But this probability space is the epistemological probability space of Beauty, and we have no axioms that justify our assuming that the probabilities will be the same in that probability space as in the independent observer's one. — andrewk
There's nothing to eliminate. This is it. — Michael
We flip a coin. If it's heads then we wake Mary once. If it's tails then we wake her twice.
What is the probability that it's her first awakening? 2/3, because two of the three outcomes are first awakenings? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)? — Michael
Because that's where we get to after we eliminate the one possibility from the aforementioned case. — BlueBanana
As I said, Tuesday and heads. — BlueBanana
From the trivial scenario where she's woken up twice regardless of the coin flip.
Where has Tuesday come from? It wasn't mentioned at all in the experiment I described here. — Michael
Then you might as well say that we've eliminated the trivial scenario of heads + heads in this experiment. — Michael
Heads + heads utterly irrelevant and doesn't change the probability at all. — Michael
Because that's the original scenario where it's already eliminated. — BlueBanana
You mean modifying that scenario so that it includes heads+heads doesn't change the probabilities in it? — BlueBanana
I'm saying that there is an eliminated outcome – heads + heads – and yet the probability is still 0.75. — Michael
With the outcome added there are certain odds, and when that is modified by eliminating that one possibility the odds are changed in a way that somehow makes sense. — BlueBanana
No, if you consider heads + tails the chances go to 0,5. Eliminating that possibility results in 0,75. — BlueBanana
We flip a coin. If it's heads then the result stands. If it's tails then we flip again and the new result stands.
What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)? — Michael
Completely different situation. There's no "eliminated" outcome. — BlueBanana
The probability of first awakening is 0.5 + (0.5 * 0.5) = 0.75. — Michael
Then there's 6 states, not 7. You're counting the tails state twice, which you shouldn't do. The two tails days need to share the probability that it's the tails state (1/6) giving each 1/12 which is the correct figure you get when you apply the probability rule:
P(A and B) = P(A) * P(B|A)
P(Tails and Tuesday) = P(Tails) * P(Tuesday|Tails)
P(Tails and Tuesday) = 1/6 * 1/2 = 1/12 — Michael
Mon Tue Heads 5/6 0 Tails 1/12 1/12
Mon Tue Heads 5/7 0 Tails 1/7 1/7
Mon Tue Heads 5/12 5/12 Tails 1/12 1/12
Beauty doesn't gain relevant new information when awakening, she knew all this before hand. If we do the experiment on you, then you have a prior belief that it is 1/3, what new information would then update that? Priors need relevant new information which would allow us to update it, not just any old information that you think happened. — Jeremiah
In the Monty Hall problem, the host gives you information that changes the probabilities that you assign to each door. That information is new to you.
— Andrew M
The host does not, that's the trick. — tom
Not as the problem was described at the top of the thread. No information is given to Sleeping Beauty beyond what she was told would happen. To her each awakening is identical, and there are three of them. — tom
Since Beauty has no available information distinguishing the three states from her point-of-view, she is simply indifferent about which state she is currently in, and so assigns a probability of 1/3 for each awake state. — Andrew M
Note that Beauty (as a halfer) does the same thing when she is told that it is Monday. For a halfer, P(Heads|Monday) = 2/3 so, on attaining that self-locating information, she updates P(Heads) to be 2/3. — Andrew M
How so? Isn't the host telling you a specific door number (that doesn't contain the prize) information? — Andrew M
The event of waking provides information about which states one can eliminate (not merely the conditional probabilities, which were previously known). BTW, do you assign 1/2 or 1/3 to P(Heads|Awake)? — Andrew M
With that preface, perhaps I misunderstood your post here. In the fifth line of the reasoning, you assert, seemingly as an axiom, that P(Heads)=0.5. While - as indicated above - I sympathise with that position, it does look like assuming one's conclusion, if the object of the exercise is to determine what probability Beauty assigns to the coin having landed Heads. — andrewk
Anyway, I've never encountered anyone who agrees with me on this. — tom
P(Awake) = 1 (as the sample space of interviews is always when awake) — Michael
Beauty (as a thirder) knew at the time before the experiment that P(Heads) = 1/2. She also knew at the time before the experiment that P(Heads|Awake) = 1/3. — Andrew M
That is not how conditionals work. If it were, there would never be a point to conditional probability. You'd just always use 1, and P(A|B) would just be a funny way of writing P(A). In how much of the total space is she awakened? Your way is just wrong. — Srap Tasmaner
3/4.
It is possible to have a heads on Tuesday, same as always. It's within the total space, just not the conditional space. — Srap Tasmaner
How does this new information alter the fact that one door has a probability of 1/3, and 2 doors has a probability of 2/3? — tom
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