**Update** This has been moved up from its original 29 Feb 2012 date on 23 March 2012 to allow additional comments.

**Update** This has been moved up from its original 29 Feb 2012 date again on 21 May 2012 to allow additional comments.

Are you awake? I copied and pasted this direct from Wikipedia (I know, I know):

Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details. On Sunday she is put to sleep. A fair coin is then tossed to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday and Tuesday. But when she is put to sleep again on Monday, she is given a dose of an amnesia-inducing drug that ensures she cannot remember her previous awakening. In this case, the experiment ends after she is interviewed on Tuesday.

Any time Sleeping beauty is awakened and interviewed, she is asked, “What is your credence now for the proposition that the coin landed heads?”

I interpret *credence* as *probability*. Think about the answer before reading further.

—————————————————————————-

Suppose she is awakened on Monday. She knows it is Monday, she knows that if the coin landed heads she would have been awakened. She also knows that if the coin landed tails she would have been awakened. In other words, regardless of the way the coin landed, she will be awakened on Monday.

Formally, her evidence is C = “A coin with two sides, labeled head and tail, will be flipped and only one side can show”, M = “It is Monday”, and also E = “The details of the experimental protocol.” She is asked to compute (or form):

Pr( head | C & E & M) = 1/2.

Since knowledge that it is Monday and of the experimental protocol tells her she will be awakened on Monday no matter what happens with the coin, the probability is numerically (but not logically) equivalent to Pr( H | C ), which everybody agrees is 1/2.

Now suppose T = “it is Tuesday.” Here the problem becomes ambiguous (at least, as Wikipedia has it). If the coin originally came up heads she will have been awakened Monday. There are then no words, i.e. no evidence, which tells us what she does Monday night into Tuesday. Perhaps she stays awake all Monday night celebrating Barack Obama’s defeat (we’re imagining this experiment taking place in November). Therefore, when it is Tuesday she will know that she was awakened Monday, because she will have recalled all the events which took place since her awakening; in particular, she will know she was not awakened on Tuesday.

So if queried on Tuesday about the coin she will have a different response than when she was queried on Monday. That is,

Pr( head | C & E & T & Up all night with memories) = 1,

and this will be the same if she goes to sleep on Monday but remembers being awakened on Monday, because of course she was not administered any drugs. So really all we need is her memory of what happened on Monday; she needn’t stay awake. That is,

Pr( head | C & E & T & Memories of Monday) = 1.

But if it is Tuesday and she is awakened and she hasn’t any memory of what happened on Monday, then

Pr( head | C & E & T & No memories of M ) = 0,

because she knows she was *awakened* on Tuesday and she knows she has no memory of Monday, whereas she would have those memories if the coin was heads. So she knows the coin was tails.

Since all that was too easy, I suspect the problem has been stated poorly. More evidence for this is that the Wikipedia discussion of other authors’ solutions appear to suggest that Sleeping Beauty has no idea what day it is. If that is so, her only knowledge is that she was awakened. Actually, not quite. She still knows that the day(s) she is awakened will be one of two, M or T.

We have already worked out the solutions conditional on her knowing the day. So we need to fit in the uncertainty in the day. She knows (and remember, we’re going on *her* information)

Pr( M | C & E ) = Pr( T | C & E ) = 1/2.

Here we don’t need memories, because if it’s M there aren’t any to be had, and if it’s T they’ve been wiped away (this is implicit in our new understanding of E). So what she wants is

Pr( head | C & E & M ) x Pr( M | C & E) + Pr( head | C & E & T ) x Pr( T | C & E) = 1/4.

Now Pr( head | C & E & M ) = 1/2 since if she assumes it is Monday, she knows by C & E that she will be awakened no matter what, just as above. But, Pr( head | C & E & T ) is different, since if she assumes it is Tuesday she knows that the coin must have been tails, so Pr( head | C & E & T ) = 0.

This is different than the other solutions, which were 1/3 and 1/2, so it’s possible that I have misinterpreted the experiment or that I have made a bone-headed mistake. What do you think?

Given the reference to the Absent Minded Driver problem I assume Sleeping Beauty doesn’t know if it is Monday or Tuesday.

IE a bleery eyed person is woken up and just asked whether she thinks the coin came up heads – it could be Monday with a probability 1/2 or Tueday with probability 0, but she doesn’t know which – is this the first time, or second time she’s been asked.

Doesn’t that make the problem what is P(heads|Monday or Tuesday).

Wiki says she’s put to sleep Monday night. Presumably not in the way one might put a pet to sleep.

On Monday, she’s awakened (in, fact, she’s awakened both days) there is no additional information conveyed other than the experiment ends. On either day, she would be awakened regardless of the coin outcome.

If she doesn’t know what day it is, she should answer 1/2 both days since being awakened conveys no information.

If she knows it’s Tuesday then she knows the coin landed Tails on Monday. But she’s being asked what the probability of it being heads

todayso knowing it’s Tuesday doesn’t change anything. So wouldn’t it still be 1/2?If she knows it’s Monday and the experiment ends (before she is asked) the the P(H|M) = 1.

Why do you claim that she knows Pr( M | C & E ) = Pr( T | C & E ) = 1/2? From the experiment description, isn’t it the case that Pr( M | C & E ) = 2/3 and Pr( T | C & E ) = 1/3? This then yields Pr( head | C & E ) = 1/3.

Unless someone tells her, she can not know the day. I think that this lack of knowledge is implied. I think that it is also assumed that she is unaware that she will be put to sleep again (or not) until after the interview. Therefore the set of events is Monday (H), Monday (T), and Tuesday (T). Therefore the probability is always 1/3.

I didn’t read the wikipedia article or your solution before posting my previous comment, but I still stand by it. I believe that this is just another variant of the Monty Hall problem.

This seems to be one of those problems that introduces all kinds of situational factors of no real relevance.

The only time the coin is flipped is on Monday…if it is fair it will come up H or T 50 percent of the time.

We know the subject knows a coin was flipped, but she does not necessarily know the day when the interview occurs. Therefore, all she has to work with is that the coin was flipped, and presumably, it is fair. Hence, 50 percent probability is the best than can be expected on Monday.

For her to have any greater insight she would have to know something else, such as the day and the conditions of the experiment–facts unrelated to the coin.

IF she knows the day, and the coin is fair, and the terms/protocol of the experiment, she can conclude the following:

Monday: 50 percent chance the coin was heads or tails.

Tuesday-Case 1: If she is awakened and has no recollection of yesterday, but is confronted by the experimenters she can conclude the coin did not come up heads (was tails, prompting the two-day interview) with 100 percent certainty. This presumes she knows the terms of the experiment & the day. It is possible she might not know the day until after the experiment ends sometime on Tuesday…after which she could check the day. Either way, early or late(r) on Tuesday she will know for certain what the Monday coin flip was.

Tuesday-Case 2: If she awakens and there are no experimenters (as the experiment ended Monday), she definately knows it is Tuesday and then she knows with certainty the Monday coin flip came up Tails — and she knows (or can know) this upon awakening.

Thus, she knows on Tuesday what the coin flip was with certainty. WHEN on Tuesday depends on the specifics of the experiment–how much of its process protocol she was privy to.

Briggs,

Demerits for missing the point.

I went back and re-read the problem. It isn’t clear if the coin was flipped again on Tuesday. It looks like it was only flipped once. She has no memory of previous awakenings and presumably unaware of the current day so I don’t understand what relevance there is in the number of awakenings. She could have awakened once or twice but she doesn’t know how many. I assume we are to answer given

herknowledge vs. ours. Since she has no idea how many times she has been awakened, why would her answer change between Monday and Tuesday?In my previous post it looked like this is a rather wordy version of the 1000 flips problem: A fair coin has been flipped 999 times and has landed heads each time. What is the probability it will land heads on the next toss?

A fair coin has been flipped 999 times and has landed heads each time. What is the probability it will land heads on the next toss?

100%!

A fair coin doesn’t come up head 999 times. Therefore, we should throw out that premise, and assume that you are filipping a 2 headed coin.

DAV, All,

My read was that the coin was flipped only once. And, of course, we must consider the evidence from only her point. This makes the analysis when she does know—or assumes she knows—the day relevant.

OK then. Assuming she’s aware that there are three situations:

1) S(Mon & heads)

2) S(Mon & tails)

3) S(Tue & tails)

From her viewpoint she could be in any one of the three situations. I don’t think there’s is any reason for her to assume anything other than all are equally probable. So I would go with 1/3 as heads appear in only 1 of the three.

Taking a slightly different approach form yours: Assume instead we draw from three slips of paper labeled with each of the situations.

There are two chances to draw Monday so

Pr(Mon|C & E) = 2/3

Pr(Tues|C & E) = 1/3

Pr(heads|C & E) = 1/3

I think Eric pointed out the Pr(Day|C & E)

How does Beauty know what day it is? When she wakes up, her last waking memory is going to bed on Sunday night.

Beauty must guess wheter it is Monday or whether it is Tuesday, and if it Monday, how did the coin flip turn out.

Given E = {all the details of the experiment}, Sleeping Beauty knows that she will be awakened on Monday no matter what happens with the coin, and on Tuesday only when the coin comes up tails. My intuition tells me that P (Monday| E) should be greater than P(Tuesday | E). Just my intuition.

Mr. Briggs,

My coin has fallen a tail. Sleeping Beauty, my friend, will stay awake on Tuesday, November 6, 2012, to hear Obamaâ€™s victory speech. Ha.

JH, it seems the outcome of our coin toss is yet more evidence of the impending demise of the world at the end of December.

I’m thinking it’s not the coin, it’s the waking up that’s driving the probability here.* Sure the coin comes up heads with a probability of 1/2, but that doesn’t mean the probability of being awakened on a certain day is 1/2:

Heads: She wakes up once.

Tails: She wakes up twice.

So, there are three possible outcomes from her perspective, she wakes up on Monday and the coin is heads, she wakes up on Monday and the coin is tails, and she wakes up on Tuesday and the coin was tails: Pr(Wake up| Heads) = 1/3, Pr(Wake up| Tails) = 2/3.

*I’m assuming she only gets put back to sleep if the coin is tails on Monday, if it’s heads the experiment is over and she goes on her merry way. If she were put back to sleep and reawakened Tuesday regardless of the outcome (what I initially thought the experiment said) the probability of heads should be 1/2.

Or, to state it another way:

There are two possible outcomes for the coin toss: Pr(Heads) = 1/2

There are three possible states of the world when she wakes up: (Monday, Heads), (Monday, Tails), (Tuesday, Tails), so the probability that it is Monday when she wakes up: Pr(Monday) = 2/3

Since she will get awakened both on Monday and on Tuesday if the coin flip was tails, she has to guess the state of the world, not just the result of the coin flip: What’s the probability of it being Monday AND the coin is heads = 2/3*1/2 = 1/3.

Think of a related question: When she wakes up, what is the probability of it being Tuesday? Is that 1/2?

I am going with the position that only 1/2 makes sense.

If am am sleeping beauty I am not getting involved in your experiment unless I know that I will be waking up on both Monday and Tuesday.

Given that I accept the protocol that you will give me amnesia and ask me what I think before I can find out what day it is, I have my answer before the experiment begins.

i.e. The next time I remember waking up you will be asking me about the probability of one coin toss coming up tails (or heads). The answer is 1/2 since I mak you use a coin I have tossed a few thousand times and decided it is close enough to equal probability as matters.

A comparison to Monty Hall is here:

http://philmat.oxfordjournals.org/content/13/2/194.full

An interesting defense of the 1/2 solution is here:

http://lesswrong.com/lw/286/beauty_quips_id_shut_up_and_multiply

I am not so sure of the answer now. It will require much more thought. A Google search produces a number of discussions and philosophical articles. A good one Briggs!

Wm

Mr Briggs,

Bone headed.

Your first attempt in which she knows what day it is is wrong because the question states that “she is given a dose of an amnesia-inducing drug”.

Your second attempt goes wrong here:

>Pr( M | C & E ) = Pr( T | C & E ) = 1/2.

>where E = “The details of the experimental protocol.”.

She should recognise that there is a 2/3 chance of waking on Monday. Monday always happens, Tuesday depends on a coin toss. (The halfer description given on wiki is wrong in the same way)

4 events have these probabilities, (assuming no second coin toss)

P(Monday, heads) = 0.5, experiment ends

P(Monday, tails) = 0.5, experiment continues

P(Tuesday, heads) = 0.0, never happens

P(Tuesday, tails) = 0.5, Tuesday has a 50% chance of occurring, and is always tails

so,

P(heads) = (0.5+0.0)/(0.5+0.5+0.0+0.5) = 1/3

Here’s a much better problem statement:

The paradox imagines that Sleeping Beauty volunteers to undergo the following experiment. On Sunday she is given a drug that sends her to sleep. A fair coin is then tossed just once in the course of the experiment to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday, given a second dose of the sleeping drug, and awakened and interviewed again on Tuesday. The experiment then ends on Tuesday, without flipping the coin again. The sleeping drug induces a mild amnesia, so that she cannot remember any previous awakenings during the course of the experiment (if any). During the experiment, she has no access to anything that would give a clue as to the day of the week. However, she knows all the details of the experiment.

Each interview consists of one question, “What is your credence now for the proposition that our coin landed heads?”

A little joke…they tried the same experiment with Prince Charming. PC woke up and looked around. The interviewers reminded him of the experimental protocol and asked him the key question. PC rubbed his chin thoughtfully and reasoned that if it were Monday, he would have to answer 50-50, but if it were Tuesday, he’d know that the coin landed tails. He peered at the interviewers intently to see if he could find a clue as to whether it was Monday or Tuesday. Then he brightened and said, “It’s Tuesday! So the coin must have come up tails!” How could he tell?

If she’s told what day it is the problem is not interesting. On Monday it’s just a normal coin toss with P(H)=.5,and on Tuesday it’s a coin toss where all the heads were discarded so P(H)=0.

The intent of the problem is presumably that she NOT know what day it is until after giving her answer and since in a sequence of repetitions of the experiment there are twice as many expected wakeups for tails as for heads the probability of head on any given wakeup is 1/3

William Sears, that less wrong discussion is indeed interesting.

The way I see it, it seems indisputable that if the credence/probability is taken to mean the proportion of times she is asked and it would be correct to guess heads were the experiment to be repeated infinitely then the answer is 1/3. It also seems fairly clear that if she is not told any more then if someone asks her out of the blue after the conclusion of the experiment, then Sleeping Beauty does not have any reason to prefer tails over heads or vice versa (i.e. p=1/2).

We must all know which of these meanings of probability Briggs is more interested in, but his post goes further to claim that knowledge of the procedure does not give the princess when woken any reason to think it is more likely to be Monday than Tuesday.

His beard had grown two days worth, sir, but that’s not important.

@WilliamSears: Thanks for the links, but it looks to me like this is a case of “less is more” (though not, I’m afraid, in a good way).

The problem with that tree diagram is that it really just amounts to giving a long calculation to prove what she already knows (or at least has been told) – namely that the coin is fair.

If so then of course the P(H)=.5 but that question is even less interesting than the cases in which she is told what day it is.

The actual question we are asked is what credence she should attach to the statement that when she wakes up she will be shown a head, and both Richard Neal’s gambling approach and Nick Bostrom’s extreme case should make it clear that the answer is 1/3.

Or, if you really like tree diagrams, think of it this way:

p=(1/2) Heads

p=(2/3) – it’s Monday <

/ p=(1/2) Tails

Wakeup< p(H)=(2/3)*(1/2)+(1/3)*0. p(T)=HW

\ p=0 Heads

p=(1/3) – it's Tuesday <

p=1 Tails

Part of the difficulty in all this is that what we mean by asking for the "probability" of something is not well defined without a clear specification of the "game" or "experiment" involved. The 1/3 solution is the answer to the question "When you get woken up like this, in what fraction of the cases do you expect to see a head?" – which to my mind its the correct interpretation of "when you get woken up like this what will you take as the probability of seeing a head?"

That is very different from the question "when you get woken up like this what will you think was the probability of getting a head when the coin was tossed?"

Examples like this are why I have little trust in people who discuss probability but toss aside questions like "What is the actual game or experiment in terms of which you are saying that the probability of event E is p?"

Sorry about the lost formatting on that tree diagram.

I have posted it also on my own blog at http://qpr.ca/blogs/2012/03/01/less-is-more/

Why does everybody keep treating the Tuesday interview as a separate event? It isn’t. It’s part of a complex event triggered by tossing tails. This is evident if we add more details to the Tuesday event. The experimenters all wear pink underwear, Germany invades Poland, who cares? If she doesn’t know it won’t change the event or the probability.

There are two

independentevents not three. One has extra detail to confuse the reader.William Sears,

I originally did that (the 1/2 solution using the tree) but then realized that P(tail,Monday)=1/2.

There are only two outcomes of the coin toss: Monday & Monday-Tuesday. P(awake|Monday,tails)=the probability of being awakened on Monday and tails is 1. P(awake|Monday,heads)=1. So P(tails,Monday) = P(Monday)*P(tails) = 1/2. Your link has it at 1/4. If the coin comes up heads, Monday and Tuesday aren’t selected with a probability of 1/2. They are guaranteed.

The branches leading from tails to Monday and Tuesday should have been labeled ‘1’. You multiply the labels along the path from the root to a leaf. The result is three leaves each with a likelihood of 1/2 (which you should normalize to get the probability)

The three situations have equal probability.

Sorry. The sentence

If the coin comes up heads, Monday and Tuesday …” is incorrect.Should have read: “If the coin comes up

tails, Monday and Tuesday arenâ€™t selected with a probability of 1/2.”I concur with 1/3. It is perfectly valid to treat Tuesday’s interview as a separate event, because from her perspective, it is.

Imagine there was never a Monday interview. The chance of the coin landing heads is still 1/2, but that doesn’t mean that – if interviewed on a Tuesday – she should answer 1/2. That would be ignorant – she should answer 0.

I don’t know whether Python code works in these comments, but maybe this demonstrates that, from SB’s perspective, at each interview the chance of the coin having been flipped heads is 1/3.

import random

num_heads = 0

num_tails = 0

def wakeup(heads):

if heads:

global num_heads

num_heads += 1

else:

global num_tails

num_tails += 1

rng = random.Random()

for i in xrange(10000):

heads = random.choice([True, False])

wakeup(heads)

if not heads:

wakeup(heads)

print float(num_heads) / (num_heads + num_tails)

Rich,

The wording of the problem could have been better. She is faced with three scenarios even though there are only two outcomes from the coin toss. Effectively, she is being asked “what is the probability you were awakened

andthe coin landed heads?” Heads only appears in one of the three scenarios. The only thing that might change is the likelihood of a given scenario. I believe they are equally likely but the link given by William Sears disagrees.If the drug wipes out Sleeping Beautyâ€™s memory of all the details of the experiment, then her answer should be Â½. If she still has the memory, the probability should be different from 1/2. Knowing the details should somehow alter her answer, unless one can reason that knowing the details of the experiment is irrelevant or adds no extra value. Which is against our life experience. Hmmm? Need to time to think about this.

DAV,

No worries. December signals the end of a calendar year, not the world (of democrats). It also tells you a new, brighter year is coming.

You have assumed that â€œMondayâ€ and â€œTailsâ€ are independent, which is not true. Based on the descriptions of the experiment, the â€œdayâ€ depends on the result of the coin toss.

Genemachine,

The fact that sum of the joint probabilities must equal to 1 indicates that your assignment of probabilities to the four events is incorrect.

Alan Cooper and DAV,

The Prince may have repeatedly kissed Sleeping Beauty 30 times to wake her up, but the coin has only been tossed repeatedly 20 times. Something is amiss in this reasoning, but I canâ€™t pinpoint what it is. That is, I canâ€™t see why the three se scenarios are equally likely either.

JH,

>The fact that sum of the joint probabilities must equal to 1 indicates that your assignment of probabilities to the four events is incorrect.

The “events”, as I have defined them, of (Monday, tails) and (Tuesday, tails) are not exclusive.

The 2 paths through the experiment are

(Monday, heads)

or

(Monday, tails),(Tuesday, tails)

The 2 paths are equally likely but the second one has 2 “events”. The expected number of “events” in an experiment run is 1.5.

JH, she is awakened on Monday regardless of the coin toss. That’s what I really meant by Monday. Apparently, Briggs has enemies who don’t discriminate by author when mangling posts. Should have said:

Maybe this will clarify it:

P(awakened Monday|heads)=1

P(awakened Monday|tails) =1

P(awakened Monday)=1

P(tails, awakened Monday) = P(awakened Monday|tails)*P(tails) = 1/2.

P(heads, awakened Monday) = P(awakened Monday|heads)*P(heads) = 1/2.

P(awakened Tuesday | heads)=0

P(awakened Tuesday | tails) =1

P(tails, awakened Tuesday)= P(awakened Tuesday|tails) * P(tails) = 1/2

P(heads, awakened Tuesday)= P(awakened Tuesday|heads) * P(heads) = 0

There are only three scenarios with P>0,

P(tails, awakened Monday) = P(awakened Monday) * P(tails) = 1/2.

P(heads, awakened Monday) = P(awakened Monday) * P(heads) = 1/2.

P(tails, awakened Tuesday)= P(awakened Tuesday|tails) * P(tails) = 1/2

All are equally likely to occur.

I tried out the PRE tag to see if it works here. I may have messed up the spacing since I can’t edit with a fixed pitched font.

I have figured it out.

Us right thinking halfers know that whenever sleeping beauty is woken up she is asked “what is the probability of heads in a coin toss?”. Nothing more.

You wrong thinking thirders think she is asked “What is the result of the experiment?” which is equivalent to asking what day of the week she thinks it is.

Since I am now studying to become a tree ring data smoother/adjuster I have decided that the above proves us halfers is right and you thirders is wrong. And since us halfers have a consensus that you thirders are wrong that makes you three times wronger!

I find the arguments of the thirders far more convincing.

How about a slight change to try to see if my understanding of the situation is correct.

After the coin is tossed it is placed into a box – and Sleeping Beauty is asked when she is woken up to give the probability that the coin is showing heads.

Will this question give the same answer as before?

Now lets change it some more and say that if the coin lands heads then she will be woken up, and asked the question (followed by her having her memory wiped) not once, but 1000 times.

In this situation I believe that it is reasonable for her to expect to be asked the question [0.5 x 1] + [0.5 x 1000] times – when she wakes up it is therefore unreasonable to presume that this is the one and only time she has or will be woken to answer heads – the answer tails will occur far more often (oh dear will Professor Briggs hate me for this frequentist phrase!) in the long run of testing.

Are halfers, still halfers, in this test? I feel being a hundredther makes more sense.

I eventually understood the Mony Hall problem by extending the number of doors to 100 from the usual 3 and saw the sense of swapping – isn’t this similar?

DAV,

DAV,

The premise E= {all details of the experiment}is important.

Sleeping Beauty would assign P(Monday | Tails, E) = 1/2. In this case, given the knowledge of E and Tails, she would wake up wondering whether today is Tuesday or Monday.

And P(Monday | E) is not 1. She could be awaken on a Monday or Tuesday. If itâ€™s one, it basically says that P(Tuesday | E) is 0. I am not sure how to assess this probability.

P(Heads, Monday |

E) = P(Monday | Heads,E)*P(Heads |E)We know that P(Heads) = 1/2, and what we want to find is

P(Heads | E). The probability P(Heads | E) equals to P(Heads) = 1/2 if E is independent of Heads, i.e., E is irrelevant, which is halfersâ€™ belief according to Wikipedia.So here is what I have. Want to find P(Heads | E).

P(Tails) = P(Heads) = 1/2

P(Monday | Heads, E) = 1; P(Tuesday | Heads, E) =0

P(Monday | Tails, E) = 1/2; P(Tuesday | Tails, E) = 1/2

P(Tails | Tuesday, E) =1; P(Heads | Tuesday, E) = 0

I shall give my students this problem as homework.

JH,

You missed an important condition. From Wiki, “If the coin comes up

heads, Beauty is awakened and interviewed on Monday, … If the coin comes uptails, she is awakened and interviewed on MondayandTuesday.”She is awakened on Monday regardless of heads or tails.

P(Awakened Monday|E,fair coin=Heads

orTails)=1P(Heads|Awakened Monday,fair coin,E)=1/2 is true also.

DAV,

Yes, P(Heads|Awakened Monday, E)=1/2 is true also, I think. I’ll add this to the problem.

The probability P(Awakened Monday|E, coin=Heads or Tails)=P(Awakened Monday|E) because that a coin lands on either head or tails, the information “coin = Heads or Tails” becomes redundant.

Anyway, P(Awakened Monday|E, coin=Heads or Tails) = 1 implies that P(Awakened Tuesday|E, coin=Heads or Tails)=0, which is clearly wrong. I can’t think of a better explanations for why it’s wrong.

(The information of fair coin is contained in E already).

JH,

Anyway, P(Awakened Monday|E, coin=Heads or Tails) = 1 implies that P(Awakened Tuesday|E, coin=Heads or Tails)=0The first happens to be true but doesn’t imply the second.

The first is really two statements:

P(Awakened Monday|E, fair coin=Heads) = 1

P(Awakened Monday|E, fair coin=Tails) = 1

Neither of which say anything about Tuesday. The particulars for Tuesday are:

P(Awakened Tuesday| Heads,fair coin,E)=0

P(Awakened Tuesday| Tails,fair coin,E)=1

Sorry about that last sentence in my previous post. I had started on a different tack and forgot to delete it.

This problem is begining to make my head hurt – I’m thinking about the example I gave in my post above where I altered the problem to give 1000 tests if tails came up rather than 2 in the Wikipedia example, and where Sleeping Beauty is asked to guess if a box contains a head or a tail. I have to admit I am still very uncertain what is the right answer and I am suddenly wondering if the halfers are correct.

I am confronted by the fact that there is a 50% chance she’s in the situation where its the first and only test due to the fact that the box contains a head. The other 50% chance will result in her being woken up 1000 times – but so what – in every case the answer is tales. She’s totally forgotten if its the 1st or the 1000 time she’s being asked, but this doesn’t change the fact she’s only got a 50% chance of being asked multiple times.

So have I flip flopped to be a halfer. I’m half convinced!

It is the 100 years sleep scenario that has turned me toward the halfer camp. Should Sleeping beauty assume that she has come up on the loosing side of the bet, just because it means that she will be interviewed so many more times in the tales scenario?

Suppose we change the expirement such that we don’t use a coin but some other random number generator such that the expiriment ends after one night 99% of the time, and continues for 100 years 1% of the time. When she wakes up the question will be, will this expirement end today? I would say, “YES!”

Why should it change her credence in the result of a coin flip if she is woken, interveiwed and drugged repetedly vs. just sleeping through the whole thing? She can’t tell the difference.

When I get these sorts of problems I often try to find an analogue for them in a different situation which presents the conundrum in a different light.

Do people think the following is an accurate analogue for an altered Sleeping Beauty test – where she is woken 1000 times if it comes up tales, or once if it comes up heads – this would make the dichotomy currently between halfers and thirders into one of halfers and thousandthers (or there abouts) – the bigger the split the easier it is to challenge our assumptions!

A person is presented with two doors – and told they will flip a coin – Heads they go to the door on the left, tails they will go to the right. They are told one door only has one coin in it which is showing heads, the other has 1000 coins all showing tales.

When they open a door they will find a random number generator which will give them a number between one and a thousand.

They will then walk into another room which, depending whether it is the left hand or right hand door, will either contain one box which will have the same random number on it containing a coin head up, or a thousand boxes numbered 1 to 1000 each of which contain a coin tale up.

They have to open the box numbered with their number.

What is the probability of the person finding a head in the box they open?

Or is a frequentist answer better – if this test is run an infinite number of times what is the proportion of tests where a head if found?

———

I am using the random number generator as an analogue to the memory loss – is this reasonable? IE if the number says 437 this is equivilent to the 437th time SB has been asked, but she has no memory of the other 436 requests.

Prof Briggs – I am sure you are much more qualified to provide some comment than many of us – you are not posting much within this thread which leaves me unsure whether the debate is making any progress, or whether it is, as Keynes* might say, heading to Bedlam.

*”Starting with a mistake a remorseless logician can end up in Bedlam”

DAV: “Effectively, she is being asked â€œwhat is the probability you were awakened

andthe coin landed heads?â€”How can the question, “What is the probability that you are awake?” have an answer other than 1? Or what have I missed?

I still think that the events on Monday and Tuesday are not independent and ergo should not be calculated separately. Let’s have her woken and anmenomicked on Wednesday, Thursday and Friday too in the event of tails coming up on Monday. Do we now have 6 independent events? I think not.

â€Starting with a mistake a remorseless logician can end up in Bedlamâ€

As we saw in the post on “post-birth termination” it is not necessary to begin with a mistake. Remorseless logic is quite enough on its own.

Chinahead,

“What is the probability of the person finding a head in the box they open?”

Effectively asked but not answered here: http://www.imdb.com/title/tt0101410/

Sorry to sound like a broken records but until Briggs’ authority convinces you deviant halfers..

Doug M,

>Why should it change her credence in the result of a coin flip if she is woken, interviewed and drugged repeatedly vs. just sleeping through the whole thing? She canâ€™t tell the difference.

She should be know from the experimental protocol that if it comes up tails she is going to be woken twice and if it is heads she will only be woken once.

So, given that she has been woken and does not know the coin result, she should be able to calculate that she is twice as likely to be in a coin=tails situation.

>Suppose we change the experiment such that we donâ€™t use a coin but some other random number generator such that the experiment ends after one night 99% of the time, and continues for 100 years 1% of the time. When she wakes up the question will be, will this experiment end today? I would say, â€œYES!â€

And you would be a fool. Or you are parsing the question differently from me.

First night, 99% of the time you would be correct.

However, in addition to the 1% of the time you got the first night wrong you can add being wrong a further 36524 nights.

Take 100 runs; you get it right on 99 of first nights but are wrong on 1 first night plus 36524 further nights.

Your odds of getting it right on any particular night = 99/(99+1+36524) = 0.0027

Chinahand,

Sorry, your doors example is a bit convoluted and I struggle to make sense of it.

I think a better doors analogue of the original experiment might be that she is in a room with 2 exits and she flips a coin to go through one. Heads opens the door which leads outside, and tails leads to a another room with another door which in turn leads outside.

She never goes through the same door twice nor does she know where she is the experiment. She is presumably some kind of amnesiac but is good with statistics.

So, heads opens 1 door, tails opens 2. In each experimental run all doors are equally likely to be opened (50% of the time).

She is about to open a door…

Rich,

Right up there with getting a phone call in the middle of the night and being asked “Are you awake?”

DAV,

This probability statement means that, given Sleeping Beauty remembers E and is told that the coin has landed on Tails when awake, she knows for sure (with a probability of 1) that today is Monday. However, she doesn’t, since she has no memory of being awakened previously.

DAV, I agree that P(Beauty is woken M, H | E) = P(Beauty is woken M, T | E) = P(Beauty is woken T, T | E) = 1/2. This conclusion sensibly leads to an answer 1/3 if your ultimate purpose is something like the expectation of the fraction of times Beauty can correctly answer H (if repeated…), but what you do with that depends on whether your question wants them as seperate events despite their lack of independence. Clearly the probability of (is woken M, H) union (is woken M, T) union (is woken T, T) is not 1.5.

Whatever you make of the halfer probability tree (clearlyl, if it is valid, then it is unecessary), the P(Monday) that it (and Briggs) speak of is clearly not intended to be P(Beauty is woken on Monday) but P(It is Monday). Then Monday and Tuesday are indeed mutually exclusive, but there is no disagreement with your claims that each scenario is equally likely to happen at all.

Just imagine the Beauty is told that she will undergo this process with one set of interviewers, be put back to sleep without seeing the coin, and then, completely independently, another interviewer will wake her on Friday regardless of the result of the coin toss, and ask her the same question about the same toss. Clearly her answer on Friday must be 1/2. Does your view of probability allow her to give a different answer on Tuesday (or Monday) as opposed to Friday?

If you are happy with the 1/2 answer, you can of course answer the “fraction of times she is woken (if repeated…)” and what bets should she take sort of questions without any contradiction, by simply using the number of wakings or payoffs corresponding to each result, and calling them that, rather than any sort of probability. If instead, like Alan Cooper, you instead think that the games/experiments are necessary for probability, then you use the word probability slightly more in coming to the same answers, and dismiss the more ‘abstract’ questions as meaningless.

(Or we can all do the thing in Briggs’ post, and answer the question “Given it’s the Monday or Tuesday, what’s the probability that SB was woken this morning as a result of heads result?)

Jonathan D,

The problem statement leaves much to be desired. In the end it’s unclear what she is being asked and what we are being asked. In fact, the Wiki version doesn’t ask us anything. What are we being asked? It rather seems we should answer for her.

The question “What is your credence

now…” rather implies she is receiving information but I’m not sure the implication is intentional.My first reaction was 1/2 based on the fact that she has no memory of previous awakenings (on Tuesday).

That she is being asked “Is it Monday?” is to me a rather bizarre interpretation unless that is short for “What’s is the probability you are in scenario #1 (Monday, Heads)?”

I now think she’s being asked the probability of being in the scenario with coin=Heads which it isn’t any different than being asked the probability of opening the door which hides the prize. If so, the experiment protocol is rather dumb.

OTOH, this problem was likely designed to illustrate the dichotomy between the Self-Induction and Self-Sampling assumptions. Depending on the one you pick leads to 1/3 and 1/2. Interestingly, the Wiki article on the each opens by summarizing the other in one sentence but doesn’t summarize the given topic.

If the problem clearly stated what was being asked (and of whom), then all of this go-around wouldn’t exist.

JH,

???

P(Awakened Monday|E, fair coin=Tails) = 1 means she’s definitely awakened on Monday based on the givens. If she knows the protocol (E) then she also knows this which the probability of being awakened on Monday given the coin outcome. The statement has nothing to do with what she is told upon being awakened.

Jonathan D,

“Clearly the probability of (is woken M, H) union (is woken M, T) union (is woken T, T) is not 1.5.”

Using P(…) for the result was misleading. I only gave the numerators in Bayes’ equation: P(H|D)=P(D|H)P(H)/P(D). To get the result you need to divide by the denominator, P(D), which happens to be the sum of the numerators.

Sleeping Beauty is the one who assesses all the probabilities.

When a premise is given, it means whoever determines the probability has the premise information. That is, Beauty somehow comes to know that the coin has landed tails. I just assumed that she was told of this information due to my lack of imagination.

Beauty remembers that she’d be awakened on both days if Tails, but when she is awakened, does she know what day it is? No.

SB can’t assess P(awakened Monday|E,coin=tails)? Really? If she knows E then she should be able to. I can and I don’t know the outcome of the coin toss. In fact, knowing it isn’t necessary. All I need to do is suppose it happened with the given value and know the experiment protocol. Same for her. Do you think she’s a blonde?

Yes, “whoever determines the probability has the premise information” but the givens are E=experiment and C=coin outcome. “Given” doesn’t mean “told is true” (well, it

could) but instead usually means “suppose it true”. That’s what “premise” implies: “fact” or statement put forth for the sake of argument. If she knows the outcome and the day there wouldn’t be a need to do a probability assessment.If she knows E then she should know that during the experiment as described she will be wakened Monday regardless of the coin outcome. IOW: P(awakened Monday|E,coin=tails)=1 just as P(awakened Monday|E,coin=heads)=1. Those are true regardless of the actual results or day. They will even be true tomorrow which is Saturday. Neither says that it

isMonday or how the coin actually landed .Are you, for whatever reason, objecting to the fact that the value is 1? I really don’t see what is confusing you.

DAV,

I didn’t say that SB couldn’t P(awakened Monday|E,coin=tails)! I am saying she is not sure whether it’s Monday or Tuesday when she is awakened.

(OK. “suppose it’s true”, it is! The value of the probability remains the same.)

So if P(awakened Monday|E,coin=tails)=1, what is P(awakened Tuesday|E,coin=tails)?

“… If the coin comes up

tails, she isawakenedand interviewed onMonday and Tuesday…”P(awakened Monday|E,coin=tails)=1

P(awakened Tuesday|E,coin=tails)=1

assuming she hasn’t run off with Prince Charming in the middle of the preceding night.

Here’s one way to look at the problem.

When Sleeping Beauty is awakened, she knows that one of three mutually exclusive scenarios has occurred:

A) It’s Monday and the coin shows heads.

B) It’s Monday and the coin shows tails.

C) It’s Tuesday and the coin shows tails.

Because of the amnesia and the randomness of the coin, she cannot know precisely which one has occurred, so she has to assess the probabilities.

From her context of knowledge (having just been woken up, but knowing the rules of the experiment), how could she prefer one of these to any of the others?

Thirders say that from her perspective these are all equally likely. Therefore, P(heads | E) = P(A | E) = 1/3. (Incidentally, it would then follow that P(It’s Monday | E) = P(A | E) + P(B | E) = 2/3 and P(It’s Tuesday | E) = P(C | E) = 1/3, which accords well with my intuition.)

Halfers, on the other hand, say that A occurs only if the coin shows heads. Therefore P(A | E) = P(a fair coin shows heads) = 1/2 and so P(heads | E) = P(A | E) = 1/2. What about P(B | E) and P(C | E)? These usually aren’t discussed, but I’m very curious about how a halfer would answer. I see no reason to prefer one over the other and therefore P(B | E) = P(C | E) = 1/4. This then implies that P(It’s Monday | E) = 3/4 and P(It’s Tuesday | E) = 1/4, two values that I find hard to fathom.

Consider this:

Alter the experiment so that SB will be awakened on both Monday and Tuesday no matter what the result of the coin toss (and she knows this before she is put to sleep). As before, she is given amnesia, so that on Tuesday she does not remember the Monday awakening.

Then, we have a rather dull experiment. Whenever she is awakened, she knows nothing about the coin status or the day of the week. Therefore, she believes:

P(heads) = P(tails) = 1/2

P(it’s Monday) = P(it’s Tuesday) = 1/2

Furthermore, her beliefs about the coin and about the day of the week are independent, i.e., P(heads & it’s Monday) = 1/4, etc.

Now, after she is awakened one day, suppose one of her interrogators tells her, “If the coin shows heads, then today is Monday.” This is logically equivalent to “the coin shows tails or today is Monday”. Given this new information, SB updates her belief:

P(heads | tails or Monday) = P(heads & (tails or Monday) ) / P(tails or Monday)

= P(heads & Monday) / [P(tails) + P(Monday) – P(tails & Monday)]

= 1/4 / [1/2 + 1/2 – 1/4]

= 1/3

Returning now to the original experiment: SB is awakened one day and asked, “What is your credence now for the proposition that the coin landed heads?”. She knows the rules of the experiment and so proceeds to tell herself, “If the coin shows heads, then today is Monday.”…

DAV, I don’t think anyone is suggesting that SB is asked “is it Monday” – all I was saying was that the Monday and Tuesday used for calculating in Bayes’ equation and anything related must indeed be shorthand for being M/T in the scenario, rather than generally being woken on that day. Otherwise we are not talking about mutually exclusive events, and at some point you are dividing by a denominator, supposedly a probability, which is greater than 1.

I didn’t multiply by the prior for each scenario, P(S). There are four scenarios (although it turns out one of them is impossible). If I had uniformly assigned P(Si)=1/4 then P(D) would be less than 1. Multiplying each numerator by the same constant while clearer is unnecessary as the constant is canceled out in the final step, i.e., when dividing by P(D). When I’m dealing with uniform priors, I usually just set each to 1. Sloppy but saves effort — not this time though.

Briggs, you did make several mistakes in interpreting the problem. But I don’t know if I’d call them bone-headed, or simply a result of over-analyzing the problem statement.

“Suppose she is awakened on Monday. She knows it is Monday.” No. Nowhere does the problem statement say she is told it is Monday. And the point of (1) the amnesia drug and (2) telling her about the amnesia drug, is so she can’t deduce what day it is simply because she only recalls one awakening.

“There are then no words (telling) us what she does Monday night into Tuesday.” Sure there are: “But when she is put to sleep again on Monday…” This happens only after Tails is rolled – if the coin landed heads, the problem statement said she was already sent home. But in a more classic statement of the problem, Sleeping Beauty is put to sleep regardless, and left asleep through Tuesday after tails is flipped. In some versions, she is left asleep indefinitely after the experiment so she can’t compare notes with herself.

“So if queried on Tuesday about the coin she will have a different response than when she was queried on Monday.” This is the point of the whole problem: to set up a situation where she has no evidence that differentiates one day from the other, so her response can’t be different.

The thirders’ (correct) argument goes like this: Since her answer can’t be different in the three possible situations, and the three answers must sum to 1, the answer must be 1/3.

The halfers make several unsupportable claims. The worst is that she has gained no information since being put to sleep on Sunday, and so can’t change her credence. It isn’t the act of adding information to her knowledge of the state variables that allows her to update a prior, it is changing how her knowledge of the state variables restricts them. (This argument was paraphrased from Nick Bostrom, whose work is incorrectly cited in the Wikipedia article.) On Sunday, she knew that both Monday and Tuesday were going to happen. When she is awakened, she does not know what day it is, or if it is possible that the day is Tuesday. So while she doesn’t have “additional” knowledge, she does have knowledge which restricts the state differently than her knowledge did on Sunday.

Everybody understands how to handle the random variable COIN={H,T}. But there is a difference between DAY_EXISTS={Mo,Tu} and TODAY={Mo,Tu} that is hard to grasp. The former isn’t a random variable at all – both days will exist over the course of the experiment. But the halfers get 1/2 by treating the random variable DAY as independent of COIN; that is, as DAY_EXISTS. With independence comes separability, so P(COIN=H|DAY)=P(COIN=H)=1/2.

If you treat the random variable DAY as TODAY, you can’t separate them that way. You must address all four combinations of the two random variables COIN and DAY; that is, the sample space is {H&Mo, H&Tu, T&Mo, T&Tu}. That fact that H&Tu won’t be observed by Sleeping Beauty does not affect its possibility of occurring, only of being observed. Each of these combinations has a 1/4 probability of corresponding to any point in time during the experiment. But the probability it will be observed by Sleeping Beauty at a random moment in time is some value Q (we do let her sleep during parts of the days she is awakened) for H&Mo, T&Mo, and T&Tu, and 0 for H&Tu. Thus, P(H&Mo|Observed)=(Q/4)/(Q/4+0+Q/4+Q/4)=1/3.

Another unsupportable claim some halfers make, including the article at lesswrong linked to by William Sears, is that P(H|Mo)=1/2 but P(T|Mo)=P(T|Tu)=1/4. This fallacy results when, after treating DAY as DAY_EXISTS to get the answer 1/2, the author changes to treating DAY as TODAY for the conditional probabilities. And the absurdity of this result can be seen by making three simple changes. First, move the coin flip to Monday night after Sleeping Beauty has been amnesia’ed and put back to sleep. This can’t affect the experiment in any way since the result of the flip isn’t used until Tuesday. Second, change the rules so that she is awakened not just once, but some variable N more times after a tails. So the halfer’s argument says P(T|Mo)=1/(2N). Finally, after Sleeping Beauty makes her initial assessment of credence on Monday, either 1/2 or 1/(2+N), ask her to pick N.

This informs Sleeping Beauty it is Monday, so P(T|Mo) becomes her credence. And if she is a halfer, that means her credence that a future coin flip will result in a heads is 1/(2N). For an N she gets to choose. How does the coin know about her choice?

I know this is very long after the events, but it may help others like me who come to the table late. The problem is built around SB’s total lack of knowledge as to which of the three awakenings she is experiencing. The argument is that these awakenings form three independent equally likely outcomes, only one of which corresponds to heads, so the probability of heads is 1/3. The fallacy lies in the assumption of independence; two of the outcomes are not independent of each other and so the maths is wrong. Consider a 3 sided die, constructed so that it either throws a 1 (50%) or else throws both a 2 *and* a 3 (50%). There are three outcomes (1, 2, or 3), each with the same probability of occurring (50%), so how likely is it to throw a 1? Is it 1/3, because there are 3 outcomes, or 1/2, because the probability of the dice throwing a 1 is 1/2? I remember this problem from the newsgroup days when there was a list put together of proofs that the answer was 1/3, all of which were built on the same false assumption of independence.