Tuesdays and Heads is clearly and obviously not in the sample space. It is not a possible outcome, she will be never be interviewed on Heads and Tuesday. — Jeremiah
It is a possible outcome; she just won't be asked about it. — Srap Tasmaner
Notice how you've decided (correctly) not to consider Heads + Tuesday and argue that in my example there's a 1/3 chance of it being heads. Why the inconsistency? — Michael
so that's not the explanation for the magic 1/6. — Srap Tasmaner
1/2+1/6 = 2/3
P+(Heads) = P(Heads)+1/6 — Jeremiah
That is P(HEADS | you told me it's Monday) = P(HEADS | you asked me) + 1/6 — Srap Tasmaner
Let H be the proposition that the outcome of the coin toss is Heads. Before being put
to sleep, your credence in H was 1/2. I’ve just argued that when you are awakened
on Monday, that credence ought to change to 1/3. This belief change is unusual. It is
not the result of your receiving new information — you were already certain that you
would be awakened on Monday.3 (We may even suppose that you knew at the start of the
experiment exactly what sensory experiences you would have upon being awakened on
Monday.) Neither is this belief change the result of your suffering any cognitive mishaps
during the intervening time — recall that the forgetting drug isn’t administered until well
after you are first awakened. So what justifies it?
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. - Elga
The problem with Elga's argument is that can't count your temporal location relevant if you don't know what it is. — Jeremiah
Wednesday is not from the same population of awakening where she is interviewed. — Jeremiah
you don't know if it's Monday or Tuesday or Wednesday, but you know you've been awakened. — Srap Tasmaner
When you are first awakened, you are here: — Srap Tasmaner
n = 10000 #Number of flips coin <- sample(c("Heads", "Tails"), n, rep = T) #The coin flip MondayHeads <- 0 MondayTails <- 0 TuesdayTails <- 0 #Loop to count the outcome for (i in coin) { if (i == "Tails") MondayTails <- MondayTails+ 1} for (i in coin) { if (i == "Tails") TuesdayTails <- TuesdayTails + 1} for (i in coin) { if (i == "Heads") MondayHeads <- MondayHeads + 1} Tails <- sum(coin == "Tails") Tails #Number of tails Heads <- sum(coin == "Heads") Heads #Number of Heads MondayHeads #Number of Monday and Heads MondayTails #Number of Monday and Tails TuesdayTails #Number of Tuesday and Tails
One might argue that all things being equal Beauty should go with the 50% credence, since 50 is greater than 33, which gives her a higher chance of being right. If the both argument are equal that is. — Jeremiah
The problem is that wagering confirms the odds are 2-1, which, duh, there are 2 tails interviews for every heads interview. If it's all about getting to give the right answer most often, there's no way to go but tails. — Srap Tasmaner
Oh right. Give the 50% answer you'd give on Monday, because there are more Monday interviews. I remember thinking about that a while ago -- you get to be right 2/3 of the time. — Srap Tasmaner
The simulation proves that both views are valid ways to look that the possible outcomes. — Jeremiah
Is there any argument for not to reporting both the 1/2 and 1/3? Seems like a perfectly valid solution to me and if I ever has a like dilemma in the real world that is exactly what I would do. — Jeremiah
If your position is that it's just a prior and you can pick whatever you want — Srap Tasmaner
However, she will be able, and she will be taught how, to distinguish her brief awakenings during the experiment from her Wednesday awakening after the experiment is over, and indeed from all other actual awakenings there have ever been, or ever will be. — Lewis
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