## Sleeping Beauty Problem

• 5.9k

I'll analyse that case if you can describe it very specifically. Like in the OP.
• 14.2k
I'll analyse that case if you can describe it very specifically. Like in the OP.

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.
• 5.9k

Can you write me it in the style of the OP? No variables, just when the coin flips happen, when sleeping beauty would wake up etc.
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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.
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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.

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.
• 5.9k
@Michael - you ruined my mind again god damnit.

Edit: now I'm a halfer. Damnit it's late.
• 14.2k
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.

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.

You're welcome. ;)
• 302

Can this not be experimentally validated using a simulation? Write a computer program simulating SB and the experiment. Run the simulation 1 million times. Each time SB wakes up make a note of whether the coin was seen on heads or tails.

Will the number of heads and tails seen on wake up be 1/2 and 1/2 or 1/3 and 2/3?

Surely someone must have thought of doing this? I might have a search.
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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.
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What should SB's credence be that this is her first (or second) interview?

Any thoughts?
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The simplest "experiment" is just to imagine yourself in Sleeping Beauty's shoes.

Wasn't that rather the Cinderella problem?

You're inviting us to imagine ourselves in Sleeping Beauty's shoes to support the halfer position. However, there are other ways to imagine her situation that can be argued to support the thirder position. Consider the following thought experiment:

Suppose we introduce a character, Sue, who signs up for a lottery study. She joins a waitlist where participants are selected one or two at a time based on a coin toss: two participants when it lands heads and one when it lands tails. Upon being selected, they're given the option to buy a ticket for $100 that could potentially be worth$180 if the coin had landed heads (or else is worth nothing).

The expected value of the ticket, and whether Sue should purchase it, depends on her credence about the coin toss. If Sue holds the halfer position, her credence is 1/2, and the expected value of the ticket is $90. Hence, she shouldn't buy it. However, if Sue holds the thirder position, her credence in the proposition that the coin landed heads is 2/3, making the ticket's expected value$120. Hence, she should buy it.

Sue could argue for the thirder position as follows: if she has been selected from the waiting list, it is twice as likely that she has been chosen (together with another participant) as a result of the coin landing heads. As a frequent participant in the study, Sue would find, over time, that she profits if she always buys the ticket (an average net gain of $20 per participation), which corroborates the thirder position. 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. The pitfall of the 'imagine-yourself-in-her-shoes' argument lies in conflating Sue's perspective with the experimenter's by focusing only on Sue's situation before the coin toss. Eventually, everyone on the waitlist will be selected, just as Sleeping Beauty is guaranteed to be awoken at least once. Her credence that the coin will land heads is 1/2 before being put to sleep, and the same is true for the study participants before they're selected. However, once the coin has been tossed and they've been chosen, their credence about the value of their tickets being$180 (and that the coin landed on heads) should be updated to 2/3. The same applies to Sue's credence that her current awakening was due to a coin landing heads up.
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You're inviting us to imagine ourselves in Sleeping Beauty's shoes to support the halfer position.

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.

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.
• 2.3k
This is a follow-up to my previous post.

How do you condition on such a thing? What values do you place into Bayes' theorem?

In the case of Sue's selection to participate in the lottery study, we have

Since on each fair coin toss, 1.5 participants are being selected on average and when the coin lands on heads 2 participants are selected, P(Selected|Heads) / P(Selected) is 2/1.5 = 4/3.

Likewise, in the case of Sleeping Beauty

• 14.2k
I think you numbers there are wrong. See this.
• 14.2k
P(Selected|Heads) / P(Selected) is 2/1.5 = 4/3.

Also this makes no sense. You can't have a probability of 2.
• 2.3k
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.

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 it more likely that a randomly selected awakening is the result of a coin having landed heads. When she if afforded the opportunity to make one singe bet on any given awakening, her expected value when making this bet is conditioned on the probability that this particular awakening is the result of the coin having landed heads. Do you agree that her expected value for this single bet (in my scenario) is $120? If she would rather judge that the probability for the coin having landed heads is 1/2, she should expect the expected value of her bet to be$90 and would be rationally justified to decline waging \$100.
• 14.2k
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.

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

This is not a probability. It's a ratio of probabilities that I have expressed as a ratio of corresponding frequencies. 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.
• 14.2k
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.

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.
• 2.3k
I think you numbers there are wrong. See this.

In the quoted post you say: "P(Awake|Heads) is just the prior probability that she will be woken up if the coin lands heads"

I think my lottery study analogy suggests a better interpretation of the P(Awoken|Heads)/P(Awoken) Bayesian updating factor. But I must go to sleep now. Thanks for engaging! I'll be back with equal probability on one of my next two awakenings.
• 14.2k
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.
• 5.9k
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

The question which has been eating me is "What is the probability of the day being Tuesday?". I think it's necessary to be able to answer that question for the thirder position. But I've not found a way of doing it yet that makes much sense. Though I'm sure there is a way!
• 14.2k
What is the probability of the day being Tuesday?

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.
• 2.3k
(I woke up early)

The question which has been eating me is "What is the probability of the day being Tuesday?". I think it's necessary to be able to answer that question for the thirder position. But I've not found a way of doing it yet that makes much sense. Though I'm sure there is a way!

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 the day being Tuesday is 1/2. The updated probability therefore is P(Tuesday|Awoken) = (2/3)*(1/2) = 1/3, as expected.
• 14.2k
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.

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.
• 5.9k

I'll illustrate my reasoning about the days.

There's an issue in calculating the marginal distribution over the days for the thirder. I reasoned something like this - if you're SB, you can wake up once or twice per week. If you've woken up once per week, it seems to make sense to say that probability that you woke up on a Monday is 1. If you've woken up twice per week... Well what's the probability that you woke up on Monday? You'd wake up "half the time" on Monday and "half the time" on Tuesday, but those "half the times" would then need to apply to the case of the flip being tails.

That would give the probability of monday as P(monday and heads) + P(monday and tails), which equals P(monday|heads)P(heads)+P(monday|tails)P(tails).

P(heads) and P(tails) for the fair coin would be 0.5, as in the prior. But then I'm needing to calculate P(monday|heads). I could say that's 1. P(monday|tails) - I'd say that's 0.5. Which would give the probability of Monday as 0.5+0.5*0.5=0.75?

P(Monday)+P(Tuesday) I think should sum to 1, since they're mutually exclusive and exhaustive events. So P(Tuesday) should be 0.25.

Marginalising over awakenings and days to get a posterior probability of heads given awakening should behave like marginalising over the number of awakenings per week. As @Pierre-Normand points out, P(Awoken) (whatever event or random variable Awoken is) should look like 0.75 for the thirder position. But I'm also thinking it looks like P(Monday) is 0.75 after you take into account the impact heads and tails have on the day frequencies.

Which isn't to say P(Monday) = 0.75, It's to say something funny goes on, as saying "The probability that I'm awoken on a Monday is equal to the probability that I'm awoken at all" makes little sense.
• 2.3k
I don't think it makes sense to say P(Awake) = 3/4. P(Awake) is just the probability that she will be woken up, which is 1.

But I'm not saying that. What I'm saying is that she is being awoken every Mondays and she is awoken half the time on Tuesdays. So, on average, on a random day, she is being awoken 3/4 times. I am then using Bayes' theorem to deduce the probability of a random awakening having occurred on a Tuesday. But the result was rather trivial.
• 14.2k
So how would your reasoning work for this situation?

Let's say that there are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.

My reasoning is that P(Awake) = 1/2 given that there are 6 possible outcomes and I will be awake if one of these is true:

1. Heads and I am 1
2. Tails and I am 2
3. Tails and I am 3

My reasoning is that P(Awake | Heads) = 1/3 given that if it is heads I will only be awake if I am number 1.

This gives the correct Bayes' theroem:

\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over3}*{1\over2}}\over{1\over2}}\\&={1\over3}\end{aligned}
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I am then using Bayes' theorem to deduce the probability of a random awakening having occurred on a Tuesday.

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, 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. The experiment doesn't work by repeating the experiment several times, randomly selecting an interview from the set of all interviews, and then dropping Sleeping Beauty into it.

This is most clear with my extreme example of 2101 interviews following a coin toss of 100 heads in a row. Any random interview selected from the set of all interviews is most likely to have followed 100 heads in a row. But when we just run the experiment once, it is most likely that the coin didn't land 100 heads in a row, and so Sleeping Beauty's credence should only reflect this fact.
• 2.3k
My reasoning is that P(Awake) = 0.5 given that there are 6 possible outcomes and I will be awake if one of these is true:

1. Heads and I am 1
2. Tails and I am 2
3. Tails and I am 3

If you mean P(Awake) to refer to the probability of your being awakened at all (on at least one day) then P(Awake) is indeed 0.5. But in that case P(Awake|Heads) should, consistently with this interpretation, refer to your being awakened at all conditioned on the case where the coin landed heads. This is (1/3+2/3)/2 = 0.5
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