Here's a halfer analogy I found convincing before:

Two urns, one with a single white marble, one with many black marbles; you flip a coin to decide which urn to draw from; even though there are more black marbles than white, the chances of getting the white marble are equal to the chances of getting one of the many blacks.

The argument I've been making lately seems to be roughly this: if you close your eyes and someone selects a marble and places it in your hand, and if you know there are more black marbles than white, then you can figure it's more likely to be black. Fair enough.

But if you know they selected the marble in your hand by flipping a coin to select which urn to draw from, you should figure it's just as likely to be the one white as any of the blacks.

That looks like trouble for the thirder position, but it's missing the repetition, and missing SB's uncertainty about her own state. It's not just that she knows someone's put a marble in her hand, she knows they'll do it more than once for tails. So SB is justified in wondering, who is this person with a marble in her hand? Who is she more likely to be?

I think.

Two urns, one with a single white marble, one with many black marbles; you flip a coin to decide which urn to draw from; even though there are more black marbles than white, the chances of getting the white marble are equal to the chances of getting one of the many blacks. — Srap Tasmaner

:up:

The argument I've been making lately seems to be roughly this: if you close your eyes and someone selects a marble and places it in your hand, and if you know there are more black marbles than white, then you can figure it's more likely to be black. Fair enough. — Srap Tasmaner

:up:

But if you know they selected the marble in your hand by flipping a coin to select which urn to draw from, you should figure it's just as likely to be the one white as any of the blacks. — Srap Tasmaner

:up:

This works with Heads for Monday alone, Tails for Tuesday alone and no Wednesday. If the chance of the coin being heads depended upon the number of marbles in the jar, we're in a closer situation to OP's SB, I think.

and missing SB's uncertainty about her own state. — Srap Tasmaner

Indeed. What I'm imagining the halfer position as is that the probability of flipping a coin is always half, and that sleeping beauty's state of awakeness has no correlation with the coin. Whereas you know it does, as the "sampling day" of SB's report depends upon the coin flip. An analogy there might be asking shoppers on a Saturday what day they shop for longest on. Asking the question that way, itself, influences the responses.

I am also confused now. God damnit. I fear I answered the wrong question.

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?

So what if the coin toss doesn’t happen until after the Monday interview? Does that affect your answer? — Michael

It wouldn't in this case I think. I'm going to commit now to writing 1 if my random number generator spits out 1, and 0 if it spits out 0. It spat out 1. If I'd computed it earlier, I would've written the same thing.

More extremely, if someone wrote down the study plan for heads and tails on specific weeks, and coin flipped them, it would behave exactly the same as if they flipped them at the appropriate times in the experiment.

It would be different if SB knew the "sampling day" AND the "sampling day" informed what the RNG did. In that case I'd need a prewritten plan of SB's random responses to the different sampling days too.

Consider what the halfer says next with my marble analogy: which marble-holding-self this is was selected by a coin toss. That's the same as saying whether the marble I have is white or black was determined by a coin toss.

But that's not good enough, because there's more than one black-marble-holding momentary self. There's just no way to select from more than two moments this might be with a single coin toss.

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.

$\begin{array} {|r|r|}\hline Flip & Day & Status \\ \hline Heads & Monday & Awake \\ \hline Heads & Tuesday & Asleep \\ \hline Tails & Monday & Awake \\ \hline Tails & Tuesday & Awake \\ \hline \end{array}$

I think that's all possible outcomes. You can calculate the probabilities by the proportion of their occurrence in the table. You can also subset to calculate conditional probabilities.

From that you can say P(Heads)=0.5, P(Awake)=3/4, P(Awake|Heads)=0.5, P(Heads|Awake)=1/3 .

It's interesting that the principle of indifference prior on 3 interview days (what I did before) gives 1/3, which is the same as the frequency table here. Though my previous calculation was silly as the logic which ties Day to Flip in the 2 interview case also ties Days to Flip in my shite 3 day interpretation. The frequency table calculation there gives 2/5 . I've flagged my previous attempt as misinformation.

So I'm still a thirder, even though my previous calculation is wrong lol.

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:

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

MacTaggart's remarks concerning the A and B series are relevant here. The premises of the SB paradox are tenseless and so refer only to the B series , whereas Sleeping Beauty's question is tensed and so refers to her ability to correlate her A series to her B series.

The problem's premises can be written

$\begin{array} {|r|r|}\hline Sample\ Space & Awoken\ Mon & Awoken\ Tue & Joint\ Probability \\ \hline Heads & True & False & 1/2 \\ \hline Tails & True& True & 1/2 \\ \hline \end{array}$

According to this parameterisation, it doesn't make sense to ask SB "what Day is Today?" for "Today" isn't a random variable of the sample-space. (Thirders implicitly ask this question). But all that can be talked about, according to this parameterisation, is the state of the coin and whether SB is awoken on both monday and tuesday, and not "what day is today?".

To ring home the point, suppose that if the coin lands tails then the experimenters change SB's location from London on Monday to Paris on Tuesday. in which case, we can eliminate the notions of time and date from the premises of the the problem, by changing "Monday" to "London" and "Tuesday" to Paris.


Recall that MacTaggart decided the A series to be "unreal" and contradictory. From the B series point of view, "today" is subjective and refers to SB's mental state that comprises a variable that is separate and independent of the "calendar" events in the B series, such as wakening a subject up in London and then in Paris. Under the assumption of SB undergoing amnesia however, her mental state is uncorrelated with her understanding of B series events, and hence uncorrelated with the state of the coin.

According to this parameterisation, it doesn't make sense to ask SB "what Day is Today?" for "Today" isn't a random variable of the sample-space. (Thirders implicitly ask this question). But all that can be talked about, according to this parameterisation, is the state of the coin and whether SB is awoken on both monday and tuesday, and not "what day is today?". — sime

I see. This one is clear. I'll have a think about it.

Isn't the confusion here in the ambiguity of the question put forward to sleeping beauty? Are not the two camps interpreting the question put to SB differently?

Camp 1 - The probability that the coin landed on heads. This is 1/2 for a fair coin regardless of anything else.

Camp 2 - The probability that the coin is showing heads on the day she awakes. This is 1/3 as it is a combination of the probability of the coin and the probability of her waking up. 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.

So perhaps SB should ask the experimenters to clarify what probability they are after, exactly.

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.

it isn't a non sequitur. The probability that see sees tails at the point she wakes up is partly dependent on how often she wakes up for each outcome.

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?

It is more likely that you wake up and not have got heads 100 times in a row. The probability that heads lands 100 times in a row is in 8^-31, while the probability that you wake up to see heads 100 time in a row is 8^-25.

So the condition of you waking up to see it still changes the probability, but since you are starting off with a very unlikely outcome, you still end up with an unlikely outcome.

The key point is that SB waking up is liked conditionally to the coin toss. They are not independent of each other.

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

In both your and SB case the waking up and coin toss are linked, just yours is in a more extreme way.

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.

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.

Could you address my counter extreme scenario that I proposed?

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?

The coin being flipped and you waking up are not independent in any of the scenarios. We are looking at conditional probabilities here.

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.

1. — Michael

And yet the probability of a heads being flipped is 0.5. So you see that the probability of you seeing a heads is conditional on the head being flipped and the criteria for you waking up. In my example the criteria of waking up is 1 for heads and 0 for tails. In the SB problem it is 1 for heads and 2 for tails. The probability of you or SB seeing a heads is not 0.5 in either case

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

I do not think it is an absurd conclusion at all. The set up in your example is absurd in the practical sense of waking someone 2^101 times, but in order to explore your though experiment we suspend that absurdness.

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.

I was talking about frequency not probability.

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.

I am saying you are wrong. And in my example, where you wake up once for heads and never for tails shows that 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.

There is an above post by fdrake that sets this out more clearly in a table for the SB example.

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.

The conditional probability is dependent on the frequency in this case. 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.

1/3 of wake up event are caused by a head flip. 2/3 by a tail flip. So the conditional probability is influenced by the frequency.

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.

That is how the conditional probability works in this instance. If I am SB and I wake up, I know it could be (Heads and Monday), (Tails and Monday), (Tails and Tuesday) all with equal probability. The probability that it is heads is therefore 1/3 and tails 2/3.

There are 3 possible wake up event with equal probability, and only 1 of them is heads.

Under the assumption of SB undergoing amnesia however, her mental state is uncorrelated with her understanding of B series events, and hence uncorrelated with the state of the coin. — sime

I'm basing this on the table you've given me, I think.

I'm not sure this is true. Here are my thoughts on it. If SB being awoken was independent of the state of the coin, you'd expect the probability that she is awake on Monday, and on Tuesday to be independent of the state of the coin. By independent there I mean "joint probability is equal to the product of the marginals".

The probability that she is awake on Monday would be 1, if it were considered as a random variable by itself it would be a constant ("Awake Mon = True"), and so independent from any other random variable. The probability that she is awake on Tuesday, however, is just a relabelling of the coin flip. Specifically, the two types of events (Awake on Tuesday) and (Coinflip) are a measurable function f of each other;

Awake on Tuesday=f(Tails)
Asleep on Tuesday=f(Heads)

Which thereby means, since compositions of measurable functions are measurable (I believe) whether they are awake on Tuesday is a random variable when the coinflip is.

Explicitly terms of your table, it looks like this:

(Awake on Tuesday, Awake on Monday) = g(Tails)
(Asleep on Tuesday, Awake on Monday) = g(Heads)

Where the function maps the coin flip results to the awake/asleep bit of the table. They're not independent, one totally determines the other.

I appreciated your example, because it's getting at something which seems fundamental about the problem - our intuitions seem to suggest not every normalised measurable function (on a space of events) can be considered a random variable. But that's just what they are mathematically.

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?