The probability of the awakenings is dependent on the coin flip (1st awakening is 1 if heads, 0.5 if tails), whereas the probability that a coin flip lands heads is independent. — Michael
So why not apply the same reasoning to Sleeping Beauty? The initial coin toss has 1/2 odds of heads, so it's 1/2 odds of heads. — Michael
This is why I suggested the alternative experiment where we don't talk about days at all and just say that if it's heads then we'll wake her once (and then end the experiment) and if it's tails then we'll wake her twice (and then end the experiment). There aren't four equally probable states in the experiment.
So we either say that P(Awake) = 1 (and/)or we say that being awake doesn't provide Beauty with any information that allows her to alter the initial credence that P(Heads) = 0.5. — Michael
Saying that P(Awake) = 1 is fine since we can calculate the other probabilities accordingly. — Andrew M
But I think the four equally probable states clarifies the mathematical relationship between the independent observer viewpoint (where the odds are familiar and intuitive) and Sleeping Beauty's viewpoint. — Andrew M
Saying that P(Awake) = 1 is fine since we can calculate the other probabilities accordingly. — Andrew M
But I think the four equally probable states clarifies the mathematical relationship between the independent observer viewpoint (where the odds are familiar and intuitive) and Sleeping Beauty's viewpoint. — Andrew M
M T H 1 0 T 0.5 0.5
M T H 1 0 T 1 1
today is M today is T if H 1 0 if T 0.5 0.5
If (upon first awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either T1 or T2. Since being in T1 is subjectively just like being in T2, and since exactly the same propositions are true whether you are in T1 or T2, even a highly restricted principle of indifference yields that you ought then to have equal credence in each. But your credence that you are in T1, after learning that the toss outcome is Tails, ought to be the same as the conditional credence P(T1| T1 or T2), and likewise for T2. So P(T1| T1 or T2) = P(T2 | T1 or T2 ), and hence P(T1) = P(T2).
This is correct when waking up just once, so why not also when possibly waking up twice? — Michael
Because when you are woken up more than twice more awakenings are caused by throwing tails. — BlueBanana
Hey look at that. He saw the Wednesday argument and slipped in a defeater! — Srap Tasmaner
If that is so then when Beauty considers the actual question that is when temporal location is relevant and at that time Wednesday is already off the table, as the asking of the question eliminates it. And speculation that Beauty was considering Wednesday as a possibility before the interview is speculation and is outside the relevant temporal location which is in consideration.The answer is that you have gone from a situation in which you count your own temporal location as irrelevant to the truth of H, to one in which you count your own
temporal location as relevant to the truth of H.
We thirders think halfers are looking at the wrong event — Srap Tasmaner
Now try this with SB: instead of asking for your degree of belief, I'm going to tell you how the coin toss landed. What is your degree of belief that I will tell you it landed heads? Is it 50%? — Srap Tasmaner
the Bayesian approach — Jeremiah
if I randomly select an outcome-telling from all the heads-tellings and all the tails-tellings, are selecting a heads-telling and selecting a tails-telling equally likely? Not if there are twice as many tails-tellings. — Srap Tasmaner
There's a 50% chance you'll tell me "at all" that it's heads, and same for tails. But there's more than a 50% chance that a random selection from the tellings you've done will be a tails telling. — Srap Tasmaner
the proper way — Michael
We can then distinguish between the case of telling someone twice that it's tails and the case of telling someone once but using a weighted coin that favours tails. — Michael
If it's 1 then P(Heads|Awake) = 0.5 * 1 / 1 = 0.5. — Michael
Then perhaps you could explain how it works with my variation where Beauty is woken on either Monday or Tuesday if tails, but not both. Do we still consider it as four equally probable states and so come to the same conclusion that P(Heads|Awake) = 0.5 * 0.5 / 0.75 = 1/3? — Michael
Mon Tue Heads Awake:1/4 Asleep:1/4 Tails-Heads2 Awake:1/8 Asleep:1/8 Tails-Tails2 Asleep:1/8 Awake:1/8
Mon Tue Heads 1/2 0 Tails-Heads2 1/4 0 Tails-Tails2 0 1/4
This is correct when waking up just once, so why not also when possibly waking up twice? — Michael
Mon Tue Heads Awake:1/4 Asleep:1/4 Tails Awake:1/4 Awake:1/4
Mon Tue Heads 1/3 0 Tails 1/3 1/3
Ignore the coin toss completely. The intention of the problem is that Beauty cannot know whether this is her first or second interview. If we count that as a toss-up, then... — Srap Tasmaner
Mon Tue Heads 1/4 0 Tails 1/4 1/2
quarterer — Andrew M
Bet H T Toss H 1 -1 T -1 2
Bet H T Toss H 1 -1 T -2 2
Bet H T Toss H 2 -1 T -2 1
Bet H T Toss H .5 -.5 T -1.5 1.5
Bet H T Toss H 3 -1 T -2 2/3
This is all stuff we've said before -- this comment summarizes the mechanism by which standard thirder wagering pays out 3:1, as andrewk pointed out, instead of 2:1. — Srap Tasmaner
You could also think of it as revenge against the halfer position, which draws the table this way:
...
Halfers, reasoning from the coin toss, allow Monday-Heads to "swallow" Tuesday-Heads.
Reasoning from the interview instead, why can't we do the same? — Srap Tasmaner
Get involved in philosophical discussions about knowledge, truth, language, consciousness, science, politics, religion, logic and mathematics, art, history, and lots more. No ads, no clutter, and very little agreement — just fascinating conversations.