Michael

To apply this to the traditional problem: there are two Sleeping Beauties; one will be woken on Monday and one on both Monday and Tuesday, determined ...

I think this is best addressed by a variation I described here: Before anyone was put to sleep, for each of the four participants the probability of b...

There are two approaches. The normal halfer approach is: P(Heads) = 1/2 P(Tails) = 1/2 P(Monday) = 3/4 P(Tuesday) = 1/4 The double halfer approach is:...

I think this is a non sequitur. That most interviews are with a winner isn't that I am mostly likely a winner. Rather, if most participants are a winn...

Well, consider the Venn diagram here (which you said you agreed with). There are two probability spaces. Monday or Tuesday is one consideration and He...

This isn't comparable to the Sleeping Beaty problem because being a participant isn't guaranteed. That makes all the difference. Compare these two sce...

Regarding betting, expected values, and probability: Rather than one person repeat the experiment 2100 times, the experiment is done on 2100 people, w...

The difference is that the unconditional probability of being called up is very low, and so just being called up at all affects one's credence. In the...

There's actually two spaces. See here.

In my extreme example she wins in the long run (after 2100 experiments) by betting on the coin landing heads 100 times in a row. It doesn't then follo...

Me being awaked at all conditioned on the case where the coin lands heads is 1/3, given that if it lands heads then I am only woken up if I was assign...

Then this goes back to what I said above. These are two different questions with, I believe, two different answers: 1. If the experiment is run once, ...

So how would your reasoning work for this situation? My reasoning is that P(Awake) = 1/2 given that there are 6 possible outcomes and I will be awake ...

I don't think it correct to say P(Awake) = 3/4. P(Awake) is just the probability that she will be woken up, which is 1. This is clearer if we forget t...

Maybe this Venn diagram helps? Of course, this is from the experimenter's perspective, not Sleeping Beauty's, but it might help all the same.

I think there are two different questions with two different answers: 1. If the experiment is run once, what is Sleeping Beauty's credence that the co...

My mistake. I think your example here is the same as the example I posted at the start? As I later showed here, it provides a different answer to the ...

This is where I think my extreme example is helpful. Place a bet on each interview whether or not the coin landed heads 100 times in a row. In the lon...

Also this makes no sense. You can't have a probability of 2.

I think you numbers there are wrong. See this.

Well yes. The very question posed by the problem is “what is Sleeping Beauty’s credence that the coin landed heads?”, or in my version “what is Sleepi...

They are equivalent. There are two, equally probable, situations that E(z) uses: 1. z = 2y and y = 10 2. z = {y\over2} and y = 20 So it's doing this: ...

I'm not redefining y, the switching argument is. I'm showing you what it covertly does.

There is. I explained it above. I'll do it again. Assume, for the sake of argument, that one envelope contains £10 and one envelope contains £20, and ...

You may have missed my edit. It doesn't require anything like that. The only premises are that one envelope contains twice as much as the other and th...

I believe it does, as I showed above. It covertly redefines y such that when it concludes E(z) = {5\over4}y, y is no longer the value of the chosen en...

This is the reasoning that leads to the switching argument: P(y = x) = P(z = 2y) = {1\over2}\\P(y = 2x) = P(z = {y\over2}) = {1\over2} E(z) = {1\over2...

This and this explain it quite clearly I think.

I can be sure it isn't as per the post immediately before yours.

The simplest "experiment" is just to imagine yourself in Sleeping Beauty's shoes. You know that if the coin lands heads 100 times then you will be int...

I don't think we need to worry about days. The traditional experiment can be simply stated as: if tails, two interviews, otherwise one interview. In m...

Sleeping Beauty is put to sleep and a coin is tossed 100 times. If it lands heads every time then she is woken up, interviewed, and put back to sleep ...

I don't know how to explain it any simpler than the above. It's exactly like the traditional experiment, but rather than two interviews following from...

What do you make of this? You know the experiment is only being run one time. When you wake up, do you follow thirder reasoning and argue that it is m...

And that's the non sequitur. That I would wake up more often if the coin lands heads 100 times in a row isn't that, upon waking, it is more likely tha...

No, the probability of you seeing heads when you wake up is conditional on how likely you wake up for heads and for tails, not on how often you wake u...

And my first comment to you was literally "that it happens more often isn’t that it’s more likely", i.e. that it's more frequent isn't that it's more ...

No it's not. It's 1 for heads and 1 for tails. A probability of 2 makes no sense.

Let y be the value of the chosen envelope and z be the value of the unchosen envelope. 1. y = x or y = 2x 2. E(z) = {5\over4}y = {3\over2}x 3. y = {6\...

I explained it in more detail in that earlier post above. The variable y is used to represent three different values, two of which are the possible va...

I think the reasoning that leads you to this conclusion is clearly wrong, given that it’s an absurd conclusion. 1.

Then if it’s heads 100 times in a row I wake you up 2101 times, otherwise I wake you up once. I don’t think it reasonable to then conclude, upon wakin...

It only depends on whether or not the single coin flip landed tails. Imagine a different scenario. If I flip a coin 100 times and it lands heads every...

That’s a non sequitur. That it happens more often isn’t that it’s more likely.

I don't think that table is how to calculate the probabilities. Consider a slight variation where there are no days, just number of awakenings. If hea...

My concern is in explaining where the switching argument goes wrong. The switching argument says that because E(z) = {5\over4}y, where y is the value ...

There are four people, each assigned a number (unknown to them) between 1 and 4. Two of them are to be put to sleep at random, determined by a single ...

Only if it’s Tuesday. She gets interviewed on Monday regardless. So what if the coin toss doesn’t happen until after the Monday interview? Does that a...

But in this case they're using a variable to represent more than one value at the same time. It's a fallacy to add y to {1\over4}y for two different v...

Even if we accept the premise (as I do) that P(z = 2y) = P(z = {y\over2}) = {1\over2}, there's still no reason to switch, so the paradox has nothing t...