Why? If the flip had a 50% chance of being heads and if heads guarantees Monday then there's a 50% chance that today is heads and Monday. You seem to just be asserting these probabilities without adequately explaining how you got there. I get to my probabilities by applying an axiom of probability. — Michael

Perhaps quote the axiom in question. I suspect it applies to something not known, such as a coin toss that has not yet been made, or which has been put under a cup without observation. Beauty has been given information about the toss, and that cannot leave the odds unaltered. She has been informed that the combination of Tuesday and Heads is not the case. That information could not have been conveyed if the coin toss was still under the cup. If she had full information ("it was tails"), then the odds would change to 100% tails, axiom of probability not withstanding.

Perhaps quote the axiom in question. — noAxioms

I've done so multiple times:

Kolmogorov definition:

$P(A|B) = \frac{P(A ∩ B)}{P(B)}$

$P(A ∩ B) = P(A|B) * P(B)$

$P(Monday ∩ Heads) = P(Monday|Heads) * P(Heads)$

$P(Heads) = 0.5$

$P(Monday|Heads) = 1$

$P(Monday ∩ Heads) = 0.5$

Also since when did a Bayesian approach become the gold standard, what about a Classical approach?

There is a chance mechanism, the coin flip, it will give us one event from the population, then the ratio of the event over the total possible outcomes will give us the probability.

Possible outcomes

M1 Heads and Monday
T2 Tails and Monday
T2 Tails and Tuesday

One could stop there or, one could argue that since Beauty is asked the credence that there is a second chance mechanism in Beauty's answer. This is not updating a prior with new information, it would be a new event and the population would be dependent on Beauty's selection.

She has been informed that the combination of Tuesday and Heads is not the case. — noAxioms

But it doesn't follow from that that each of the other three outcomes are equally likely.

How do you get from

$P(Tuesday∩Heads) = 0$
to
$P(Monday∩Heads) = P(Monday∩Tails) = P(Tuesday∩Tails)$
?

You are not a very honest person.

I've done so multiple times:

Kolmogorov definition:

<sorry, the formatting rendered the quote unreadable> — Michael

OK, I didn't know that's what that was. I'm actually not much up on the notation of it all, so I have a helluva time following it.
In the linked page, B is listed as unconditional probability. So yes, unconditional odds of heads is 50%, but Beauty is not working from unconditional, and I don't see how you are applying the additional information of 'it isn't Tuesday/heads' into your computation.

I don't see how you are applying the additional information of 'it isn't Tuesday/heads' into your computation. — noAxioms

That's the P(Monday|Heads) = 1.

So yes, unconditional odds of heads is 50%, but Beauty is not working from unconditional — noAxioms

So what's the condition? P(Heads|Awake)? Well, let's apply the Kolmogorov definition again:

$P(Heads|Awake) = \frac{P(Heads ∩ Awake)}{P(Awake)}$

$P(Heads|Awake) = \frac{0.5 * 1}{1} = 0.5$

But it doesn't follow from that that each of the other three outcomes are equally likely. — Michael

Respond then to my post about her getting to wake up on Tuesday.Heads as well. It spells that out.

All four of those things were equal probability. If odds were 50 Monday heads, 25 each Monday tails and Tuesday Tails, then there would be 75% chance that it is Monday, despite the day also being the same odds as the coin toss. I say it is 66% Monday because 2 of the 3 remaining options (A and B) are Mondays.

I don't see how you are applying the additional information of 'it isn't Tuesday/heads' into your computation.
— noAxioms

That's the P(Monday|Heads) = 1. — Michael

Excuse me if I am new to the notation. I read this as the probability of it being at least one of Monday or Heads is 1, but since it might be Tuesday/Tails, this is wrong. I would think the probability of Monday or Tails is certain.
Maybe I just don't know how to read the notation.

Excuse me if I am new to the notation. I read this as the probability of it being at least one of Monday or Heads is 1, but since it might be Tuesday/Tails, this is wrong. I would think the probability of Monday or Tails is certain.
Maybe I just don't know how to read the notation. — noAxioms

P(Monday|Heads) means "the probability that it's Monday given the fact that it's heads". If I know that today is heads then the probability that today is Monday is 1. Whereas if I know that today is tails then the probability that today is Monday is 0.5 (because it equally could be Tuesday).

So, P(Monday|Heads) = 1 and P(Heads) = 0.5, therefore P(Monday and Heads) = 0.5.

So what's the condition? P(Heads|Awake)? Well, let's apply the Kolmogorov definition again:

P(Heads|Awake)=P(Heads∩Awake)/P(Awake)
P(Heads|Awake)=0.5/1=0.5 — Michael

As you see, the quote is getting altered. Sorry.
I think you are mixing the probabilities that something will occur (or is the case, but unknown) with probabilities that something known has occurred. You seem to assign 1 to P(Awake) which is not the probability that you will be away, but rather the probability after the information about being awake has been completely (not just partially) conveyed.

Not sure if a computation of P(awake) is going to yield what we want. It seems undefined. Of course we will be awake at some point. She's not being asked if she's awake, but being awake is information nevertheless.

P(Monday|Heads) means "the probability that it's Monday given the fact that it's heads" — Michael

Oh crap. OK then. I really don't know how to read this stuff then. '|' means 'or' in my world, but they have that intersection symbol to mean that here. Union for 'and'.

Sorry again, but the posts are coming faster than I can actually absorb that web link that seems to assume (reasonably) that you already know the rudiments of the notation. For instance the 'unconditional probability' hyperlinks to a page that makes no mention of the term. Not helpful.

Also since when did a Bayesian approach become the gold standard, what about a Classical approach? — Jeremiah

The classical approach only works if each outcome is equally likely. As I would argue each outcome isn't equally likely it would be wrong to apply the classical approach. It would be like using the classical approach to a weighted coin toss.

OK, I think I've found the disagreement between halfers and thirders:

$P(Heads|Awake) = \frac{P(Heads ∩ Awake)}{P(Awake)}$

To get a result of $\frac{1}{3}$, $P(Awake)$ needs to be $\frac{3}{4}$. This only happens if $P(Heads ∩ Asleep) = \frac{1}{4}$.

To get a result of $\frac{1}{2}$, $P(Awake)$ needs to be $1$. This only happens if $P(Heads ∩ Asleep) = 0$.

Given that:

$P(Heads ∩ Asleep) = P(Heads ∩ Tuesday),$

What is $P(Heads ∩ Tuesday)$?

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.

Similarly, in the Sleeping Beauty problem, awakening provides information that enables you to rule out one of the four states. However since you have no information distinguishing the remaining states, you should be indifferent about which state you are currently in. — Andrew M

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.

Let's imagine the situation repeated 150 times. 75 occurrences of each situation means the chances are 1/3. There's no reason to assume the probabilities are different.

There's 1/2 chance of heads and Monday happening, but that doesn't mean the chances of the Sleeping Beauty being in that situation are 1/2.

Thinking of the sailor variant, for any outsider the chances of heads or tails are 1/2, but if you're then told the sailor is your father, the chances of tails double because then he's had two children which means the odds of you being his child are higher.

If the odds were 1/2 and the Sleeping Beauty got to bet 1€ each time she was woken up, she should break even no matter what, but we notice that betting tails wins her 2€ with tails and loses her 1€ with heads.

If the odds were 1/2 and the Sleeping Beauty got to bet 1€ each time she was woken up, she should break even no matter what, but we notice that betting tails wins her 2€ with tails and loses her 1€ with heads. — BlueBanana

I've gone over this:

If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely.

You're conflating "more likely to win if tails" and "more likely that tails".

It's just a 2:1 payout on a 50/50 chance, but only if you bet a certain way.

If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely. — Michael

But she's only given 1€ with tails. The reason she wins by guessing tails is because she's likelier to be in a situation where tails has been thrown.

You're conflating "more likely to win if tails" and "more likely that tails". — Michael

I'm not conflating, I'm drawing a conclusion. She's likelier to win with tails because tails is the likelier outcome of the bet.

But she's only given 1€ with tails. The reason she wins by guessing tails is because she's likelier to be in a situation where tails has been thrown. — BlueBanana

She's given it twice: once on Monday and once on Tuesday.

You don't even have to go through the hassle of putting her to sleep. Just tell her that if it's tails she'll get £1 on Monday and a £1 on Tuesday, and if it's heads she'll just get £1 on Monday. She's going to pick tails, because there's twice the payout.

Putting her to sleep, waking her up, and then asking her to bet doesn't make tails more likely.

That is not at all how it works.

A simple random sample is a subset from the population where n members of the population have a chance of being selected by random chance, such as a fair coin flip. The actual distribution of the population doesn't matter. This subset is also called an event, and probability is the ratio of the event over the total number of possible outcomes from the population.

She's given it twice: once on Monday and once on Tuesday. — Michael

Shouldn't matter if their odds are 25%. Then the expected value would be (1/4)*1+(1/4)*1+(1/2)*(-1)=0.

Because of the amnesia different people can be used to compare. Heads, I randomly choose a person from the street to ask the question from. Tails, I choose a hundred. Should they guess I threw tails or heads?

Or if the amount of people doesn't matter, why not wake up the Sleeping Beauty 0 times with heads.

Because of the amnesia different people can be used to compare. Heads, I randomly choose a person from the street to ask the question from. Tails, I choose a hundred. Should they guess I threw tails or heads? — BlueBanana

For any given person there's a 50% chance that they're right, so it doesn't matter if they pick heads or tails. It's just that if it's tails and they pick heads then there's a greater number of losers and if it's tails and they picks tails then there's a greater number of winners.

But you don't say that if there's more winners under tails then tails is more likely. That's a non sequitur.

There aren't more winners because it's more likely but because you asked more people.

For any given person there's a 50% chance that they're right, so it doesn't matter if they pick heads or tails. It's just that if it's tails and they pick heads then there's a greater number of losers and if it's tails and they picks tails then there's a greater number of winners. — Michael

Of course that's true if they choose randomly whether to guess heads or tails. If they make the sensible choice of tails there're more winners because it's the likelier choice to be the correct one.

But you don't say that if there's more winners under tails then tails is more likely. That's a non sequitur. — Michael

It doesn't make tails more likely to be the result of my throw, but it makes it likelier for the correct answer to be tails. That's not non sequitur, that's a textbook example of what probability means.

There aren't more winners because it's more likely but because you asked more people. — Michael

It's more likely because I asked more people.

If they make the sensible choice of tails there're more winners because it's the likelier choice to be the correct one. — BlueBanana

No. There are more winners because you asked more people.

It's more likely because I asked more people. — BlueBanana

This doesn't make any sense.

Same thing. It's the likelier choice because I asked more people.

There are more winners because you asked more people. — Michael

It doesn't make tails more likely to be the result of my throw, but it makes it likelier for the correct answer to be tails. — BlueBanana

All you're saying is:

Iff 100 people asked then it was tails [the rule of the game]
100 winners [the outcome]
Therefore, it was tails

But that's obvious, and not relevant.

All you're saying is:

If 100 people asked then it was tails [the rule of the game]
100 winners [the outcome]
Therefore, it was tails

But that's obvious, and not relevant. — Michael

I nowhere said the last two parts.

If the test is repeated 150 times, in 75 out of 225 questions the correct answer is heads. That's 1/3. What definition of probability makes the chances 1/2?

I’ve showed the reasoning multiple times. It’s the Kolmogorov definition of conditional probability.