will be questioned if the coin lands heads is 1. The probability that I will be questioned is 1. — Michael
She is therefore being asked "What is P(C | M) , where M is your current mental state?" — sime
The thirder view is that only the current slice that you might be is relevant, and there are more being-interviewed slices in the tails partition, so you're more likely one of those. — Srap Tasmaner
the correct analysis is that the coin flip partitions SB's future slices into a heads set and a tails set, just two, equal chances of being in each set. — Srap Tasmaner
No, the question has nothing whatsoever with her mental state.
She is being asked, given that she is awakened, what is the probability of heads. If she is awakened 1000 times for every tails and once with heads, given enough coin flips you can see that it is overwhelmingly likely to be tails, even though the probability of heads remains 50%. This is independent of her mental state. — hypericin
This relies on the intuition of repeating the experiment over and over. If so, then there are unconditionally more tail slices. But the coin is flipped exactly once. Therefore, even though there are more tail slices, they both exist only upon a tails flip. — hypericin
The problem vexing the minds of experts is as follows: Sleeping Beauty agrees to participate in an experiment. On Sunday she is given a sleeping pill and falls asleep. One of the experimenters then tosses a coin. If “heads” comes up, the scientists awaken Sleeping Beauty on Monday. Afterward, they administer another sleeping pill. If “tails” comes up, they wake Sleeping Beauty up on Monday, put her back to sleep and wake her up again on Tuesday. Then they give her another sleeping pill. In both cases, they wake her up again on Wednesday, and the experiment ends.
The important thing here is that because of the sleeping drug, Sleeping Beauty has no memory of whether she was woken up before. So when she wakes up, she cannot distinguish whether it is Monday or Tuesday. The experimenters do not tell Sleeping Beauty either the outcome of the coin toss nor the day.
They ask her one question after each time she awakens, however: What is the probability that the coin shows heads?
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