
The sleeping beauty paradox - xtacy
http://stats.stackexchange.com/questions/41208/the-sleeping-beauty-paradox
======
Bartweiss
I'm having trouble finding the paradox here. As with many probability puzzles,
the problem seems to be a hidden conditional in the probabilities.

There's no change in belief happening, only two different beliefs. First is
"1/2 of the times it is flipped, the coin will be heads". Seconds is "1/3 of
the times I am awakened, the coin will be heads". The coin flip is fair, but
the decision to ask the question is biased.

This is extremely simple to demonstrate by resorting to the absurd case. If we
change the awakening ratio away from 1-2, the absurdity of saying "1/2"
becomes increasingly clear. At 1-9, tails will be the correct guess 90% of the
time the question is asked. At 0-1, heads isn't even a possible outcome on
awakening.

~~~
jules
Both answers are valid because the question is ambiguous. Asking "what is your
belief" doesn't work because there are multiple ways to interpret that. You
can clarify such situations by putting them in a game. Probabilities are about
how much you would bet on it.

Game 1: You get $10 every time you guess correctly. Best strategy: Guess tails
because if it's correct you get $20. If you guess heads correct you get only
$10. If you got $3 for guessing tails correctly and $6 for guessing heads
correctly then it would not matter what strategy you used.

Game 2: If you guess correctly then you get $10 by the end of the experiment.
Best strategy: it doesn't matter what you guess, if you guess tails correctly
you get the $10, and if you guess heads correctly you also get $10.

Thirders think the question is like game 1: every time they guess correctly
they win correctness points. Halfers think the question is like game 2: they
either guess correctly or they don't; you don't get additional correctness
points by guessing right twice.

~~~
aaron695
> Game 2: If you guess correctly then you get $10 by the end of the
> experiment.

The question is clear you get a reward each time you awaken, not at the end of
the game.

The reward is correctness (Humans like being correct = reward, else things
like 'should believe' becomes meaningless hence the question becomes
meaningless Q.E.D. it's not meaningless and correctness is a reward)

"When you are awakened, to what degree should you believe that the outcome of
the coin toss was Heads?"

~~~
Bartweiss
This fits with my understanding of what's happening here. The SO discussion is
a bit unclear as you go down the page, but the initial statement of problem up
top clearly claims that you're questioned each time you awaken.

I think you're quite right to point out that _being correct_ is the only fair
heuristic for this question. Several of the defenses of a 'halfer' belief
amount to "you ought to believe this so that your beliefs obey this type of
decision theory". Actively believing something with a (reliably) worse payout
to conform to a rule about belief seems like a terrible form of epistemology.

On a related note, are you familiar with Newcomb's Paradox and the LessWrong
debates over it? They get to a similar idea of "correctness above all".

------
jamesrom
Can someone explain the half position clearly? I don't understand how anyone
could think half is the correct answer.

Let me frame the question a different way. You are one of three volunteers in
separate rooms. I flip a coin and if it's heads I ask one volunteer (at
random) to guess the outcome. If I flip tails I ask two of the volunteers
(again, at random) to guess the outcome.

You know the rules I will follow, but you cannot tell if anyone else has been
asked before you. I open the door and ask you to guess the outcome of the
flip. What do you guess?

~~~
CamperBob2
The fact is, there _aren 't_ three volunteers, there's only one. Upon
awakening, she has no information she didn't have when she went to sleep,
which at first seems to be an unassailable argument in favor of the "halfers."

To me, the problem with the "no new information" argument is that there's
another popular paradox in which it also seems to apply, but fails
spectacularly. In the Monty Hall paradox
([https://en.wikipedia.org/wiki/Monty_Hall_problem](https://en.wikipedia.org/wiki/Monty_Hall_problem)),
you know in advance that the host is going to open a door at random and offer
you a chance to change your initial guess. You're aware that the host knows
what's behind the doors. You know that no matter what door you pick, one of
the remaining ones has a goat behind it. And you know that the omniscient host
will pick that door. So what new information do you gain when the host does
what you knew he was going to do?

One way to resolve the Monty Hall paradox is to rephrase it such that there
are a million doors and the host opens all of the remaining ones except one.
Of course it makes sense to switch doors in that scenario, since you knew that
the host would avoid opening your initially-chosen door as well as the one
with the car behind it. Your initial door has 1:1000000 odds of winning, while
the only other closed door has odds of 1/2.

So if you could find a way to transform the Sleeping Beauty problem into a
Monty Hall problem, that could be a good way to settle the argument in favor
of the "thirders."

Or, if with some logical sleight-of-hand, you could show that SB is _not_
isomorphic to MH, that could potentially be used to strengthen the no-new-
information argument made by the "halfer" side. I'm leaning that way, because
SB does not know upon awakening, and will not remember, whether the experiment
will result in her being awakened twice. _For all she knows, she 's playing
the same Monty Hall game for the second time._ It's this lack of an unbroken
timeline of conscious awareness that settles the question in favor of the
halfers.

To answer the thirders' rebuttal that SB would lose money if she were gambling
on the coin toss with any odds other than 1:3, I'd say that's a completely
different experiment. We aren't being asked to speculate on the set of events
behind multiple awakenings, just one in isolation. The lack-of-new-information
argument carries the day. Any arguments that involve repetition, clones, or
gambling fail because they are responding to a different question.

TL,DR: Don't gamble if the game involves erasing the players' memories. You
will lose a lot of money in a hurry, even if your strategy is mathematically
optimal.

~~~
URSpider94
Here's an amplification that might clarify it. If you flip heads, I'll wake
you up zero times. If you flip tails, I'll wake you up once. Now what's the
probability that you flipped tails. 100%, right? Why? The coin still has a 50%
chance of giving either heads or tails.

Here's another -- if you flip heads, I'll wake you up once. If you flip tails,
I'll wake you up a million times. When I wake you up, is it more likely that
you flipped tails or heads, remembering that you don't know which day it is.

You are right, by the way, that there is no new information -- the tendency to
answer "50%" the night before is the logical error. You are conflating the
odds of flipping heads or tails with the odds of WHETHER you flipped heads or
tails in light of the known outcome (which is not 1:1 with flipping heads or
tails).

Also, if you believe in statistics, the right answer in the continuous case
(infinite repeats) is also the right answer in the singular case (one repeat).
Your point about "there aren't three volunteers, only one" is fallacious.

~~~
CamperBob2
_Here 's an amplification that might clarify it. If you flip heads, I'll wake
you up zero times. If you flip tails, I'll wake you up once. Now what's the
probability that you flipped tails. 100%, right? Why? The coin still has a 50%
chance of giving either heads or tails._

I do like that, it's a strong argument. But it still relies on monkeying with
the participant's awareness in a way that is likely to (or in this case,
_will_ ) keep them from making the right decision over time. You use the term
"known outcome," but there isn't a known outcome if you're dead or asleep
forever.

 _Your point about "there aren't three volunteers, only one" is fallacious._

But so is treating it as any other problem (like gambling) that only converges
on an answer after multiple trials. The strength of your argument is that it
doesn't rely on answering a different question, just changing a parameter in
the question that _was_ asked. But it's an awfully important parameter.
Implicit in the question is the premise that you will be alive/aware at some
point in the future, so that you can be asked what you think.

Edit: another way to rebut your particular argument is by sticking with the
original terms of the problem, except that if the coin lands on heads, they
shoot you after you answer. That doesn't help you answer the question.

~~~
URSpider94
I think you're starting to grasp it. The key here is that if you are dead or
asleep forever, _then you are not waking up to answer the question_.
Therefore, the simple fact that you are waking up brings new information to
the situation -- you know for sure that you didn't flip heads.

~~~
Retra
But in the original question, you were already guaranteed to wake up, so how
does it give you any information?

------
mcphage
I don't see the problem in their problem:

> Whenever SB awakens, she has learned absolutely nothing she did not know
> Sunday night. What rational argument can she give, then, for stating that
> her belief in heads is now one-third and not one-half?

The argument that she can give is clear: she _might_ have learned something,
but the memory was taken from her. And she knew that they would take the
memory from her. So yes, she hasn't learned anything new, but that's a cop-
out—the problem explicitly prevents it.

I mean, from that perspective, she doesn't even need to go to sleep! They
could ask her, "when you wake up, how likely will you think that the coin was
heads?" and get the same (correct) response of 1/3\. That result is not based
on her "learning new information" (since the situation forbids it), it's based
purely on the situation as described.

~~~
IkmoIkmo
> They could ask her, "when you wake up, how likely will you think that the
> coin was heads?" and get the same (correct) response of 1/3.

Exactly. Let's stretch the example to the extreme, if you get heads you get
woken once, if you get tails you get woken 1 million times. Now say you run
this experiment 10 times and you happen to get 5 heads and 5 tails. You'd be
woken up 5 million and 5 times. And with each wakening, the chance of the coin
having been heads isn't 50%, if you'd say and guessed it was heads every time
you wouldn't get half of them right, you'd be statistically wrong about 5
million times out of the 5m and 5 times. The correct answer is 1 in a million,
and if you'd guessed tails on every awakening instead you'd be wrong on
average only once in a million.

It's this chance to be correct or incorrect in your answer when waking that
would inform the question that was posed 'When you are awakened, to what
degree should you believe that the outcome of the coin toss was Heads', which
then would be 1/3.

~~~
daxfohl
First of all, how often you're correct doesn't matter; say the tails option
wasn't "you'll be woken up a million times", but instead "you'll be asked _the
same stupid question_ a million times" (and have to answer as if you'd
forgotten the previous answer). _Of course_ you'll be wrong more often in the
tails situation.

Second, the experiment isn't run 10 times. It's run once. A single coin flip.
With a multitude of flips, you've got no idea where you are in the sequence so
it pushes your averages toward the 1:1000000. But a single coin flip is a
single coin flip, 50/50.

------
millstone
Here is a surefire way to win the lottery.

Start by picking numbers. Your chances of winning naturally are very small, so
I will make an arrangement with you. You go to sleep before the winning
numbers are read. Afterwards, if you have won, I will wake you up N times,
administering our trusty forget potion. But if you have not won, I will wake
you up only once.

As we have established in this thread, by increasing N, your chances of having
guessed the lotto numbers correctly upon awakening approach 100%.

I wake you up, you apply Bayes' Theorem, and then rejoice, for you almost
certainly have won the lottery!

------
colanderman
"[…] to what degree should you believe that the outcome of the coin toss was
Heads?" is a terribly unrigorous way to phrase this question. I suspect it's
where all the confusion sets in.

Are we adjusting the natural prior of a coin toss based on information we now
have? We have no information; 50%.

Are you placing a bet each time you are awake? You get to bet _twice_ if the
coin comes up Tails; 2:1 (33%) odds against Heads balances the tables.

Are we judging how often we'd be correct if we guessed Heads every time we
were awoken? Depends on the number of trials. With only 1 trial; 50%. With 2
trials; 42%. With infinite trials; 33%.

Nothing begets a paradox like an ill-posed question.

~~~
ikeboy
>We have no information

We do, we know that we are now in the experiment. In the case where you don't
wake up on heads at all, have you learned new information upon waking? If yes,
why don't you learn something anytime the relative probabilities differ?

~~~
colanderman
That's not new information. You know that you will be in the experiment before
the coin toss. Information does not flow acausally.

To contrast with Monty Hall, after the MC opens the first door, you _do_ have
new information: you glean information from the MC's choice, which he made
based on his knowledge of where the prize is.

In this problem, the researchers take no action visible to you after the coin
flip, and your memory is wiped before each new observation you can make. So
your knowledge about the flip's outcome is exactly the same after it occurs as
before: 50% chance.

~~~
ikeboy
If that doesn't count as new information, does waking up if you were to only
wake on tails count as new information?

If yes, what's the justification for distinguishing the two?

What happens if you are told it's your first awakening as soon as you get up?
Bayes theorem would imply you can't think the chances of heads stay the same
before and after hearing this, so at least one must not be 50%.

~~~
colanderman
I might be misunderstanding your conditions, but I would say no, waking up
only on tails would not count, unless of course you got to make your choice
_after_ the researcher indicates they're about to put you back to sleep. Then
you obviously know that the coin flip was tails! But in any case where you
have just awoken after having had your memory wiped, and you have not been
told anything by the researcher, you have been given no new info.

I think, if you are told it's your first awakening, then you still must assume
50% heads or tails: in both cases, you will always have a first awakening, so
there is no new info there. _However_ , since you will _not_ be told this in
your second awakening, if that occurs, you know immediately the coin must have
come up tails.

~~~
ikeboy
>I think, if you are told it's your first awakening, then you still must
assume 50% heads or tails: in both cases, you will always have a first
awakening, so there is no new info there.

How is this not a direct violation of Bayes theorem? The probability of
learning that it is the first awakening differs based on what the coin flip
was, so it requires an update.

~~~
colanderman
Once you know it is the first awakening, the probability of it being the first
awakening is 100%.

~~~
ikeboy
When you wake up, you don't know which awakening it is. You claim the
probability of heads is 50%.

Now, you are told it is the first awakening. The probability of learning that
if heads is twice as much as the probability of learning that if tails, so
your Bayes factor is 2:1. You therefore cannot still believe that the
probability of heads is 50%.

------
Gravityloss
Maybe it's really an argument about this:

Upon awakening:

Halfer:

    
    
        P(monday,heads)=0,5
        P(monday,tails)=0,25
        P(tuesday,tails)=0,25
    

Thirder:

    
    
        P(monday,heads)=0,33
        P(monday,tails)=0,33
        P(tuesday,tails)=0,33

~~~
Gravityloss
I think it boils down to the thirder position.

P(monday,heads)=0,33 since P(heads)=0,5 and P(monday)=0,67.

This is because we are sampling awakenings, not coin tosses or beauties.

It's like betting on a coin toss, but with a side twist that the bet is
evaluated twice (no new toss) when the coin is tails. Hence a majority of
evaluations will be on tails:

    
    
      Guess, Coin, Profit
      H,     H,    +1
      H,     T,    -2
      T,     H,    -1
      T,     T,    +2
    

Analysis for betting heads: average is +1 + -2 = -1 Betting tails: average is
-1 + +2 = +1.

I'd bet tails!

------
baddox
A similar problem is God's Coin Toss as described in Scott Aaronson's
excellent lecture series (and book) "Quantum Computing Since Democritus":

[http://www.scottaaronson.com/democritus/lec17.html](http://www.scottaaronson.com/democritus/lec17.html)

------
blahblah3
Not a paradox...here is a proof that it is 1/3

Let P(T) = probability that coin landed on tails, P(H) = probability that coin
landed on heads.

Let "1st" denote the event that it is the first time you are awaken, "2nd" the
second time.

Note that P(T|1st) = 1/2 (if you were told that this is the first time you
were woken up, it's equally likely that the coin landed heads or tails). And
of course, P(T|2nd) = 1

By the definition of conditional probability, 1/2 = P(T|1st) = P(1st|T) _P(T)
/P(1st) = (1/2)_P(T)/P(1st) Hence P(T) = P(1st) = P(T) _1 /2 + P(H) =
P(T)_1/2+(1-P(T)) = 1 - P(T)/2 -> P(T) = 2/3

------
daxfohl
Both are right: the two camps posit completely different things.

Halvers stipulate a single experiment. Thirders stipulate infinite
experiments.

It is pretty straightforward math to show these are not inconsistent, and
there are even options in between.
[https://stats.stackexchange.com/questions/41208/the-
sleeping...](https://stats.stackexchange.com/questions/41208/the-sleeping-
beauty-paradox/169582#169582)

------
personjerry
From my understanding, the answer is 1/2.

Let us assert that the probability of the coin flip being heads is 1/2.

Now, you have awoken. Regardless of if you've awoken to the Heads flip, or the
first time to the Tails flip, or the second time to the Tails flip, the
original flip's chance was still 1/2.

The possible misunderstanding comes from the fact that you will awake TWICE to
Tails, which is more than to Heads! But the thing is this is irrelevant to the
question, because you don't know whether this is the first or the second time,
and you only need to wonder whether the original flip (recall it is
probability 1/2) was heads or not. Potentially you could think that you'd be
"wrong" more often, but we are only looking at one specific instance of you
waking up in isolation, for which you have no additional information.

For example, consider waking up 1000 times if the flip is tails. Upon waking
up, do you think the probability heads becomes 1/1001? I think regardless of
waking up the first time or the 1000th time, you have no more information so
it might as well have been the first time, and hence the probability is 1/2.

~~~
hanoz
> ...Upon waking up, do you think the probability heads becomes 1/1001?

Of course, do you really think otherwise? If I promised to put a pound in your
moneybox every time you correctly guessed the coin state on each awakening,
would you really fall to sleep confident in a 50/50 guess strategy in your
1000 awakenings to 1 scenario?

~~~
lxw
the betting argument is not the same as what is being discussed.

to a halver, the likelihood of heads and tails are still each 1/2, but the
expected winnings of betting tails would of course be larger.

it would be like saying i'll flip a coin, and if you guess correctly that it's
tails i'll give you a million dollars. obviously i will guess that it's tails,
even though i still believe each outcome is equally likely.

------
fenomas
It seems to me that the paradox/confusion here comes from asking someone to
consider a distribution over a bound variable - i.e. you are asked to examine
the odds of X happening in a situation where there are already side effects of
whether or not X happened.

In this sense, it reminds me of the puzzle where a man gives you a choice
between two envelopes, one of which is specified to contain twice as much
money as the other, with the paradox centering on a bystander's argument that
you should then switch envelopes, since the other one must have either half or
twice your value, giving you a 1.25x higher expected value for switching.

As I understand it, in both cases the "traditional" solution to the problem is
to recognize that probability doesn't work that way, and you can't consider
distributions over bound variables, but the more interesting solution is to
rephrase things in Bayesian terms, in which case the analysis is reasonably
straightforward.

I'm a dabbler though; experts please tell me if I'm spouting gibberish.

------
ucaetano
You're flipping a fair coin again and again. Every time you flip heads, you
add 1 black ball to a (initially empty) bag. Every time you flip tails, you
add 2 red balls to the same bag bag.

After a large number of flips, you pick a ball randomly from the bag.

What are the odds that the ball was added when a heads was flipped?

What are the odds that the ball is black?

~~~
millstone
The question specified one coin flip. In your thought experiment, after one
flip, the answer to both your questions is 50%.

------
staz
I'm a layman in probability but isn't it the same thing as the Inspection
paradox? [http://allendowney.blogspot.be/2015/08/the-inspection-
parado...](http://allendowney.blogspot.be/2015/08/the-inspection-paradox-is-
everywhere.html)

------
jsnell
I remember reading rec.puzzles occasionally in the day when the Sleeping
Beauty question was first posed. Coming back after not reading the group was
for a couple of weeks was incredibly confused because it was full of minor
variants of this single question constructed by people to support their
position. I'd not only missed the original question and was thus lacking the
context completely, but also had missed the megathread that followed and
didn't understand the political undercurrents which would have been obvious if
I'd only known who was halfer and who a thirder. It was just crazy.

Here's a great account of how it unfolded:
[http://www.maproom.co.uk/sb.html](http://www.maproom.co.uk/sb.html)

------
acchow
This reminds me of the Monty Hall problem, but in reverse - you were playing
the Monty Hall game and just won the car! You forgot whether or not you
switched doors during the second step - what is the probability you switched
doors?

~~~
Bartweiss
There's actually a strong analogy to an extended Monty Hall here.
Specifically:

1\. Monty offers the challenge, and I pick door B.

2\. Monty opens door C, showing a goat.

3\. Monty asks if I want to switch. My odds of winning are now P(A) = 2/3,
P(B) = 1/3.

4\. My mom flips on the TV to see me play, but has missed my answers and only
sees the state of the doors. What are her odds of guessing right?

Result: my mom's odds of winning are 1/2, even though the odds of a given door
winning are not. There's a bias between the two doors, but her perspective is
neutral - she's guessing which door has better odds, and that guess is
unbiased.

As you say, this question is the reverse: the odds on the coin flip are
unbiased, but my odds of being asked the question are biased towards one of
the two outcomes.

~~~
CamperBob2
See my other post; I actually think that a Monty Hall comparison argues in
favor of an unbiased 1:2 guess, because of how the terms of the two problems
differ.

In the SB question, your odds of being asked the question are 1:1. You just
don't know how many times you'll be asked. You didn't know the night before
(and neither did the researchers), and you still don't know, even though the
researchers now do.

In the MH game, you give the host some new information when you make your
initial choice. He already knew what door _not_ to open, but now he knows what
door he _must_ open, and the rules require him to communicate that to you.
That's when you receive the new information (as you point out with the example
of your mom walking into the room).

In the SB problem, you don't get any new information before the question is
asked, including whether or not you're going to be put back to sleep. So the
only answer you can rationally give is 1:2.

If the researchers used a d20 instead of a coin to determine how many times to
wake you up, you could safely guess that any given awakening wasn't your first
or your last. But you still can't give any answer about the number on the die,
other than a random guess from 1-20. You need to store some information for
later recall, and they're not letting you do that.

------
JohnLeTigre
I'm not sure this is a paradox. There are 2 expected outcomes. All the added
outcomes rely on the experimenter to disrespect the set rules.

This is a conditional probability with a hard to predict condition (ie. human
factor) portrayed as a non-conditional probability. This looks more like a bad
representation of a problem rather than a paradox.

This reminds me of my youth. Regardless of the project I was trying to
accomplish, as soon as my little brother got involved, all bets where off. He
was a hard to predict little bugger.

------
nicholas73
Even if you get awakened 99 times with Tails, you have equal probability going
down the Heads (1 awakening) or Tails (1/99 awakenings) paths.

So if you awaken with no knowledge of other awakenings, you are equally likely
to be on either path, with the coin being Heads or Tails.

You are NOT equally likely to guess that your awakening was due to Heads or
Tails however, and that's the paradox.

~~~
ikeboy
This is basically just asserting SSA.

One problem with the halfer claim is that if you learn it is the first
awakening, you then place much more probability on it landing heads (basic
Bayes theorem application), even when the coin is yet to be flipped. How would
you resolve that?

------
drblast
I'm not sure I understand this given the way the question is posed.

We're doing a single experiment, and I am put to sleep without remembering
either once or twice, and then awaken (potentially a third time) after that?

Or could this keep going indefinitely until heads comes up?

And am I guessing each time I'm woken, or only after the final time?

------
stephengillie
How do the researchers get you back to sleep for the second (tails) awakening?

You're woken once or twice and given the potion. Or you're woken and given the
potion each time. This riddle wants it both ways, and that is part of the
problem with it.

~~~
ikeboy
You're only given the potion right before going to sleep, after you need to
state your probability estimates.

