A Sleeping Beauty paradox

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?

29 Responses to A Sleeping Beauty paradox

  1. Mathaniel says:

    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.

  2. 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…

  3. Julia Galef says:

    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


    • Scott says:

      “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.

  4. Andrew says:

    (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.

    • Andrew T says:

      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.

  5. 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.

    • Max says:

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

  6. 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!

  7. Max says:

    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.

  8. David Schreier says:

    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.

  9. 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.

  10. siodine says:

    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.)

  11. edkins says:

    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.

    • Katja Grace says:

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

    • Mathaniel says:

      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.

  12. JP says:

    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.

  13. edkins says:

    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.

    • Katja Grace says:

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

      • edkins says:

        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.

  14. 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).


    “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)

  15. Scott says:

    “So, is that the answer?”

    What exactly is the question?

  16. David says:

    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 😉 ….


    • edkins says:

      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:


      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?

  17. Zak Gillman says:

    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?

  18. Stephen Tashiro says:

    This bogs sets the problem as solving for P(coin landed heads | SB is awake). So we sidestep the question posed by the current Wikipedia article on this problem, which states that Sleeping Beauty is asked “What is your credence that the coin landed heads”. and this implicitly leads to considering some wagering scenario. Good! Let’s sidestep it.

    The fact that both the “thirder” and the “halfer” can be derived from the given information (with differing additional assumptions) indicates that P(coin landed heads | SB is awake) is not a well defined quantity.

    Both the “thirder” and “halfer”answer have probability models that are consistent with all the information explicitly stated in the problem. The ambiguity arises because the problem only informs us about a stochastic process that generates sequences of events. It doesn’t define any distribution for selecting one of those events at random to represent the situation that applies when “Sleeping Beauty is awakened”.

    A “halfer” model for selecting the event is 1) Randomly select from the events H,T giving each an 0.5 probability of being selected. 2) If H is selected, choose the event (heads, Monday). If the event T is selected then randomly select one of the two events (tails, Monday), (tails, Tuesday) given each a probability of 0.5 of being selected.

    “Thirders” dislike the various conditional probabilities produced by that “halfer” model, but its results don’t contradict any information given in the problem. They may contradict some assumption make by “thirders” – such as the assumption that events must be selected by probabilities proportional to their expected relative frequency in a large number of experiments.

    If gambling is introduced and Sleeping Beauty is required to state a fair price for the bet: The bet owner receives $1 if the coin landed heads” then she (whether “halfer”, “thirder’ or agnostic) should reason without considering P(coin landed heads | SB is awake). The fair price for the bet can be figured out from the specifications of the experiment, which Sleeping Beauty knows. If she always pays X for the bet, her expected expense is (1/2)X + (1/2) 2X = (3/2)X. Her expected gain is (1/2)1 + (1/2)0 = 1/2. So a fair price for the bet is 1/3.

    However, this does not imply that she believes P(coin lands heads | awake) = 1/3. The result 1/3 is computed from a betting strategy. It is not a fair price for the “pure” bet “The bet owner gets $1 if the coin landed heads”. Instead 1/3 is fair price for the bet “The bet owner gets $1 if the coin landed heads and loses twice what is paid for the bet if the coin landed tails”. That is because Sleeping Beauty (being rational) must decide on a single price for bet that applies every time the bet is offered. If the coin lands tails, the bet is offered twice.

  19. Jeff Joy says:

    There is no paradox here. This is simply a poorly stated problem as it is ambiguous as to the exact event you are attempting to measure. Is your desire to define the probability of the coin turning up heads vs tails or are you trying to define the probability of a specific result being true by measurement frequency? Clarify your question and the ambiguity disappears along with any arguments. I do understand how this attempts to explain the complexity of tackling certain problems due to measurement bias, but being aware of and accounting for that bias by clearly and unambiguously defining the question at hand, as opposed to other possible interpretations of the question, will go a long way toward avoiding such issues. As with many problems, poorly defined questions or requirements leads to improper answers or solutions. The onus is on the person defining the problem at hand to provide unambiguous clarity.

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