The Most Controversial Problem I’ve Seen

The Sleeping Beauty Problem is a highly debated problem in probability theory. Although the sleeping beauty problem simulation is an easy situation to describe, the answer is very controversial. (i.e What is the probability the coin flip is heads?)

In this video I'll give you my take on the Sleeping Beauty Problem and solution.

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#math #sleepingbeauty #probability

Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

[BriTheMathGuy]

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[patriciahawthorn]

What is the probability that Sleeping Beauty is kissed by a handsome Prince, marries him, and then is HAPPY after marrying a man who is a TOTAL STRANGER.

[harshsinha]

If she says that it was heads everytime she's woken up then she'll only be correct a third of the time.

[Zxymr]

Essentially:
- The probability that you would observe Heads when flipping the coin is 1/2 (i.e. the coin flip is the event).
- The probability that the coin is on Heads at some point SB is questioned is 1/3 (i.e. the questioning is the event).
It just depends on what you mean by "probability".

[BriTheMathGuy]

I feel like every time I think about this problem I go back and forth about what the solution "should" be. What do *you* think?!

[SlimThrull]

Well, I thought it was best to test this. So I set up a computer program to give me the answer. Turns out it's 42.

[andrewgiorgi]

I feel like the issue is here is that P(Sleeping Beauty Guesses Talk Correctly)≠P(Tails). It's two different probabilities that are being measured. It's still 50/50 for the coin flip, but her guess has a 33/66 split

[Jotakumon]

Imagine your roommate is playing a video game. And he's grinding for some experience, or loot, or whatever. And he always enters the same door, and that door gives him either of two dungeons to explore with equal change. The first dungeon consists of only three rooms. The second dungeon consists of 100 rooms. The rooms are indistinguishable to you. It takes about a minute to finish a dungeon, sometimes two, sometimes 10.

You now go outside and take a walk. You completely lost track of time, and after a while, you just feel tired and go home. You come back and your roommate is still grinding through those dungeons. You see him in a room, and before you even think about calculating how long you were gone, etc. you just ask him what dungeon he's crawling through right now, number 1 or number 2. What do you think the answer will be?

This pretty much should give you the intuition behind the problem without considering some strange hypotheticals. It's true that your friend would be transported into either dungeon with a chance of 1/2 at the start of a new dungeon, but once he's in that second really long dungeon, he will stay there for a while. If you had to guess where he is right now, it would be the second one, and maths would verify that.

Each dungeon is equivalent to each outcome of the coin flip, and the number of rooms in the dungeon are equivalent to the number of days that the princess could be woken up in. You entering the room and wondering where he is right now is equivalent to wondering what outcome the coin flip had.

A cool way of calculating this (not that the calculation in the video was wrong, it was actually really clever), is to take Markov chains to simulate doing the experiment over and over again, as if the princess was put to sleep for the rest of the week and being woken up again on Monday (and possibly Tuesday) for infinitely many weeks. After all, a probability is meant to represent the proportion that of successes if an experiment is repeated often enough.

Each state is the pair of coin flip and a day. So you have the three states Mh, Mt, Tt for Monday-heads, Monday-tails, and Tuesday-tails. At states Mh and Tt there is each an arrow with probability 1/2 going to Mh and Mt each to simulate the start of a new week, from Mt there is an arrow to Tt with probability 1 as throwing tails on Monday inevitably leads to the second Tuesday where tails was wrong that week. Now you calculate the absorption states (or whatever they call in the case where they're not really absorption states) and you obtain 1/3 for each. This means that in the long term you'll spend a third of a time in each combination, and since two of those combinations (Mt and Tt) are tails, the probability that it was tails given that you get woken up at any point in this experiment is 2/3.

This Markov chain also easily shows that if the princess is woken up n days in a row when it's heads and m days in a row when it's tails (maybe ignore that the week only has 7 days), then the probability that it heads was thrown if she wakes up is n/(n+m). Same with the video game dungeon analogy.

[Jop_pop]

I think you're right. I came to the exact same conclusion even though at first I thought it was 1/2.

[gardlanghammer]

Damn. Im never going to flip a coin on a monday ever again.

[MuffinsAPlenty]

I haven't thought about this particular problem yet, but I like what you mentioned at the end! Much confusion can be cleared up when we think of these probability paradoxes not as "how likely is something to be true", but rather as, "based on the information given, what is your best guess of how likely something is to be true in this scenario?" These are all about the _information_ someone has available to them!

[davidguthary]

I say that the probability of the coin having been heads is still 50% even in this situation. Assuming she is fully informed about the nature of the experiment, Sleeping Beauty will realize that the experiment continuing to Tuesday is contingent on the coin being tails, not the other way around. As the day she is woken up does not impact the outcome of the coin flip, the probability that the coin was heads is still 1/2. If she were to be asked the probability that the date is Monday, she would then deduce that, as her being awoken on Monday and Tuesday given the coin was tails are indistinguishable to her, these two incidents would have the same probability, and that these two probabilities must add to 1/2. As she will only be awoken on Tuesday if the coin was tails, the probability that it is Tuesday is 1/4, and the probability that it is Monday is 3/4.

[ScorpioneOrzion]

there is no coin flip between Monday and Tuesday so the coin is only flipped once and the probability for heads is 1/2.
in all 3 situations you're considering the coin flip and the current day, not what the coin flip alone was.

[angelmendez-rivera]

The Sleeping Beauty Problem is not a probability theory question, contrary to how it was presented in the video. The Sleeping Beauty Problem is an epistemic question, and what the problem asks Sleeping Beauty is "What is your degree of belief that the coin is in heads?"

Given that the coin is a fair coin, Sleeping Beauty can conclude that P(Heads) = P(Tails) = 1/2.

Sleeping Beauty also knows that P(Monday | Tails) = P(Tuesday | Tails) = 1/2, P(Monday | Heads) = 1, P(Tuesday | Heads) = 0, because she is told the procedures of the experiment before the experiment, and the erasing of her memory only erases the memory of her awakening, not of the experiment procedures.

Calculating the other probabilities is trivial based on the definition of conditional probability, and on Bayes' theorem. As for the calculation, P(Monday & Tails) = P(Monday | Tails)·P(Tails) = 1/2·1/2 = 1/4, P(Tuesday & Tails) = P(Tuesday | Tails)·P(Tails) = 1/2·1/2 = 1/4, P(Monday & Heads) = P(Monday | Heads)·P(Heads) = 1·1/2 = 1/2, P(Tuesday & Heads) = P(Tuesday | Heads)·P(Heads) = 0·1/2 = 0. Given that P(Heads & Tails) = P(Monday & Tuesday) = 0, P(Something) = P(Heads or Tails) = P[(Monday & Heads) or (Tuesday & Heads) or (Monday & Tails) or (Tuesday & Tails)] = P(Monday & Heads) + P(Tuesday & Heads) + P(Monday & Tails) + P(Monday & Heads) = 1/2 + 0 + 1/4 + 1/4 = 1. There is no contradiction here, the video is simply mistaken in the claim that P(Monday & Heads) = P(Monday & Tails), because P(Heads | Monday) = P(Monday | Heads)·P(Heads)/P(Monday) = 1·1/2/P(Monday) = 1/[2·P(Monday)], while P(Tails | Monday) = P(Monday | Tails)·P(Tails)/P(Monday) = 1/2·1/2/P(Monday) = 1/[4·P(Monday)] = P(Heads | Monday)/2.

However, as the question is not asking for a probability, these calculations do not solve the problem, because the point of the problem is to pose a question concerning how we use these probabilities as information in order to conclude that the coin is in heads, not calculating the probabilities themselves. For example, there is debate as to whether one should use P(Heads) = 1/2 as the basis of degree of belief, or one should use an expectation value provided by the average probability of non-null events, meaning [P(Monday & Heads) + P(Monday & Tails) + P(Tuesday & Tails)]/3 = 1/3, or just the average probability, given by 1/4. This is what the Sleeping Beauty Problem really is about.

[Krunschy]

2:40 Me when I get a wrong answer on the test, but fill it in regardless.

[shkabeeenxd]

the coin flip literally has a 1/2 chance of landing heads, and nothing changes that, forthermore, sleeping beauty will experience waking up on any day the same, so there is nothing she can use to potentially deduce something

[coreyostrander]

This reminded me of the math problem in 21. Id say sleeping beauty's correct answer would have to be 50%.

[Eknoma]

The question really is about how often you would be right if you always say heads.

Half the time, it is heads, so you are woken up once, and are correct in guessing heads. Half the time, it is tail, so you are woken twice, and are wrong.

So in total you would be right 1 in 3 times you are woken up, thus the probability of 1/3

[TanmaiKhanna]

Your math doesn't consider the fact that (Monday-tails) and (Tuesday-tails) are not independent events. If one occurs, the other HAS to/to have occurred. I don't think you can apply the conditional probability formula like this here, based on what I've learnt.

[ameerunbegum]

Give the sleeping beauty a kiss and ask her yourself