# A Sleeping Beauty paradox

November 10, 2011 27 Comments

Imagine that one Sunday afternoon, Sleeping Beauty is taking part in a mysterious science experiment. The experimenter tells her:

“I’m going to put you to sleep tonight, and wake you up on Monday. Then, out of your sight, I’m going to flip a fair coin. If it lands Heads, I will send you home. If it lands Tails, I’ll put you back to sleep and wake you up again on Tuesday, and then send you home. But I will also, if the coin lands Tails, administer a drug to you while you’re sleeping that will erase your memory of waking up on Monday.”

So when she wakes up, she doesn’t know what day it is, but she does know that the possibilities are:

- It’s Monday, and the coin will land either Heads or Tails.
- It’s Tuesday, and the coin landed Tails.

We can rewrite the possibilities as:

- Heads, Monday
- Tails, Monday
- Tails, Tuesday

I’d argue that since it’s a fair coin, you should place 1/2 probability on the coin being Heads and 1/2 on the coin being Tails. So the probability on (Heads, Monday) should be 1/2. I’d also argue that since Tails means she wakes up once on Monday and once on Tuesday, and since those two wakings are indistinguishable from each other, you should split the remaining 1/2 probability evenly between (Tails, Monday) and (Tails, Tuesday). So you end up with:

- Heads, Monday (P = 1/2)
- Tails, Monday (P = 1/4)
- Tails, Tuesday (P = 1/4)

So, is that the answer? It seems indisputable, right? Not so fast. There’s something troubling about this result. To see what it is, imagine that Beauty is told, upon waking, that it’s Monday. Given that information, what probability should she assign to the coin landing Heads? Well, if you look at the probabilities we’ve assigned to the three scenarios, you’ll see that *conditional on* it being Monday, Heads is twice as likely as Tails. And why is that so troubling? *Because the coin hasn’t been flipped yet.* How can Beauty claim that a fair coin is twice as likely to come up Heads as Tails?

Can you figure out what’s wrong with the reasoning in this post?

A good read if you’re interested in this topic, http://meteuphoric.wordpress.com/2011/01/09/against-the-hybrid/

Fun post.

If it is Monday, then Tuesday is not a factor and the heads-tails probabilities are 1/2-1/2. Also, the coin is not tossed on Tuesday at all, anyway. If it were Tues. the coin HAD been tails, not that it may be tails.

I feel like I must have missed something here. Prior to Sleeping Beauty being told it is Monday, her reasoning stands. As soon as she’s told it is Monday, then she has a different decision tree to evaluate, one where 50% of the time a coin is heads, and 50% tails.

Unless she’s about to question whether the coin is a fair one after a single flip, I’m not sure I get the issue here…

timswheelbarrow, you don’t even have to assume that she’s actually told it’s Monday. Just imagine Beauty lying there, not knowing what day it is, and calculating the probabilities as I did in the post. She still has to admit: “If it is Monday, then Heads is twice as likely as Tails.”

She can’t simultaneously say that P(Heads) = 1/2 and that P(Heads | Monday) = 1/2.

Heads: SB wakes up on Monday

Tails: SB wakes up on Monday and Tuesday

50% of the time its Heads (Fh)

——One Monday wake up (Wm) half the time

50% of the times its Tails (Fh)

—— One Monday wake up (Wm) half the time, One Tuesday wake up (Wt) half the time

Fh + Ft = 100%

50% * (1 Wm) + 50% * (1 Wm + 1Wt) = 1 Wm : .5 Wt

2/3 Monday wake ups

1/3 Tuesday wake ups

.

“She can’t simultaneously say that P(Heads) = 1/2 and that P(Heads | Monday) = 1/2.”

Of course she can, since P(Heads | ~Monday) = 0.

(Heads, Monday) 1/3 or all wakings.

(Tails, Monday) is 1/3 of all wakings.

(Tails, Tuesday) is 1/3 of all wakings.

If you run it 1,000,000 times there will be 500,000 waking on monday with a heads result. There will be 500,000 wakings on Monday with a tails result. And those 500,000 people will wake up on Tuesday again. That’s 1,500,000 wakings. So given that it’s monday, there is a 1/2 chance that the coin to be flipped will come up heads.

Yah, the key is to define what population you’re actually selecting from; in this case it’s “all wakings before going home”, in which case the numbers break down as Other Andrew stated.

Setting up the problem in the 1/2,1/4,1/4 way seems logical but isn’t really answering any clear question; it’s mixing P(random event) with P(random selection from a subet). Which isn’t valid/applicable since it mixes events Beauty experiences (waking up on a day) with ones she doesn’t (coin flip). I think.

I would walk through it like this:

First flip:

– if you run this scenario 100 times, then in 50 of them you’ll end up at home on Monday

– the other 50 you go back to sleep

Second flip:

– in 25 of them you will be sent home Tuesday with your memory

– in the other 25 you will be sent home Tuesday without the memory of having woken before

One can eliminate Tues with memory, because in those 25 scenarios you know what’s what, so out of the remaining 75 scenarios where you wake up and go home without being sure of what day it is, on 50 of them it’s Monday and 25 of them it’s Tuesday.

So as such when you arrive at home there’s a 2/3rds chance its Monday and 1/3rd that it is Tuesday. The key is in that elimination of the subset where you know what’s going on.

I thought there’s only one flip on Monday. If it’s tails, she goes back to sleep AND has her memory erased.

Her options are;

It’s Monday, a coin is going to be tossed

It’s Tuesday, and a coin was tossed

As it happens, we know that the chances of each are based on a fair coin toss (50:50). The timing of the coin toss doesn’t seem to be a dealbreaker here.

If it’s Tuesday, then 100% of the time, that information will be relevant to knowing the outcome of the toss, while if it is Monday, then 0% of the time will that provide useful information as to the outcome of the toss. Neither suggest that knowing what day it is within this situation will affect the likelihood of the coin toss.

This all sounds concerningly like the many nights I’ve spent at the pub talking philosophy with various philosophy phd students, but is a refreshingly new side to it (they largely deal with ethics). As per usual, I spend as much time making sure I understand the question as I do worrying about an answer.

Thanks for giving me something interesting to think about!

Heads: Monday

Tails: Monday, Tuesday

Out of every 3 times she wakes up, 2 are on Monday. Out of those, the coin is heads half the time.

Lost about 3 hours of my life on this one, and a question really wasn’t even posed here. But if there was one, maybe the best bet would be for SB to wait it out until the experimenter drops back in the room. If the good doctor heads for the IV drip, she should be sure its Monday. If not, AND she is also starving, it’s Tuesday.

The coin only gets factored once- in the written logic, it gets done twice. She wakes up Monday, and either goes home or goes to sleep, forgets, and goes home Tuesday. Since she’s always waking up on Monday, the odds are that she will- half the time she’ll wake up and go home, the other half she’ll wake up and go to sleep. In other words, she wakes up 100% of the time on Monday, and 50% of the time on Tuesday. Given M is twice T, M+T=1, P(M)=2/3 P(T)=1/3.

If that doesn’t make sense, look at it this way: In a given “run” she has the chances of waking up 2 times (M, T) or waking up 1 time (M). Out of 3 possibilities, M comes up 2 times, giving 2/3 M, 1/3 T.

I think the thought experiment is more easily understood if you replace Sleeping Beauty with yourself.

So, I know exactly how this experiment will be carried out, but I don’t know what day it is. I’m supposed give my credence for the coin having landed on heads or tails.

Knowing how the experiment is designed, I can find the probability that will lead me to the correct answer the most amount of times before the experiment is ever run. That will be my credence (1/3 every time).

If I didn’t know how the experiment was designed, my credence would have to be 1/2.

(In the version you outlined, it seems Beauty does know when it’s Tuesday and you seem to be examining the experiment from the experimenter’s perspective instead of Beauty’s. I’m assuming those are mistakes.)

Thank you so much for posting this when you did! I’m right in the middle of thinking about this kind of stuff, and this post has made me realize that I’m still more confused about this topic than I thought.

At the risk of looking silly, I think that SB should conclude:

P=1/2: reality corresponds to the “heads” timeline

P=1/2: reality corresponds to the “tails” timeline

The notion that there is a particular point in time called “today” is merely a convenient fiction. So anything involving P(Today = Monday) is undefined.

Is it impossible then for SB to learn that it is Monday?

I agree with you. Whether it is Monday or Tuesday is independent to whether the coin will be heads or tails. Rather, the result of the toss determines if SB is sent home on Mon /Tues. I wonder if this is less a statistical problem and more a philosophical issue.

This may repeat some of what has been already said, but here’s my take on this:

There are two timelines, with equal probability:

(1) Wakeup Monday (A), coin is Head

(2) Wakeup Monday (B), coin is Tail, Wakeup Tuesday (C)

If we were to repeat the experiment, say, 2N times (with 50-50 results), each of A, B or C will come up the same number of times (N). So, waking up, SB will find herself in one of three equally probable wakeup events, giving:

Probability of being in case A, B or C: 1/3 each.

Probability of being Monday: 2/3.

Probability of being in timeline (1): 1/3

Probability that the coin has been or will be Tail: 2/3. This is not paradoxical as tails are observed twice as often as heads.

Suppose that each night the bed is made with sheets of a random color. There are a very large number of colors to choose from, so the probability that it’s a particular one is, say, 0.00001.

Let Red be the name of an arbitrary color. Then:

P(Heads) = 0.5

P(Red sheets on Monday | Heads) = 0.00001

P(Red sheets on Monday | Tails) = 0.00001

P(Red sheets on Tuesday | Tails) = 0.00001

Now suppose SB wakes up and observes Red sheets. Then the intuitive notion of “Today = Monday” can be rephrased as “Red sheets on Monday”. (The probability that they’re Red on both days is so small as to make no material difference here).

Glancing at the probabilities, we see that P(Heads | Red sheets on Monday) = 0.5.

And the thing is, I think that my definition of the problem is essentially equivalent to the original. Instead of bedsheet color we could have used the temperature of SB’s coffee or the exact pattern of wrinkles on her pillow – *something* will be unique about her pattern of observations each day.

So I think we can use Julia’s original probabilities but still say that P(Heads | “today is monday”) = 0.5. We only get into a mess if we implicitly assume that “today” is a constant.

http://philsci-archive.pitt.edu/2888/ Radford Neal uses that kind of setup to argue for 1/3.

You know… you’ve got a really good point. SB might wake up, observe red sheets and update on “Red sheets on Monday or Red sheets on Tuesday”. Glancing at the probabilities again, we get P(Heads | R on M or R on T) = 1/3.

So I was still confused, because I feel about 80% sure that 1/3 is the wrong answer. Updating on irrelevant evidence shouldn’t change P(Heads).

I think we need to reason in terms of Bayesian updaters which can cope with copies of themselves being made (the SB problem seems very similar to thought experiments involving copyable brain emulations).

If Heads, everything proceeds as normal. If Tails, then SB-Sunday-evening is copied, and one copy is placed at Monday-morning, the other at Tuesday-morning. Neither is allowed to assign a probability to “I am the Monday branch”, but if they observe something (e.g. bed sheet color), they can give assign probabilities to that event happening on Monday, Tuesday or both. If one of the SB’s observes a sequence of events, she knows it’s impossible that some of them only happened only on Monday and some of them only on Tuesday.

Anyway, this comment is getting too long so I’ll think about it some more and maybe write it up properly. Thanks for the link.

You’re treating the ‘Tails, Tuesday’ condition as though it is a ‘Tails, Tuesday & ~Monday’ condition (which, assuming Beauty is being told the truth about the protocol, will never occur).

This:

“Heads, Monday (P = 1/2)

Tails, Monday (P = 1/4)

Tails, Tuesday (P = 1/4)”

Should be this:

Heads, Monday (P = 1/2)

Tails, Monday (P = 1/4)

Tails, Monday & Tuesday (P = 1/4)

Tails, ~Monday & Tuesday (P = 0)

“So, is that the answer?”

What exactly is the question?

Perhaps one lesson to draw from this is that we need to be very very careful what we mean when we use the word “probability” P. Specifically, we need to pay attention to how P is calculated, and who is doing the calculating. I’d make an analogy with relativity: if you ask for the “distance” D between two space-time locations A and B, the question is ambiguous / incomplete until you specify the reference frame of the observer, and Einstein was able to figure this out by paying close attention to how D is calculated and who is doing the calculating (observing).

So, perhaps it is the same thing with probability. Suppose you tell SB that you are going to show her the coin and want her to predict whether she will observe heads or tails. Given she is woken twice for each tails flip but only once for each heads flip (see Andrew’s comment earlier), she will observe tails 2/3 of all awakenings but heads 1/3 of all awakenings. So from her perspective “the probability” P of tails is 2/3. Yet, she knows it is a fair coin that has a probability P of producing tails 1/2 of the time – but this would be calculated in a different manner. The key is that these two probabilities are not the same, just as distance between A and B measured by one observer is not the same as that measured by another observer.

I hope that makes SOME sense and is not totally opaque ;) ….

David

Less Wrong has been mulling over a decision theory where agents don’t attempt to calculate probabilities, but are still able to make good decisions:

http://lesswrong.com/lw/334/another_attempt_to_explain_udt/

This sort of captures my current feelings towards the Sleeping Beauty problem, but it’s counterintuitive that (in toy problems at least), expected utility maximization would be a more robust notion than probability.

Is this something like what you have in mind?

I really don’t understand the overall reasoning to this post (nor the use of the word “paradox” in “Sleeping Beauty Paradox”…), as the problem seems to propose the questions “What day is it? What is my probability that I am waking up on Monday versus waking up on Tuesday?” and not the question of “What was the result of the coin flip?”

Sleeping Beauty’s experience is going to be the same regardless of the coinflip — she will be woken up and sent home. The only question is whether it happened on Monday (due to a Heads result on the coinflip) or on Tuesday (due to a Tails result and subsequent memory wipe). As the probabilities are based on the coinflip, which has P(Heads) = 50% and P(Tails) = 50%, it then translates directly to P(WokeMonday) =50% and P(WokeTuesday)=50%.

Ummm, but at some point, I’m supposed to answer your question and point out the flaw in your example, aren’t I? … … … (^_^’)

As far as I can tell, the problem is in breaking up your potential situations as “It’s Monday, and the coin will land either Heads or Tails.” and “It’s Tuesday, and the coin landed Tails.” to get the possibilities “Heads, Monday”, “Tails, Monday” and “Tails, Tuesday.”

This is not a good breakup of the possibilities, as “Tails, Tuesday” directly implies “Tails, Monday” (as “Tails, Tuesday” is conditional on “Tails, Monday” — you can’t have had a Tail on Tuesday unless the coinflip was a Tail on Monday, because there was only one coinflip).

So, when you created the dual situation of P(Tails|Monday) = 25% and P(Tails|Tuesday)=25%, there was the flaw in the logic: P(Tails|Tuesday) requires P(Tails|Monday), so really, P(Tails|Tuesday) = P(Tails|Monday) = 50%.

And so, then P(Heads|Monday) = 50% and P(Tails|Monday) = 50%, making Beauty’s knowledge moot and removing the odd problem that telling her what day it is would cause the Head to be more likely.

Anyway, to summarize (couldn’t I have just done that from the beginning?), the problem is that you created 3 possibilities (“Heads, Monday”, “Tails, Monday”, and “Tails, Tuesday”) and missed that two of the possibilties were the exact same thing. In reality, there were only two possibilities.

… … … Right?