P(Tuesday|Awoken) = (P(Awoken|Tuesday) / P(Awoken)) * P(Tuesday)

Sleeping Beauty is awoken with probability 3/4 on an average day (Monday or Tuesday). On Tuesdays, she is awoken with P = 1/2. Therefore, P(Awoken|Tuesday) / P(Awoken) = (1/2)/(3/4) = 2/3.

This (2/3) is the Bayesian updating factor. The unconditioned probability of her being awoken on Tuesday is 1/2. The updated probability therefore is P(Tuesday|Awoken) = (2/3)*(1/2) = 1/3, as expected. — Pierre-Normand

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 the days. It is just the case that if it is heads then she is woken up once and if it is tails then she is woken up twice. It doesn't make sense to say that she's not awake for her second awakening if heads.

What is the probability of the day being Tuesday? — fdrake

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 coin landed heads 100 times in a row?

2. If the experiment is repeated 2¹⁰⁰ times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads 100 times in a row?

I don't think it's rational for Sleeping Beauty to use the answer to the second question to answer the first question. I think it's only rational for Sleeping Beauty's credence that the coin landed heads 100 times in a row to be $1\over2^{100}$.

And so too with the original problem:

1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?

2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads?

Thirders answer the second question, which I believe is the wrong answer to the first question.

This is not a probability. It's a ratio of probabilities. The updated probability P(Heads|Awoken) is 2/3. The quoted ratio being larger than one just reflects the fact that Bayesian updating results in a probability increase in this case. — Pierre-Normand

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 original problem.

It makes it twice as likely that individual bets are winning bets. Right? Likewise in Sleeping Beauty's problem, the fact that she is being awoken twice when the coin lands heads makes is more likely that a randomly selected awakening is the result of a coin having landed heads. — Pierre-Normand

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 long run (e.g. after ~2¹⁰⁰ experiments) you’ll start making money, but it just doesn’t follow that it having landed heads 100 times in a row is more probable.

And also the premise of both the original and my problem is that the experiment is just run once.

So this goes back to what I said above:

After being woken up, which of these is the most rational consideration?

1. The coin almost certainly didn't land heads 100 times, and so this is most certainly my first and only interview, or

2. If this experiment was repeated 2¹⁰⁰ times then the total number of interviews after the coin landed heads 100 times is greater than the total number of interviews after it didn't, and so if I was to pick an interview at random from that set then there is a greater probability that that interview would have followed the coin landing heads 100 times.

I think the first is the most (and only) rational consideration.

[Although] the second is true ... given that the experiment isn't conducted by picking an interview at random from that set and dropping Sleeping Beauty into it, it's also irrelevant.

P(Selected|Heads) / P(Selected) is 2/1.5 = 4/3. — Pierre-Normand

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

I think you numbers there are wrong. See this.

You're inviting us to imagine ourselves in Sleeping Beauty's shoes to support the halfer position. — Pierre-Normand

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 Sleeping Beauty’s credence that the coin landed heads 100 times in a row?”

To make this scenario more directly analogous to the original problem, let's imagine that Sleeping Beauty, upon each awakening, can not only express her belief about the coin toss but also place a bet on it. In the long run, she would profit from taking the bet as a thirder, further reinforcing this position. — Pierre-Normand

Being able to bet twice if it lands tails, and so make more money, doesn’t make it more likely that it landed tails; it just means you get to bet twice.

You might as well just say: you can place a £1 bet on a coin toss. If you correctly guess heads you win £1; if you correctly guess tails you win £2.

Obviously it’s better to bet on tails, but not because tails is more probable.

Perhaps this is more evident with my extreme example. Place a bet on each interview whether or not the coin landed heads 100 times in a row. In the long run (e.g. after ~2¹⁰⁰ experiments) you’ll start making money, but it just doesn’t follow that it having landed heads 100 times in a row is more probable.

And also the premise of the problem is that the experiment is just run once.

Your two definitions of E[z] aren't equivalent. — sime

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:

$E(z) = {1\over2}2y[where\space y=10] + {1\over2}({y\over2})[where\space y=20]$

I'm just making this more explicit:

$E(z) = {1\over2}(2\cdot10) + {1\over2}({20\over2})$

Which leads to:

$E(z) = 15 = {5\over4}12$

Hence the redefinition of $y$. It no longer stands for the value of the chosen envelope given that the value of the chosen envelope isn't 12.

Yes, I see that. So why are you redefining y? — sime

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

No covert redefinitions of y are happening — sime

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 that I pick an envelope at random.

$P(y = 10) = P(z = 2y) = {1\over2}\\P(y = 20) = P(z = {y\over2}) = {1\over2}$

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y\\E(z) = {1\over2}(2\cdot10) + {1\over2}({20\over2}) = {5\over4}12$

Notice that in $E(z)$ the variable $y$ stands for 3 different values. In the first case it stands for the value of the smaller envelope (10), in the second case it stands for the value of the larger envelope (20), and in the third case it stands for a different value entirely (12).

That third value (12) isn't the value of the chosen envelope, given that the chosen envelope contains either £10 or £20.

This is true for every possible pair of values and is true even when we don't assume the values of the two envelopes (it's just harder to see).

And it should be obvious that the probability assignments are correct. If one envelope contains £10 and one envelope contains £20 and I pick one at random then the probability that I will pick the envelope that contains £10 (the smaller amount) is $1\over2$ and the probability that I will pick the envelope that contains £20 (the larger amount) is $1\over2$.

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 that I pick one at random. The probability that I will pick the envelope with the smaller amount is 1/2.

It's contradictory expectation values don't appeal to faulty reasoning given acceptance of the premises. — sime

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 envelope.

Rather the switching argument is unsound, for among it's premises is an improper prior distribution over x, the smallest amount of money in an envelope. And this premise isn't possible in a finite universe.

Intuitively, it's contradictory conclusions makes sense; if the smallest amount of money in an envelope could be any amount of money, and if the prior distribution over the smallest amount of money is sufficiently uniform, then whatever value is revealed in your envelope, the value of the other envelope is likelier to be higher. — sime

It doesn't require anything like that. The only premises are that one envelope contains twice as much as the other and that I pick one at random. The probability that I will pick the envelope with the smaller amount is $1\over2$.

This is the reasoning that leads to the switching argument:

$P(y = x) = P(z = 2y) = {1\over2}\\P(y = 2x) = P(z = {y\over2}) = {1\over2}$

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y$

But then by the exact same logic (or by performing the appropriate substitions):

$E(z) = {1\over2}2x + {1\over2}x = {3\over2}x$

Therefore:

$E(z) = {5\over4}y = {3\over2}x$

I don't understand. — SolarWind

This and this explain it quite clearly I think.

You can be sure that the expected value for the other envelope is 5/4 of that of the one you have. — SolarWind

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 interviewed 2¹⁰¹ times, otherwise you will be interviewed once, and you know that the experiment will not be repeated.

After being woken up, which of these is the most rational consideration?

1. The coin almost certainly didn't land heads 100 times, and so this is most certainly my first and only interview, or

2. If this experiment was repeated 2¹⁰⁰ times then the total number of interviews after the coin landed heads 100 times is greater than the total number of interviews after it didn't, and so if I was to pick an interview at random from that set then there is a greater probability that that interview would have followed the coin landing heads 100 times.

I think the first is the most (and only) rational consideration.

Your proposed simulation would certainly prove that the second is true (although the math alone is enough to prove it), but given that the experiment isn't conducted by picking an interview at random from that set and dropping Sleeping Beauty into it, it's also irrelevant.

If I understand right, if the coin is heads 100 times, she wakes up on Monday and is not woken up on Tuesday. If the coin is not heads 100 times, she wakes up on Monday and Tuesday? Then the experiment ends. — fdrake

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 my experiment it is: if 100 heads, 2¹⁰¹ interviews, otherwise one interview.

But if thinking about it in days helps then: if the coin is heads 100 times, she wakes up on Day 1, Day 2, Day 3, ..., and Day 2¹⁰¹. If the coin is not heads 100 times, she wakes up on Day 1. Then the experiment ends.

What I think this variation shows is that it is wrong to determine the probability by imagining that we randomly select from the set of all possible interviews (weighted by their probability), and then "dropping" Sleeping Beauty into that interview. The experiment just doesn't work that way; it works by tossing a coin 100 times and then waking her up. I think it's clear at a glance that these will give different results, and I think that the second is the correct one, even from Sleeping Beauty's perspective.

So I suppose this is a reductio ad absurdum against the self-indication assumption that guides thirder reasoning.

Michael - you ruined my mind again god damnit. — fdrake

You're welcome. ;)

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 2¹⁰¹ times, otherwise she is woken up, interviewed, and put back to sleep once.

When being interviewed, is her credence that the coin landed heads 100 times in a row greater than her credence that it didn’t?

If we accept thirder reasoning then it is, which I think is wrong. It is a mistake to use the number of times that she would wake up were it to land heads 100 times in a row to determine the probability that it did land heads 100 times in a row.

Any reasonable Sleeping Beauty would understand that it almost certainly didn’t land heads 100 times in a row, and so that her current interview is almost certainly her first and only.

Whichever reasoning applies to this experiment must also apply to the traditional experiment.

I'll analyse that case if you can describe it very specifically. Like in the OP. — fdrake

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 a $P = {1\over2}$ coin toss it's 2¹⁰¹ interviews following from a $P = {1\over{2^{100}}}$ coin toss (with just 1 interview otherwise).

Thirder reasoning would entail that, after waking, it is more likely that the coin landed heads 100 times in a row, and I think that's an absurd conclusion.

I would say that it doesn't matter how many times you will wake me if the coin lands heads 100 times in a row. When I wake up the only reasonable conclusion is that it almost certainly didn't land heads 100 times in a row.

What do you make of this?

if it’s heads 100 times in a row I wake you up 2¹⁰¹ times, otherwise I wake you up once

You know the experiment is only being run one time.

When you wake up, do you follow thirder reasoning and argue that it is more probable that the coin landed heads 100 times in a row?

Or do you follow halfer reasoning and argue that the number of times you would wake up were it to land heads 100 times in a row is irrelevant, and that it almost certainly didn't land heads 100 times in a row?

Because SB wakes up more on tails, a given wake up event is more likely to be caused by a tail flip that a head flip. — PhilosophyRunner

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 that the coin landed heads 100 times in a row.

the probability of you seeing heads when you wake up is conditional on how often you wake up for heads and how often for tails — PhilosophyRunner

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 up.

In your case, the reason it matters is because the probability of waking on heads is 1 and tails is 0.

In the ordinary case it doesn't matter, because the probability of waking on heads is 1 and tails is 1.

I was talking about frequency not probability. — PhilosophyRunner

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 probable.

Waking on tails is twice as frequent but equally probable.

In the SB problem it is 1 for heads and 2 for tails. — PhilosophyRunner

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. $y = {6\over5}x$ (solving ${5\over4}y = {3\over2}x$ for $y$)

3 contradicts 1.

This is clearer if we assume the value of $x$ for the sake of argument:

1. $y = 10$ or $y = 20$

2. $E(z) = {5\over4}y = 15$

4. $y = 12$ (solving ${5\over4}y = 15$ for $y$)

3 contradicts 1.

So the conclusion of the switching argument, that $E(z) = {5\over4}y$, where $y$ is the value of the chosen envelope, is false.

Despite the initial definition, the $E(z)$ formula covertly redefines $y$.

And for the case where we know that $y = 10$:

1. $x = 5$ or $x = 10$

2. $E(z) = {3\over2}x = {5\over4}10$

3. $x = 8.\bar{3}$ (solving ${3\over2}x = {5\over4}10$ for $x$)

3 contradicts 1.

The paradox arises because the same variable ($x$ and/or $y$) is used to represent more than one value. It's a disguised fallacy, unrelated to any probability assignments.

Where have I gone wrong? — Srap Tasmaner

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 values of the chosen envelope, and the third (the one in ${5\over4}y$) isn’t the value of the chosen envelope, and so the conclusion that the unchosen envelope has a greater expected value than the chosen envelope doesn’t follow.

In that case it is more likely that given an instance I wake up I will see the coin has been flipped heads 100 times in a row. — PhilosophyRunner

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

I flip a coin and if it lands heads I wake you up tomorrow, if it lands tails you never wake you up. If you wake up and are asked the probability the coin landed heads, what would you say? — PhilosophyRunner

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 waking, that it is more likely that it landed heads 100 times in a row. The fact that you would be woken up far more times if it did happen just doesn’t make it more likely to have happened.

It is only reasonable to understand that it landing heads 100 times in a row is so unlikely that it almost certainly didn’t.

The only thing that matters is the coin flip(s). The rest is a distraction.

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 time I will wake you up a million times, otherwise I will wake you up once.

After waking up, is it more likely that I got heads 100 times in a row?

She is more likely to wake up and see a coin showing tails, as she will wake up more often if the coin lands on tails. — PhilosophyRunner

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 heads then woken once, if tails then woken twice. And consider perhaps that the coin isn't tossed until after the first awakening.

P(Heads) is just the prior probability that the coin will land heads, which is 0.5, P(Awake) is just the prior probability that Sleeping Beauty will be woken up, which is 1, and P(Awake|Heads) is just the prior probability that she will be woken up if the coin lands heads, which is 1.

That gives us:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake| Heads) * P(Heads)} \over P(Awake)}\\&={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

Given a set {10, 20}, the expected value of a number selected from that set is 15. There's nothing wrong with your first set of equations, and it gives the right answer. You don't have to go through all that; you just need the average. — Srap Tasmaner

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 of the chosen envelope and $z$ the value of the unchosen envelope, it is rational to switch.

But given the set $\{10, 20\}$, $E(z) = 15 = {5\over4}12$, and $12$ isn't the value of the chosen envelope.

The same reasoning holds for all $\{x, 2x\}$ sets (where $x \gt 0$).

What this shows is that the $y$ in $E(z) = {5\over4}y$ isn't the value of the chosen envelope, and so it does not suggest to switch.

(1) What are the chances that y = x and the chances that y = 2x if y is chosen randomly from a set {x, 2x}? (You may, if you like, write it backwards as x = y and x = y/2.)

(2) What are the chances that a y chosen randomly from a set {x, 2x} was chosen from a set {y, 2y} and the chances it was chosen from a set {y/2, y}? — Srap Tasmaner

I think I explained it best here:

We can only consider this from the perspective of the participant, who only knows that one envelope contains twice as much as the other and that he picked one at random. His assessment of $P(y = x)$ and $P(y = x|y =10)$ can only use that information.

Is it correct that, given what he knows, $P(y = x) = {1\over2}$?
Is it correct that, given what he knows, $P(y = x|y =10) = P(y = x)$?

If so then, given what he knows, $P(y = x|y = 10) = {1\over2}$.

Perhaps this is clearer if we understand that $P(y = x|y = 10)$ means "a rational person's credence that his envelope contains the smaller amount given that he knows that his envelope contains £10".

And if the above is true then so too is $P(y = 2x|y = 10) = {1\over2}$.

Learning the value of the chosen envelope is an uninformative posterior, and so the prior probability of picking/having picked the envelope containing the smaller amount is maintained.

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 coin toss: if heads then 1 and 2 are put to sleep; if tails then 3 and 4 are put to sleep.

After the coin toss one of them is put to sleep first. For each of the remaining three, what is the probability that they will be put to sleep? Is it 1/3, because there are three of them and only one is to be put to sleep, or is it 1/2, because that was the probability before the first was put to sleep?

I think this basically is the Sleeping Beauty problem.

Whereas you know it does, as the "sampling day" of SB's report depends upon the coin flip. — fdrake

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 affect your answer?

Eh, probability modelling also includes assigning random variables. It has a lot to do with what random variables you put in play. — fdrake

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 values of $y$.

This is most obvious when we assume two values for the sake of argument, e.g. £10 and £20, where $y$ is the value of the chosen envelope and $z$ the value of the unchosen envelope:

$P(y = 10) = P(z = 2y) = {1\over2}\\P(y = 20) = P(z = {y\over2}) = {1\over2}$

$E(z) = {1\over2}2y + {1\over2}({y\over2}) = {5\over4}y\\E(z) = {1\over2}(2\cdot10) + {1\over2}({20\over2}) = {5\over4}12$

Given that, assuming one envelope contains £10 and the other £20, the chosen envelope doesn't contain £12, it is false to assert that $E(z) = {5\over4}y$, where $y$ is the value of the chosen envelope. $y$ is in fact a different value entirely.

This is true for all $x$ and $2x$ pairs.

That's all the paradox is.

With respect to the case where we open the chosen envelope to find £10:

$P(y = x|y =10) = P(z = 2y|y =10) = {1\over2}\\P(y = 2x|y =10) = P(z = {y\over2}|y =10) = {1\over2}$

$E(z) = {1\over2}2y + {1\over2}({y\over2})\\E(z) = {1\over2}2x + {1\over2}({2x\over2}) = {3\over2}x\\E(z) = {1\over2}(2\cdot10) + {1\over2}({2\cdot5\over2}) = {3\over2}8.\bar{3}$

Given that the chosen envelope contains £10, the smaller envelope doesn't contain £8.333..., and so once again $E(z)$ commits a fallacy in using the same variable to represent more than one value at the same time.

I could imagine using it for teaching probability modelling. Get students to analyse the problem. Then do it IRL with both sampling mechanisms. Should be a cool demonstration of "physical" differences between what's seen as a merely "epistemic" probability assignment! — fdrake

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 to do with probability at all.

Yes

The $x$ in that case wasn't referring to the smaller value but to the value of the other envelope. I'll rephrase it to make it clearer:

Let $x$ be the amount in one envelope and $2x$ be the amount in the other envelope.

Let $y$ be the amount in the chosen envelope and $z$ be the amount in the unchosen envelope.

$P(y = x) = P(z = 2y) = {1\over2}\\P(y = 2x) = P(z = {y\over2}) = {1\over2}$

$E(z) = {1\over2}2y[where\space y={1\over2}z] + {1\over2}({y\over2})[where\space y=2z]\\E(z) = {1\over2}({2z\over2}) + {1\over2}({2z\over2}) = {z\over2} + {z\over2} = z$

How do you get 1? — Dawnstorm

Because Sleeping Beauty is certain to be questioned during the experiment.

Compare with my alternative scenario here where the probability of being questioned is 1/2.

How do you interpret P(Questioned)? — Dawnstorm

"My credence that I will be questioned."