Monday, February 11, 2008

An approach to the Sleeping Beauty problem

Experiment 1:
A fair coin is flipped on Sunday, without you seeing the result, and then you go to sleep.
Tails: You get woken up Monday and Tuesday and each time shown a red flag. Your memory is erased each time, and you don't know whether it's Monday or Tuesday when you wake up.
Heads: You get woken up Monday and shown a red flag.
So, you're awake and see a red flag. What probability should you assign to heads? The two most common options are 1/2 and 1/3.

Experiment 1 is equivalent to the Sleeping Beauty problem, with a red flag added, which changes nothing.

Experiment 2:
Same as Experiment 1, except that on heads, you get woken up on both Monday and Tuesday, but on Tuesday you see a white flag.

It seems clear to me that in Experiment 2 you should assign 1/3 to heads. It's a standard Bayesian thing—you start with 1/2, and update on the additional information of a red flag: given heads, the chance of a red flag is 1/2, and given tails, the chance of a red flag is 1, so the red flag is evidence for tails, and the numbers work out to a probability 1/3 for heads.

I now claim that you should give the same answer for both experiments. This implies you should assign 1/3 to heads on Experiment 1.

Two arguments for equivalence. First, we can imagine a continuum of cases between Experiments 1 and 2, where the amount of awareness you have on Tuesday given heads varies continuously from zero (Experiment 2) to full human awareness (Experiment 1). Where exactly you would be on this spectrum on a white flag Tuesday does not seem to me to affect what credence is rational when you see a red flag and are fully humanly conscious.

Second argument. You can imagine that you're not told ahead of time whether you're in Experiment 1 or 2, but when you wake up and see a red flag you can press a button that you know has the following effect. If the coin was tails, the button has no effect. If the coin was heads, in which case it's Monday (since on Tuesday you don't see a red flag, if you're awake at all), and you press the button, then you'll wake up on Tuesday and see a white flag; if you don't press the button, you won't wake up on Tuesday in that case.

If you press the button, you're effectively in Experiment 2. If you don't, you're effectively in Experiment 1. If different credences of tails are appropriate in the two experiments, then how you decide about the button press after the coin toss affects what credence you should assign to tails. That's weird. (Is there a better choice of button press, one that will provide me with a better credence?)

Variant on second argument: You either do or do not see an independent random process depress the button when you wake up—you don't control the button. Should the outcome of this random process affect your credence about the initial coin toss? Certainly not: the outcome of this process is ex hypothesi independent of everything else. So, you should assign the same credence to tails in Experiments 1 and 2, and this credence should be tails.

I've been told that the claim that 1/3 is the right answer for Experiment 2 would be controversial. If so, then the argument only shows the credence is the same in the two cases, not that it's 1/3. But I think 1/3 is the right answer for Experiment 2.

12 comments:

Anonymous said...

Hi Alexander,

I think the problem with this puzzle is the ambiguity over what the sleeper is trying to do. Here are two versions of the problem using money:

Version 1 - A fair coin is flipped on Sunday, without you seeing the result, and then you go to sleep.
Tails: You get woken up Monday and Tuesday and each time you are given one pound. Your memory is erased each time, and you don't know whether it's Monday or Tuesday when you wake up.
Heads: You get woken up Monday and you are given one pound.

Version 2 - A fair coin is flipped on Sunday, without you seeing the result, and then you go to sleep.
Tails: You get woken up Monday and Tuesday and each time you are asked if the coin came up heads or tails; if you get it right you get a pound, if not, you get nothing. Your memory is erased each time, and you don't know whether it's Monday or Tuesday when you wake up.
Heads: You get woken up Monday and you are asked if the coin came up heads or tails; if you get it right you get a pound, if not, you get nothing.

In version 1 you get the money whatever you say. The probability you should assign to the coin being tails is the same as the probability you should assign to you winning one pound instead of two. This is clearly 1/2.

In version 2 you only get the money if you utter a truth. In this case it is clearly preferable to say it was tails because that will get you more money, but it won't change the probability.

In the original problem, when you are woken up you should assign a probability of 1/2 to both heads and tails. However, if you wish tomake more true utterances that you should always say tails.

Alexander R Pruss said...

Here's an argument that 1/3 is the credence in my Experiment 2. In Sleeping Beauty cases, the only two plausible credences are 1/2 and 1/3. Now, let's suppose you are in Experiment 2. You wake up. Before you open your eyes, however, you think about your credence. It's clearly 1/2. You open your eyes and see a red flag. Clearly this is evidence--you might have seen a white flag instead. It would be absurd to say that your credence should remain the same upon seeing the red flag.

Here's a more rigorous argument. If you were to see a white flag after opening your eyes, that would change your credence in heads to 1. But:

Theorem: If P(H|E)=1 and 0 < P(E) and 0 < P(H), then P(H|~E)< P(H).

So, not seeing a white flag after opening the eyes must decrease your credence (since 0 < P(white flag) and 0 < P(heads)).

Hence, after seeing the white flag, one's credence has to be less than 1/2.

f|35h said...
This comment has been removed by the author.
Zsolt Nagy said...

But what is exactly P(white flag)?!?
Is it P(white flag)=P(white flag|heads)•P(heads)+P(white flag|tails)•P(tails)
=0.5•0.5+0•0.5=0.25?
Or is it P(white flag)=P(white flag|heads)•P(heads)+P(white flag|tails)•P(tails)
=0.5•1/3+0•2/3=1/6?
Is it then P(red flag)=0.75 or P(red flag)=5/6?
Is it then P(heads)=P(heads|white flag)•P(white flag)+P(heads|red flag)•P(red flag)
=1•0.25+1/3•0.75=0.5
or =1•1/6+1/3•5/6=4/9≈1/3?!?

You said for experiment 2, that "... the numbers work out to a probability 1/3 for heads". And HOW EXACTLY do those numbers work out a probability for heads being 1/3 in experiment 2?!?
I guess, that this is the eternal and internal struggle of philosophy only capable of providing ambiguous and cryptic proofs of their claims.
I'm so said.

Zsolt Nagy said...

"If different credences of tails are appropriate in the two experiments, then how you decide about the button press after the coin toss affects what credence you should assign to tails. That's weird."

Yes, Sherlock. Relevant Information should always be accounted for by credence assignments.
In Experiment 1 seeing a red flag is an irrelevant information, while in Experiment 2 it is relevant and if we are to further complicate this scenario with a butten, which relevantly changes the situation/Experiment, then of course we are also entitled by given information about that switch being pulled or not to adjust our credences accordingly to that. What is so difficult to understand about this matter of fact?

William said...

Here is a simulation suggesting thirding for the base problem and "halfing" for the red flag on waking case (red flag Monday, white flag if Tuesday, wake with red flag):

from random import choice

def sleeping_beauty_experiment(repetitions, whiteflag):
"""
Run the Sleeping Beauty Problem experiment `repetitions` times, checking to see
how often we had heads on waking Sleeping Beauty. Sleeping Beauty is shown
a red flag on Monday and a white flag on Tuesday if whiteflag is true. the
count is of the times ske wakes if whiteflag is False and of the times she sees
a red flag if whiteflag is True
"""
gotheadsonwaking = 0
wakenings = 0
redflagwakenings = 0
for _ in range(repetitions):
coin_result = choice(["heads", "tails"])

# On Monday, we check if we got heads.
wakenings += 1
redflagwakenings += 1
if coin_result == "heads":
gotheadsonwaking += 1

# If tails, we do this again, but of course we will not add as if it was heads..
if coin_result == "tails":
wakenings += 1
if coin_result == "heads":
gotheadsonwaking += 1 # never done


# Show the number of times she was wakened.
print("Wakenings over", repetitions, "experiments:", wakenings)

if not whiteflag:
# Return the number of correct bets SB made out of the total number
# of times she is awoken over all the experiments with that bet.
return gotheadsonwaking / wakenings
else:
# whiteflag is True so show the proporthion of red flag days
return gotheadsonwaking / redflagwakenings

CREDENCE = sleeping_beauty_experiment(1_000_000, False)
print("Results of experiment (no flags): Sleeping Beauty should estimate a credence of:", CREDENCE)
CREDENCE2 = sleeping_beauty_experiment(1_000_000, True)
print("Results of experiment (with flags): on red flag wakenings, Sleeping Beauty should estimate a credence of:", CREDENCE2)

Results are 0.333 and 0.5. Of course we only see a white flag if it was heads.

Alexander R Pruss said...

William:

I don't think this is a simulation of my Experiment 2. In Experiment 2, you wake up on Tuesday even if it's heads. Moreover, on tails, you get a red flag on both days.

Alexander R Pruss said...

Zsolt:

It's been a long time since I looked at this, but I think your first calculation of P(white) is the right one because P(heads) and P(tails) are the PRIOR probabilities, so they are both 1/2.

Then the calculation goes like this: P(heads|red) = P(red|heads) P(heads) / P(red) = (1/2)*(1/2) / (3/4) = 1/3.

Zsolt Nagy said...

Yeah, William, I also don't think, that you got Alexander's proposed Experiment 2 correctly, since the number of "wakenings" is redundant there in Experiment 2 (- here sleeping beauty is certainly awakened on Monday AND on Tuesday for both cases of the coin flip made on Sunday resuling either in heads or in tails).
Also how exactly are the ratios "gotheadsonwaking"/"wakenings" or "gotheadsonwaking"/ "redflagwakening" relevant here?!?
I don't think, that those are the relevant metrics or the relevant measurements for any credence here.

Apropos, what is supposedly the relevant metric or relevant measurement for credence here!
Now, that I red this Blogpost a couple of more times, I get how you might have gotten that "1/3" as a result. But thank you for your response and clarification on that (- also don't bother all too much with your wrong simulation there, William. I needed also quite a few attempts only understanding these Experiment and point/blogpost of Alexander here).
I know, that P(heads)=P(tails)=1/2 is or are the "PRIOR" probabilities here as you said it.
But isn't meant to be the "CREDENCE" here, since the question "What is yours, Sleeping Beauty's, CREDENCE of the coin flip made on Sunday resulting in heads or in other words, what do you, Sleeping Beauty, think, is the PRIOR probability for the coin flip made on Sunday resulting in heads?"?
At least that's what I get from it as a frequentist and as a frequentist it is also quite clear, what the answer to that question should be.
So what I really want to know from you, Alexander:
In what sense is P(heads|red flag) relevant for the sleeping beauty problem or for the question of "What the CREDENCE of Sleeping Beauty should be regarding the coin flip made on Sunday resulting in heads?"?
And in what sense exactly is P(heads|red flag)=1/3 in Experiment 2 the same as P(heads|red flag)=P(heads)=CREDENCE=1/2 in Experiment 1?!?
I can see reasons my self for my first question.
But rational reasons for my second question?!?
No, I can not come up with some rational reasons for that to be the case myself.

Zsolt Nagy said...

I wonder about, which kind of money one could make by dutch book bets in Experiment 2 here.
Let's suppose, one makes a bet in Experiment 2 by following rules:
(I) One bets its credence c amount of dollar for 1$ on the coin flip made on Sunday being heads in Experiment 2, if and only if one sees the white flag upon awakening.
(II) One bets its credence c amount of dollar for 1$ on the coin flip made on Sunday being tails in Experiment 2, if and only if one sees the red flag upon awakening.
Expected Value of gained money on average: E[gained money] = (1-c)•p-c•(1-p)
= p-c

with p: probability of winning the made bet.
⇒ E[gained money] = p-c
=3/4-1/2=0.25$

with the probability of winning the made bet being here p=3/4=75%, since out of these four equally probable instances {(Monday|red flag|heads), (Monday|red flag|tails), (Tuesday|red flag|tails), (Tuesday|white flag|heads)} one wins in these three instances {(Monday|red flag|tails), (Tuesday|red flag|tails), (Tuesday|white flag|heads)} with the set up rules (I) and (II) for making bets in Experiment 2
and with credence c=1/2=50%, since of course my credence for that coin flip made on Sunday being heads should be that given that coin to be a "fair" coin and according to the principle of maximum entropy.

My mother's credo is and advice for me was, that you should never make bets or gamble, since it is never or usually not rational to do so.
My own credo on the other hand is, that if it is rational to make some bets or gamble, then why the heck wouldn't you make those bets or gamble?
"How Not to Be Wrong: The Power of Mathematical Thinking - with Jordan Ellenberg" by The Royal Institution

Zsolt Nagy said...
This comment has been removed by the author.
Zsolt Nagy said...

Complement and sidenote: We get the following for the standard deviation σ:
σ = ((1-c-(p-c))²•p+(-c-(p-c))²•(1-p))^0.5
= ((1-p)•p)^0.5 = ((1-0.75)•0.75)^0.5 ≈ 0.43$

It's interesting, that in this sense the standard deviation σ (average deviation of gainable money from the expected value of gained money) is independent of any credence c here apparently.
In my opinion an expected value derivation or calculation is not much worth without the appropriate standard deviation derivation or calculation alongside.
So all in all we get here:
Average of gained money from those bets per bet
= (p-c)±((1-p)•p)^0.5 ≈ (0.25±0.43)$

which is very good as for making such dutch book bets based upon a reasonable credence and a reasonable strategy or rules, playing such dutch book bets based upon someone's credence in Experiment 2 here.