## Mathematical Conundrum or Not? Number Five

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Probability is about expectations. Success should be measured not by what proportion of your predictions were accurate, but by what proportion of outcomes you successfully predicted. That won't usually make much of a difference, but SB is skewed so that you get twice the credit when you're right about tails but are only singly penalized for being wrong about heads. If you guess heads all the time, in our guessing version of SB, you get exactly the same proportion of outcomes right as guessing tails, but you get doubly penalized for being wrong so that only 1/3 of your predictions were right.

Because the SB scenario doubles tails outcomes, it is difficult, or at least unnatural, to express your confidence about the outcome through wagering. For instance, suppose the coin is biased 2:1 in favor of heads. You have inside information and are thrilled to be given even money odds. Then we get these results:
100 tosses, 67 heads, 33 tails, 133 interviews.
Betting heads consistently on a biased coin at even money, you break even. WTF?
The sucker who was betting tails, who didn't know the coin was biased? He breaks even too.

Here's the spontaneous version of guessing-SB:
Suppose I'm going to teach Andy & Michael a little about probability. I'm going to flip a coin a bunch of times, but before each flip, they each guess. When they're right they get an M&M, and when we're done we'll count the M&M's and stuff. Now suppose before one toss, Michael guesses "Heads! Heads heads heads heads heads!!!!" If the coin lands heads, do I give him 1 M&M or 6?
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100 tosses, presumed result of 50 heads and 50 tails, 150 interviews.
If the Beauties all guess tails all the time, they will get 100 right out of their 150 answers.

That looks like a 2/3 success rate, right? But is it?

Out of the 100 tosses, they got 50 of them wrong. Looked at this way, that's a 50% success rate.

Yes that's the nature of the experiment. There are two ways of looking at it.

In my view, probability is a measure of the state that the agent is in. Unconditionally, there is a 1/2 chance that Beauty will be in a state associated with heads.

However when conditioning on being awake and interviewed, there is a 1/3 chance that Beauty will be in a state associated with heads.

So the issue between double-halfers and thirders is whether conditioning is valid here.

Here's a variation of the experiment. Suppose that for Tuesday and Heads Beauty is also awakened and interviewed. At every interview she is informed whether or not it is a Tuesday and Heads interview. She knows these rules prior to the experiment. Naturally if she is informed that it is Tuesday and Heads at the interview, she can conclude with certainty that she is in a state associated with heads.

However if Beauty is told that it is not Tuesday and Heads at her interview, should she condition on that information or not?
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Here's the spontaneous version of guessing-SB:
Suppose I'm going to teach Andy & Michael a little about probability. I'm going to flip a coin a bunch of times, but before each flip, they each guess. When they're right they get an M&M, and when we're done we'll count the M&M's and stuff. Now suppose before one toss, Michael guesses "Heads! Heads heads heads heads heads!!!!" If the coin lands heads, do I give him 1 M&M or 6?

Normally you would just give 1 M&M. But it is ultimately a question about what sample space and rules are appropriate in the circumstances.
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What happens on Tuesday&HEADS is a part of the HEADS protocol, so you excluded part of it. — JeffJo

(a) No it isn't. From the OP:

A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends.
Yes, it is. The bolded text tells the lab techs what to do - or more accurately, what not to do - on both days. It defines two protocols for TAILS: interview Monday, interview Tuesday. It defines two protocols for HEADS: interview Monday, sleep Tuesday. Even if they send her home that day, that would still be SOTAI.

My point is that it can't matter what SOTAI is. There is a protocol for Tuesday&HEADS, and in an interview Beauty knows that it is not the protocol that is currently in progress.

The only thing that matters is one for heads and two for tails
What matters is that there is a protocol on both days for both HEADS and TAILS. And that one of these four protocols is inconsistent with Beauty being interviewed. You keep treating the fact that she sleeps through a day as if that makes the day nonexistent,or that it is not something the lab techs have to have included in their protocol.

(b) If it were part of the heads protocol, by eliminating it, you would be eliminating heads as an outcome. Simply being interviewed would tell you the coin landed tails.
?????
In general, I dislike the use of the word "eliminate." People forget that it means "An outcome which was possible has been shown to be incompatible with the current information state."

On Sunday Night, Beauty knows that her information state during the experiment will be limited to a single day's experiences. She knows that there are four possible such states:

1. The single day is Monday and HEADS flipped.
2. The single day is Monday and TAILS flipped.
3. The single day is Tuesday and HEADS flipped.
4. The single day is Tuesday and TAILS flipped.
If she is interviewed, she knows that one has been shown to be incompatible with her current information state.

If that seems like a tendentious interpretation, consider what happens as you increase the number of tails interviews: whatever the ratio, that's your odds it was tails. Do a thousand tails interviews, and it's a near certainty -- according to thirders -- that a fair coin lands tails.
Yep. Get two thousand volunteers. Order them randomly from #1 to #2,000. House #1 thru #1,0000 in the HEADS wing of your lab, and #1,001 thru #2,000 in the TAILS wing. Then flip your fair a coin.

On each of the next 1,000 days, wake all of the volunteers in the wing that corresponds to the coin result, and one - the one whose number corresponds to the day - from the other wing. Ask each of the 1,001 awake volunteers for her confidence that her wing is not the one that corresponds to the coin that was flipped.

Each of these women is in an experiment that is identical - except for the labels you put on coins and days - to what you just described. Each knows that 1,000 awake women came from one wing, and only 1 from the other.

Yes, it is a fair coin flip. But Beauty is not asked about its flip in an information vacuum. She knows that there is a 1/N chance that she would have been interviewed today under one result, but a 100% chance under the other. Her confidence in the first result must be 1/(1+N).
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Here's the thing: it sure does look like the design of the experiment involves conditioning heads on ~Tuesday, so you get (1/4)/(1/2) = 1/2 for heads -- heads ends up by definition (heads & Monday). I'm tempted to say that since this is baked into the design of the experiment, this bit of conditioning has the status of background knowledge, more or less. At any rate, I consider it an open question whether this bit of restricting the space can or should be treated differently from the conditioning that Beauty might do in considering her personal situation.

Lots more to say, but first I want to ask you two about another quickie alternative experiment, which we might have done before, I've lost track.

I toss fair coin twice. I ask for your credence that the first toss landed heads only on {HH, TH, TT}.

My question is this: do you think this is equivalent to SB? And why or why not?
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Sorry, I'm not getting your experiment, or its equivalence to SB.

One thing I'm generally uncertain about is how strongly to lean on "what day today is" being random. There are some things we can say about their equivalence for Beauty, but Elga and Lewis are both pretty cautious about that. I don't think we can just throw a big principle of indifference at "what day today is" and be done.

Here's how I converted from thirderism to halferism. Heads interviews are red marbles, tails interviews are blues.

If you select a marble from an urn with 1 red and 2 blues, sure, chances of getting the red are 1/3. But that is not Beauty's situation. Instead we have two urns, red in one and blue in the other. A coin is tossed to determine which urn to select from. It doesn't even matter how many marbles are in each; your chances of getting red are 1/2. Beauty cannot tell the difference between one interview on heads and any number of interviews on tails, but she knows that each procedure has a 1/2 chance of being followed.
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I toss fair coin twice. I ask for your credence that the first toss landed heads only on {HH, TH, TT}.

My question is this: do you think this is equivalent to SB? And why or why not?

Yes, in the sense that one should condition on being interviewed.

Sorry, I'm not getting your experiment, or its equivalence to SB.

He's simply saying that if you interview different people for each permutation instead of just one person, then the thirder-style (1/(1+N)) result follows. Which is equivalent to the SB scenario.

In terms of your M&M example, if two people guess tails and they are correct, they both get an M&M. To reflect Sleeping Beauty, the experiment is set up such that only one person gets to guess when the outcome is heads. If the person conditions on the fact that they are getting to guess at all, then they will know that they are more likely to be in the tails track.
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Here's a variation of the experiment. Suppose that for Tuesday and Heads Beauty is also awakened and interviewed. At every interview she is informed whether or not it is a Tuesday and Heads interview. She knows these rules prior to the experiment. Naturally if she is informed that it is Tuesday and Heads at the interview, she can conclude with certainty that she is in a state associated with heads.

However if Beauty is told that it is not Tuesday and Heads at her interview, should she condition on that information or not?

I really like this argument. I meant to ask about it myself -- I saw a variation of it on StackExchange a few days ago while I was digging around for other approaches -- but I forgot.

Maybe an even cleaner version is for the additional rule to be: if and only if it's a (Tue & Heads) interview, you will be told at the start that it's a (Tue & Heads) interview; then when Beauty is not told this, she infers ¬(Tue & Heads). Now when you delete the (Tue & Heads) interview, absolutely nothing else changes. You could even run standard Sleeping Beauty by having the experimenter lie, tell her it's Informative-SB, but then never do the (Tue & Heads) interviews. This looks like the perfect way to solve SB by treating it as a special case of something more obviously solvable.

Still some things to puzzle through, but I'm convinced. My sojourn in the land of halferism is at its end.

I do still disagree about how to interpret this thing though. The failure rate of my tails-guessing Beauties is still 1/2, no matter how much they pat themselves on the back. The argument you give here totally justifies conditioning on being interviewed, so the epistemic issue isn't there; it's in this conflict between the two ways of measuring success. Michael (this is my 8 year old, not TPF's Michael) gets 1 M&M and a fatherly lecture on how to measure the success of predictions.

Thanks for hanging so long, @Andrew M (weird, the Andy in my story is my 10 year old, not you). Think I learned some things. Going to take a long break now from Sleeping Beauty.
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Still thinking about how to properly score this thing.

The Lewis table is what you get if you try to compensate for SB's structure by treating the coin itself as biased 2:1 heads:tails. You start with this table
   Mon  Tue
H  1/3  1/3
T  1/6  1/6

drop (H & Tue) and conditionalize on P(H1 ∨ T1 ∨ T2) = 2/3 to get
   Mon  Tue
H  1/2
T  1/4  1/4

So it's true that the Lewis does represent an attempt to "discount" the overabundance of tails, but it does it in the wrong place. You can't mess with the coin.

The only thing to do, to get a better measure of success rate, is to score the results at 2:1, so that you can get a payoff table like this:
    H   T
H   2  -1
T  -2   1

I'm not sure this is sophisticated enough though. What if instead of going all heads/tails, you use a mixed strategy? The payout events (W and L) have the right ratio, but the payout values are still screwy.

Not sure I even need to worry about mixed strategies here though. The coin being fair gives a lower bound to failure of 50% and an upper bound to success of 50%.

Thought I was done, but I'm going to keep thinking about the scoring problem.
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In terms of your M&M example, if two people guess tails and they are correct, they both get an M&M. To reflect Sleeping Beauty, the experiment is set up such that only one person gets to guess when the outcome is heads. If the person conditions on the fact that they are getting to guess at all, then they will know that they are more likely to be in the tails track.

One more point.

If you take a step back, SB looks a bit like a fucked up way of doing two trials of a single experiment. (No worries about the single coin flip -- the trial is asking different subjects for their credence.) But whichever way you split, by toss outcome or by day, it's not two trials: it's one trial each for two different experiments and which experiment is being run is determined by the coin toss, and is thus the source of Beauty's uncertainty.
• 6
Probability is about expectations.
No, it isn't.

Statistics is about expectations. Statistics uses Probability Theory to calculate expectations. In conventional experiments, we find that the value of the expectation is the value of the probability. The concepts of "probability" and "expectation" are very different.

The Sleeping Beauty Problem bends conventionality, and makes statistics inapplicable, by making the number of trials depend on the result for which you are trying to assess a probability. So arguments about frequency and expectation are meaningless until the "bent" issues are resolved.

Specifically: people disagree about whether there are two, three, or four disjoint events that comprise Beauty's sample space. On Sunday Night, the answer is clearly "two" since how the concept "today" applies at that time is different than when she is awakened during the experiment.

Halfers want the number to be "three" inside the experiment. They treat it like Tuesday doesn't happen if Heads is rolled. The problem with the rationale they use, it that it only supports "two," which nobody thinks applies within the experiment. The inconsistency in their argument is that Monday is different than Tuesday if TAILS was flipped, but not if HEADS was flipped. In that case, Tuesday does not exist.

You reject the existence of Tuesday yourself, when you say things like:
it sure does look like the design of the experiment involves conditioning heads on ~Tuesday,
That is not what the condition is. It is {AWAKE}, which is can also be written as ~{HEADS&Tuesday} or {HEADS&Monday,TAILS&Monday,TAILS&Tuesday}.

I've presented an alternative problem that unequivocally demonstrates the solution. You give no reason for "not getting .. its equivalence to SB." It is my opinion that you "do not get it" because doing so would make you change your answer. To "get it," all you need do is let Beauty be wakened on HEADS+Tuesday, but do SOTAI.

One thing I'm generally uncertain about is how strongly to lean on "what day today is" being random.
"Random" is not a property of what you are looking at in an experiment. It is a property of what you know about it, but can't see. Either because the experiment hasn't happened yet, or it has but you can't see what happened.

Here's an example that I've used in this thread, that directly addresses your uncertainty: Forget the coin, and the two possible awakenings. Put Beauty to sleep, and roll a six-sided die. Call the result of the roll R. Leave her asleep on five days during the following week, waking her only after R nights (so R=1 means Monday, R=2 means Tuesday, etc.). When she is awake, ask her for her confidence that today is Wednesday.

Is the answer 1, meaning "what day today is, is not random," or is the answer 1/6, meaning "what day today is, is random" ?

The reason it is random, is because Beauty has no evidence of what day it could be. Does this sound anything like the Sleeping Beauty Problem to you?
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Thanks. We've moved on here, but I appreciate your thoughts. You've addressed a lot of the gaps in my understanding of this stuff, and I especially appreciate you taking the time to do that.

Your scheme for removing the time element is pretty cool, and I'm going to spend some more time looking at it.
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Still some things to puzzle through, but I'm convinced. My sojourn in the land of halferism is at its end.

Welcome back!

I do still disagree about how to interpret this thing though. The failure rate of my tails-guessing Beauties is still 1/2, no matter how much they pat themselves on the back. The argument you give here totally justifies conditioning on being interviewed, so the epistemic issue isn't there; it's in this conflict between the two ways of measuring success.

I regard them as two distinct epistemic perspectives. One is Beauty's on Sunday (or Wednesday) who doesn't condition on being interviewed and the other is Beauty's on Monday or Tuesday who does.

If you take a step back, SB looks a bit like a fucked up way of doing two trials of a single experiment. (No worries about the single coin flip -- the trial is asking different subjects for their credence.) But whichever way you split, by toss outcome or by day, it's not two trials: it's one trial each for two different experiments and which experiment is being run is determined by the coin toss, and is thus the source of Beauty's uncertainty.

Yes. You could say the setup is that there is one of two experiments which is randomly selected. In one experiment, only one random person gets interviewed. In the other, everyone does. You find yourself getting interviewed. So what is the probability that you're in the single interview experiment? 1/(1+N) where N is the number of people.

As a result, it is structurally unlike a biased coin or regular betting scenario. There is no one-to-one mapping when the coin toss outcome itself has a different number of interviews (or agents) associated with it. Instead the sample space must properly account for those degrees of freedom with probability being the measure of which state an agent is currently in.
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Welcome back!

It was worth the wandering just for the return. I'm still kind of stunned by the elegance of the argument that convinced me. The symmetry of it. In standard SB, on (H & Tue) our Beauty receives no information at all, is not even conscious; in Informative-SB, she receives nearly all the possible information. We gather all of it together into one box -- and then close the lid. Just like that, transforming one into the other.

It's quite beautiful. And a fine reminder to look for the general problem of which the one at hand is only a special case.
• 6
In standard SB, on (H & Tue) our Beauty receives no information at all.
But she isn't asked for her confidence in that situation, so this argument is a red herring. A very appealing red herring, as you go on to describe, but irrelevant nonetheless.

The issue is, does she receive information on ~(H & Tue)? And the answer can't depend on how, or even if, she would receive information on (H & Tue).

So, instead of the (H & Tue) protocol being "let her sleep," make it "take her to DisneyWorld. Just before she finds out where she is going, the Law of Total Probability then says:

Pr(H) = Pr(H|Interview)*Pr(Interview) + Pr(H|DisneyWorld)*Pr(DisneyWorld)

Here,

Pr(H) = 1/2 is the probability of Heads.
Pr(Interview) = X is the probability she will be taken to an interview.
Pr(DisneyWorld) = 1-X is the probability she will be taken to DisneyWorld.
Pr(H|Interview) = Y is the probability of Heads in an interview (what the original problem asks for).
Pr(H|DisneyWorld) = 1 is the probability of Heads at DisneyWorld.

So we can now say that:
1/2 = Y*X + (1-X)
(1-Y)*X = 1/2

If Y=1/2, as you believe, that means X=1. This is a contradiction, since it means she cannot be taken to DisneyWorld. Whether or not you accept that Y=1/3, the fact that there is a chance of going to DisneyWorld proves that Y<1/2.
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