It does. Dice roll 1-5 (safehouse #1): day 1, 0.01% opportunity to escape Dice roll 6 (safehouse #2): day 1, 0.01% opportunity to escape; day 2, 0.01%...
Yes, but it's only P(Dice roll 6|opportunity to escape) > P(Dice roll 1-5|opportunity to escape) because of the opportunity to escape (and where the %...
Can you give actual numbers? Because that determines the answer. If there's a 90% opportunity to escape on day 1 in safehouse #1 but a 1% opportunity ...
To set out the scenario: Dice roll 1-5 (safehouse #1): day 1, 50% opportunity to escape Dice roll 6 (safehouse #2): day 1, 50% opportunity to escape; ...
It does change the epistemic situation. It's exactly like the scenario with the coin tosses and the prizes, where in this case the prize is the opport...
Introducing the concept of escape possibilities simply changes the answer. You're more likely to have an opportunity to escape in safehouse #2, and so...
Are you referring to the safehouse and escape? That's a different scenario entirely. I flip a coin. If heads then I flip again. If heads you win a car...
That's precisely the point, and why I suggested ignoring days and just saying that if heads then woken once and if tails then woken twice. P(W) = 1. T...
If you're going to reason this way then you also need to account for the same with blue. You reach in after the second blue and pull out nothing. So r...
It doesn't. You're dismissed after red or the second blue. It is still the case that if I don't know whether this is Monday or Tails then I reason as ...
I didn't properly address this but I actually think it illustrates the point quite clearly. I'll amend it slightly such that the blue balls are number...
Equally likely to happen, such that P(Monday & Heads) = P(Monday & Tails) = P(Tuesday & Tails) = 1/2, as per that earlier Venn diagram, but not equall...
Well, that’s the very thing being debated. A halfer might say that a Monday & Heads awakening is twice as likely as a Monday & Tails awakening, and so...
So this goes back to what I said before. Either we reason as if we’re randomly selected from the set of all participants, and so P(10) = 1/10, or we r...
Yes, it is rational to believe that if you repeat the game enough times then you will win more than you lose, but it is still irrational to believe th...
I did mention this. There are two ways to reason: 1. I should reason as if I am randomly selected from the set of possible participants 2. I should re...
As Elga says: Sleeping Beauty's "epistemic situation" is only that her current situation is relevant. She doesn't learn anything new. All she knows is...
In fact there's an even simpler way to phrase Bayes' theorem, even using days (where "Mon or Tue" means "today is Monday or Tuesday"). P(Heads | Mon o...
I think this is a better way to consider the issue. Then we don't talk about Heads & Monday or Tails & Monday. There is just a Monday interview and th...
Elga's reasoning has its own unusual implication. In his own words: I'm inclined towards double-halfer reasoning. P(Heads) = P(Heads | Monday) = 1/2, ...
I don't see how this entails that P(A|A or B) = P(B|A or B) entails P(A) = P(B). My example proves that this doesn't follow where P is the credence fu...
There is a difference between these two assertions: 1. P(R|R or B1) = P(B1|R or B1) 2. P(R) = P(B1) The first refers to conditional probabilities, the...
I think the above in fact shows the error in Elga's paper: There is a red ball in one bag and two numbered blue balls in a second bag. You will be giv...
I've been thinking about this and I think there's a simple analogy to explain it. I have one red ball in one bag and two blue balls in a second bag. I...
I wouldn't say that the outcome H occurs one third of the time. I would say that one third of interviews happen after H occurs, because two interviews...
This goes back to what I said before. There are two ways to reason: 1. I should reason as if I am randomly selected from the set of all participants 2...
2/3 bets are right, but that’s because you get to bet twice if it’s tails. That doesn’t prove that tails is more likely. With 4 participants, 1/2 of p...
Except the experiment is only conducted once. Either all her interviews follow one hundred heads or all her interviews (one) follow not one hundred he...
This is heading towards a betting example, which as I've explained before is misleading. There are three different ways to approach it: 1. The same pa...
And with this variation, do you not agree that the probability of it being heads is 3/8? Would you not also agree that the probability of it being hea...
Then this is a different scenario entirely. If we consider the traditional problem, it would be that after the initial coin toss to determine which da...
Also, as an aside, if you correctly reason that it's tails then you escape on the first day, and so you can rule out today being the second day (assum...
That's the very point I disagree with, and is most evident with the example of tossing a coin 100 heads in a row. The possible outcomes have no bearin...
I don't think this is relevant to the Sleeping Beauty problem. It's a different experiment with different reasoning. In this case you're in safehouse ...
Actually that’s not right (starting third day). Need to think about this. First two days are right though. Not sure how this is at all relevant though...
Consider my extreme example. There are two ways to reason: 1. 2\over3 of all interviews are 100 heads in a row interviews, therefore this is most like...
They ruled that out before the experiment begun. You might as well say that they can rule out it being the case that the coin landed heads and that th...
Why? Yes, if you are randomly assigned an interview from the set of all interviews then the probability of it being a tails interview is greater than ...
It's not exactly comparable as in my example he can only put to sleep one of the two people who will be put to sleep; he cannot put to sleep someone w...
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