It's self-locating information that she can update on. — Andrew M

However, it is not. When awakened Beauty does not know if it is Monday or Tuesday.

One should update one's probabilities when given new information. — Andrew M

The problem with a Bayesian approach is that in order to update you need a prior to update, which in Beauty's case could be based on either the coin flip or the possible times she will be awakened. That's the issue with Bayesian probability and what makes it so controversial, it is subjective.

Variation 2

Suppose I toss a fair coin; if it comes up heads, I ask you if it was heads or tails; if it comes up tails, I don't. When asked, you should always guess "heads".

Now move the likelihood that I'll ask in each case.

Variation 3

If the coin comes up heads, I ask you once to guess; if it comes up tails, I ask you twice.

This is different from our case because for each round you know the first question is the first question. That one's 50-50, but once I ask again, you know to answer "tails". — Srap Tasmaner

These cases are different because being asked (again) provides you with additional information, whereas it doesn't in the original case. You're going to be asked regardless, and you have no idea if you've been asked before.

Now suppose there were a way to fix it so you didn't know how many times you were being asked or whether a question was a first or a second. All you know is that on tails you'll be asked twice. What do you guess? — Srap Tasmaner

There's a way to do this. If it's heads then we ask one person to guess. If it's tails then we ask two people to guess (independently). Nobody knows if they were the only person asked (because it was heads) or if they were the first or second of two (if it was tails). Given that the participants know the rules, what should they guess? The thirder response is that they should guess tails. But does that actually make any sense? I don't think it does. It was a fair coin toss, so it's equally likely to be heads or tails.

But let's imagine that we actually carry this out. We flip the coin 100 times, with it landing heads 50 times and tails 50 times. We'd have asked 150 people to guess. If they all applied the thirder reasoning and picked tails then 2/3 are guaranteed to be wrong (and the same if they all picked heads). But if they flipped a coin to make the guess for them? Assuming 75 heads and 75 tails then, what, anywhere between 1/3 and 2/3 will be right, averaging at 1/2?

The halfer reasoning gives you a better chance of being right.

Edit: But what if the thirders weight the choice, picking a ball from a bag containing two blue and one red, choosing tails if it's a blue ball? We'd expect 100 tails and 50 heads. That's anywhere between 1/3 and 2/3 being right, also averaging at 1/2.

Second edit: I think my math here is wrong.

It needs two blue balls and one red ball.

Listen I am not doing the example gambit; you people come up with all these examples and some can be representative of the problem, while others are not and at that point they become a straw-man, whether it was intentional or not. And if I argue a straw-man then I give false creditiably to it. The best response I can have here, is to suggest we stick to the main problem of interest.

I know many of you think I am just being a troll when I call straw-man, but I know where these numerous examples lead. They are fine in sort order, but people have a tendency to take them off the central path and if you follow them then you end up in La La Land and the discussion has steered way off track.

Let's just stick with the OP, then we all know we are debating the same problem and not something else.

It needs two blue balls and one red ball. — Jeremiah

No it doesn't. It's a blue ball on Tuesday and a black ball on Wednesday (if tails) and a red ball on Monday (if heads).

If you're given a box then there's a 50% chance that it contains a red ball, a 25% chance that it contains a blue ball, and a 25% chance that it contains a black ball.

So, if you've been woken up then there's a 50% chance that today is Monday, a 25% chance that today is Tuesday, and a 25% chance that today is Wednesday.

50% chance of heads, 50% chance of tails.

But even if it's a blue ball on both days with tails, it's still a 50% chance of being a red ball and so a 50% chance of being heads (given that a red ball is only given if heads).

There's never a reason to prefer tails over heads.

Let's just stick with the OP, then we all know we are debating the same problem and not something else. — Jeremiah

The OP version is clearly confusing you, hence the need of the above variation. The point of analogies like these is to make explicit problems that are hidden in the original formulation; they're not strawmen.

The OP version is clearly confusing you, hence the need of the above variation. — Michael

Your "above variation" is confusing you and honestly the colors of the balls would not make any difference. Monday and heads is still only one out of the three possible outcomes.

Monday and heads is still only one out of the three possible outcomes. — Jeremiah

I know, but it has a 50% chance of happening, which is exactly what I'm saying. 50% chance of red ball. 50% chance of Monday and heads. 50% chance of heads.

25% chance of blue ball. 25% chance of tails and Monday.
25% chance of black ball. 25% chance of tails and Tuesday.

50% chance of tails.

Would you consider the 2/3 debacle an argument against the 1/3 argument and for the 1/2 argument?

However, it is not. When awakened Beauty does not know if it is Monday or Tuesday. — Jeremiah

She knows it must be either Monday or (Tuesday and Tails). Do you agree that P(Heads|Awake) = 1/3?

Would you consider the 2/3 debacle an argument against the 1/3 argument and for the 1/2 argument? — Jeremiah

Don't know what you mean.


The multiple person argument is really interesting and might persuade me. (One thing that occurred to me is that if there's betting, splitting the payoffs among multiple people is wrong.)

((Sadly it'll be a little while before I can come back to this.))

She knows it must be either Monday or (Tuesday and Tails) — Andrew M

Sure, she knew that would be the case before even starting, but when she is awaken she has no new information on if it is Monday or Tuesday. You are not exactly describing a condition that is new to her knowledge.

Do you agree that P(Heads|Awake) = 1/3? — Andrew M

No, I have said many times this problem has more than one rational and valid answer.

These cases are different because being asked (again) provides you with additional information, whereas it doesn't in the original case. You're going to be asked regardless, and you have no idea if you've been asked before. — Michael

One more quick note.

The point was that being asked doesn't have to be a simple fact/non-fact, but can have a probability, and we're told that the probability of being asked is driven by the results of the coin toss.

You might still be right that this doesn't matter, I'm not sure.

One thing that occurred to me is that if there's betting, splitting the payoffs among multiple people is wrong. — Srap Tasmaner

Do you mean that if both people who are asked if it's tails correctly guess tails then that should only be counted as 1 success rather than 2?

She knows it must be either Monday or (Tuesday and Tails). Do you agree that P(Heads|Awake) = 1/3? — Andrew M

Assume she's woken on Monday if it's heads or Tuesday and Wednesday if it's tails.

Do you agree that P(Monday|Awake) = 1/2?

The same idea really. One should update one's probabilities when given new information. — Andrew M

Except in this, and in the Monty Hall problem, there is no new information.

Slow reply, I know. Been away.

This one is begging a different answer. From Sleeper's perspective, this has not been established.
— noAxioms

She already knows that it was a fair coin toss and that a fair coin toss has a 50% chance of landing heads. Nothing can change that. — Michael

Information about the toss can very much change that. If she is able to actually see the result of the toss, the odds become a certainty one way or another, not 50/50. So information does change the odds, and she has information beyond the simple fact that a coin was tossed.

To assert 50/50 odds here is to use a perspective other than that of the sleeper.

Another simple way to compute this, objective, but without knowledge:
There are four equal probabiltiy states (each equally has 50% chance of being visited eventually, all depending on the coin toss. Monday and Tuesday are eventual certainties):
A Monday Heads
B Monday Tails
C Tuesday Heads
D Tuesday Tails

The sleeper wakes up and knows not which of the four it is, but she has the additional knowledge (new information) that it is not C, so 33% chance of each of the other choices, and only one of those is heads. Odds are 33%

Don't know what you mean. — Srap Tasmaner

That was Lewis' argument in that paper I linked.

Elga rejects my premiss: his (E7) contradicts my (L1). I reject Elga’s
premiss: my (L6) contradicts his (E1).

Starting on page 3 he lays both arguments out.

http://fitelson.org/probability/lewis_sb.pdf

Gotcha. I'm at work so I'll read later. Maybe much later since I'm enjoying fighting through this "on my own".

Another good thread.

So since there are three possible awakenings and only one is when the coin comes up heads, then won't that mean she has a 33% chance of it being heads? — Jeremiah

Two of the three awakenings always happen. Only one is variable. And it depends on the coin toss. Beauty can only guess, regardless of when she is awoken, whether the coin did (or will - she is woken on Monday) come up heads.

This would follow Bayesian philosophy on probability which suggest we should update our probability models when we get new information. — Jeremiah

Yes, and the only time we receive new information is at the start, when the experiment is explained to us and to Beauty. No new information is provided thereafter, therefore reappraisal of probabilities will lead to the same answer we got originally. Not like the Monty Haul problem, where new information is provided.

Do you mean that if both people who are asked if it's tails correctly guess tails then that should only be counted as 1 success rather than 2? — Michael

Not exactly. If you're calculating wagers and payoffs, you'd need to add up all the gains and losses for the odds to make sense.

When awakened Beauty does not know if it is Monday or Tuesday. — Jeremiah

On this issue of whether there's new information: it's true that all she gets is that she's been awakened. She knows no other new facts, but she knows two things:

• She's more likely to be awakened on Monday than Tuesday;
• she's more likely to be awakened if the toss came up tails.

In the absence of new (certain) knowledge of what has happened, shouldn't she go with the probabilities of what has happened?

One difference between the days and the coin toss is that the days are just clues to the coin toss; it's the toss that determines whether she's awakened a second time.

If I'm making a mistake, it's somewhere around here.

I understand a no spoilers approach, after you exhaust it on your own you should give them a read.

I am not convinced it contradicts the 1/3, but instead just leads back to the initial problem of the inherited subjectivity in the notion that you should start with a prior belief and update that prior into a posterior belief when you get new information. This carries that subjectivity into any posterior, and as we see here those priors can look very different based on the viewer. Not really objective science.

She's more likely to be awakened on Monday than Tuesday — Srap Tasmaner

Yes, she is. Because she is certain to be woken on Monday. So you could say "It's more likely that I am called Srap Tasmaner than that Beauty will be woken on Tuesday." Just select any certainty, and it is more probable than Beauty being woken on Tuesday. I'm not sure what you're getting at here. :chin:

There are four equal probabiltiy states (each equally has 50% chance of being visited eventually, all depending on the coin toss. Monday and Tuesday are eventual certainties):
A Monday Heads
B Monday Tails
C Tuesday Heads
D Tuesday Tails

The sleeper wakes up and knows not which of the four it is, but she has the additional knowledge (new information) that it is not C, so 33% chance of each of the other choices, and only one of those is heads. Odds are 33% — noAxioms

A. Box with red ball
B. Box with blue ball
C. No box
D. Box with black ball

She wakes up and is given a box with a ball in it. What are the odds that it's red? You're saying 33%? I think this is wrong. She knows that if the original toss was heads then there's a 100% chance that her box contains a red ball and that if the original toss was tails then there's a 50% chance that her box contains a blue ball and a 50% chance that her box contains a black ball. From that she can infer that there's a 50% chance of a red ball, a 25% chance of a blue ball, and a 25% chance of a black ball.

There are three outcomes but one is twice as likely as each of the other two. They're not weighted equally.

I noticed a discrepancy.
The description of the problem does not make it clear that sleeping beauty knows the procedure. If she doesn't, then she has no knowledge of anything other than their asking about what is a random coin toss. The odds would be 50% then.
The 33% comes from the sleeper knowing that there will not be a waking on Tuesday if the result is heads, but there will be a waking on the other three scenarios. In this case, new information is gained (it is not Tuesday/heads), and the odds are not 50/50

The 33% comes from the sleeper knowing that there will not be a waking on Tuesday if the result is heads, but there will be a waking on the other three scenarios. In this case, new information is gained (it is not Tuesday/heads), and the odds are not 50/50 — noAxioms

I still don't see how that makes it 33%.

She knows that P(Heads) = 0.5 and that P(Monday|Heads) = 1. So she knows that P(Heads ∧ Monday) = 0.5.

She knows that P(Tails) = 0.5 and that P(Monday|Tails) = 0.5. So she knows that P(Tails ∧ Monday) = 0.25.

She knows that P(Tails) = 0.5 and that P(Tuesday|Tails) = 0.5. So she knows that P(Tails ∧ Tuesday) = 0.25.

Knowing that P(Heads ∧ Tuesday) = 0 might be new information but it doesn't entail that the other three outcomes are equally likely. In fact it's that information that allows her to know that P(Heads ∧ Monday) = 0.5 (rather than 0.25), and so twice as likely as either of the other two.

Not exactly. If you're calculating wagers and payoffs, you'd need to add up all the gains and losses for the odds to make sense. — Srap Tasmaner

So, I ran 100,000 games and gave 1 point for successfully guessing heads and 0.5 points for successfully guessing tails (because you get two opportunities). It doesn't matter if you always select heads, always select tails, select tails 1/2 the time, or select tails 2/3 of the time. The average score is 0.5 in every case.

Although actually I think these analyses miss the point of the puzzle.

1 point for successfully guessing heads and 0.5 points for successfully guessing tails (because you get two opportunities — Michael

I don't understand this. What you're doing there is giving even money on heads and 2-1 for on tails. You can't do that.

My policy is to wager $1 on tails whenever I'm asked. You're paying even money.
If the toss is heads, I lose $1; this happens half the time, so my expected loss is $0.50.
If the toss is tails, I make $1 each time I'm asked; this scenario happens half the time, so I have an expected profit of $1.
So I make at least $0.50 on average each time I play, no matter how the toss goes.
I have made a Dutch book against you.

Suppose instead you're offering 2-1 against heads, and I still wager $1 on tails each time I'm asked.
On heads, it's the same: expected loss of $0.50.
On tails, I only make $0.50 each time, for an expected profit of $0.50.
My profit and loss cancel out, so 2-1 is a fair book, representing the true odds.

At least that's how I think it works.

Suppose instead you're offering 2-1 against heads, and I still wager $1 on tails each time I'm asked.
On heads, it's the same: expected loss of $0.50.
On tails, I only make $0.50 each time, for an expected profit of $0.50.
My profit and loss cancel out, so 2-1 is a fair book, representing the true odds. — Srap Tasmaner

How is this any different to awarding 1 point for successfully guessing heads and 0.5 points for successfully guessing tails?

1 point for each would also be a Dutch book wouldn't it?

You should be paying out $2 on heads, for a wager of $1. That's what 2-1 (against) means. On the favorite, you only pay $0.50, because the odds are 1-2 against.