Undergrad The Sleeping Beauty Problem: Any halfers here?

[stevendaryl]

I may well be, but then you cannot even explain your statement that she can test her amnesia or "update her information". A simple question, that you cannot even provide any answer for.

I gave you the formula:

P(heads | awake) = P(heads | awake & Monday) P(Monday | awake) + P(heads | awake & Tuesday) P(Tuesday | awake)

That formula is a theorem of conditional probability (together with the assumption that it's either Monday or Tuesday). What else do I need to explain about it?

 

[Stephen Tashi]

Oh my, there is so much misinformation and misunderstanding in his thread. I am amazed.

The main source of confusion is that "the" Sleeping Beauty Problem has not been stated as a specific mathematical problem. People are offering solutions without stating which mathematical interpretation of the problem they are solving.

Questions about "What should sleeping beauty think?" are not well posed mathematical problems because what a person "should" think is subjective.

A question of the form "What is the probability that...?" has some chance of being a well posed mathematical problem if the event in the question is precisely defined.

 

[PeterDonis]

The exact answer P(heads | awake) = 1/3 requires additional justification, but concluding that it's less than 1/2 doesn't seem to.

Yes, I agree with that: unless Beauty is certain that it's Monday when she's awakened, the conditional probability P(Heads|Awake) must be less than 1/2.

 

[stevendaryl]

What if the alternative to the if ? Is she is asked question in her sleep ?

You don't have to have an alternative in order for a statement of the form "If A then B" to make sense. A statement of the form "If A then B" is true in case B is true or A is false.

In our case, A is "Sleeping Beauty is awake". B is "It is either Monday, or it is Tuesday and the coin toss result was tails"

But you're saying "If A then B" is false in this case. By the laws of logic, that is only possible if A is true and B is false. So you are saying that it is false that "It is either Monday or it is Tuesday and the coin toss result was tails"?

 

[PeterDonis]

Questions about "What should sleeping beauty think?" are not well posed mathematical problems because what a person "should" think is subjective.

I wouldn't quite say it's "subjective" (although the term "subjective credence" is indeed used in the Wikipedia article's statement of the problem). The issue is that it depends on which (objective) thing is the one the person being asked the question is supposed to use to determine their answer.

For example, the interpretation of "subjective credence" as which bets the person would or would not take means that the person has to know the payoffs, as I pointed out in response to @Dale . And the original problem statement does not include payoffs, so it is incomplete on this interpretation. But the conditional probabilities, which are also involved in a rational determination of which bets to take, are still perfectly objective--unless one takes there to still be some uncertainty possible about P(Monday|Awake). And once the payoffs are specified, the rational calculation of which bets to take is also perfectly objective.

 

[Stephen Tashi]

But the conditional probabilities, which are also involved in a rational determination of which bets to take, are still perfectly objective-.

Can we clarify things by having a statement of which conditional probabilities are to be computed?

In particular, what precisely are the "given" events involved in the conditional probabilities?

 

[JeffJo]

I'm going slow to prevent any possibility disagreement. Consider these variations:

1) The day doesn't matter. So roll a die at the same time you flip the coin (don't show it to her). If they are (Heads, Even), let her sleep through Tuesday, as in the original experiment. If (Heads, Odd), she sleeps through Monday, and awakened Tuesday. Does anybody think her answer changes? I hope not. Call it Z.

2) The coin result only affects the question. So in addition to adding the die, flip two coins instead of one. Say, a dime and a quarter. Wake her once if they are the same, and twice if they are different; the day she sleeps is still determined by the die. Then ask her for her confidence that the coins showed the same face.

2A) If you show her the dime, and it is heads, we have variation (1). And its answer, Z.

2B) If you show her the dime, and it is is tails, we have a problem with an equivalent solution, so the same answer, Z.

2C) If you don't show her the dime, she can treat it as a random variable. The law of total probability says her answer is Z*Pr(DIME=H)+Z*Pr(DIME=T)=Z.

3) Make it simpler. Write (H,Mon), (H,Tue), (T,Mon), and (T,Tue) on four cards. Pick one at random when you flip the coin. Don't show it to Beauty, but she sleeps through the day on the card, if the coin matches the card. Ask her for her confidence in that match. This is really the same problem as (3), so the answer is, again, Z.

But this problem can be solved, because an awake Beauty does have new information. One card is ruled out, she just doesn't know which. But she can identify it with a change of variables. Instead of "Mon" and "Tue", use "Today" and "Other Day". Instead of "H" and "T", use "Up" and "Down".

Now the four cards say (Up,Today), (Up,Other), (Down,Today), and (Down,Other). She was dealt one at random, and now knows it wasn't (Up, Today). She is asked for her confidence that the card she was dealt says "Up", and that is unambiguously 1/3.

 

[stevendaryl]

Can we clarify things by having a statement of which conditional probabilities are to be computed?

In particular, what precisely are the "given" events involved in the conditional probabilities?

There are three (interdependent) boolean variables that completely describe the situation on any day:

1. heads (if heads is false, that means the result was tails)
   
2. monday (if monday is false, that means today is tuesday)
   
3. awake (if awake is false, that means she is asleep)

I would say that the number to be computed is:

P(heads | awake)

If you're willing to formulate the problem in terms of those three variables, then you can immediately write down the equation:

P(heads | awake) = P(heads | monday & awake) P(monday | awake) + P(heads | tuesday & awake) P(tuesday | awake)

The real dispute seems to be over the conditioning on Sleeping Beauty being awake. As far as Sleeping Beauty is concerned, she never consciously experiences not being awake, so there is no reason to mention that variable at all. I think that's pretty silly. In conditional probability, there is no harm in conditioning on an always-true condition: P(X | true) = P(X). So it can't hurt anything to include the "awake" condition.

But the halfers would say not to mention that variable at all. In that case, then the formula becomes:

P(heads) = P(heads | monday) P(monday) + P(heads | tuesday) P(tuesday)

At this point, the halfers are facing a mathematical contradiction: They would say:

P(heads) = 1/2
P(heads |monday) = 1/2
P(heads | tuesday) = 0 (it's impossible)

So the formula boils down to

1/2 = 1/2 P(monday)

which implies that P(monday) = 1; it's impossible for it to be Tuesday. That's a nonsensical result. Of course, it's possible for it to be Tuesday.

 

[PeterDonis]

I would say that the number to be computed is:

P(heads | awake)

This is the one we have been focusing on, but note that unless we interpret "subjective credence that the coin came up heads" to mean this conditional probability, it is not necessarily one that we need to compute

For example, if we interpret "subjective credence" to mean which bets would or would not be taken, then we have to know the payoffs, as I said before, and we have to compute every conditional probability associated with a payoff, in order to compute a weighted expectation. Keeping the payoff definitions as general as possible, we would have

Payoff(Bet Heads) = P(Heads|Monday) P(Monday|Awake) Payoff(Bet Heads|Monday & Heads) + P(Tails|Monday) P(Monday|Awake) Payoff(Bet Heads|Monday & Tails) + P(Tails|Tuesday) P(Tuesday|Awake) Payoff(Bet Heads|Tuesday & Tails)

where no "Tuesday & Heads" term appears because that possibility is ruled out by the problem statement. Note that, because of the weighting by payoffs, this computation is not the same as the computation of P(Heads|Awake). A similar formula would apply for Payoff(Bet Tails).

 

[Stephen Tashi]

There are three (interdependent) boolean variables that completely describe the situation on any day:

Ok, but we should really start by defining the "probability spaces" that are involved.

Presumably, combinations of values of the variable define events in a probability space. But talking about value of a variable on "any day" seems to imply the events in the probability space involve days or sequences of days.

 

[Dale]

Just the prior of heads doesn't answer the question, though. We can write:

P(heads|awake) = P(heads | awake & monday) P(monday | awake) + P(heads | awake & tuesday) P(tuesday | awake)

The second term is zero (since it's impossible for it to be heads if Sleeping Beauty is awake on a Tuesday). So we have:

P(heads | awake) = P(heads | awake & monday) P(monday | awake)

At this point, I would say that P(heads | awake & monday) = P(heads). Knowing that you are awake and that it is Monday doesn't tell you anything about whether the coin is heads or tails. So we have finally:

P(heads | awake) = P(heads) P(monday | awake) = 1/2 P(monday | awake)

So the two conditionals I think are closely related.

I do agree that they are closely related, but they are not the same. One of the problems in this thread is the wide variety of alternative scenarios proposed, which seem to be having the opposite effect as intended regarding clarifying the original scenario. So I suggest sticking to the scenario as specified in the Wikipedia article. There Beauty is asked her credence about the coin being heads. So the clear appropriate prior would be the 0.5 prior probability that a fair coin toss is heads. The prior probability to which day it is is not as clear and not necessary for the problem.

 

[Boing3000]

I gave you the formula:

P(heads | awake) = P(heads | awake & Monday) P(Monday | awake) + P(heads | awake & Tuesday) P(Tuesday | awake)

That formula is a theorem of conditional probability (together with the assumption that it's either Monday or Tuesday). What else do I need to explain about it?

You could simple have answered the question "explain your statement that she can test her amnesia or "update her information", instead of running in circle a get back to this formula which does not apply to the Sleeping Beauty problem.
We are not asked about probability of her being awake any day. She is ask only when she is awake, some questions about a coin flipped once. Why bother if her answer may not change from day to day ?

You would not even have taken the time to read the problem, if Peter hasn't forced you to (thanks to his credence). Read the problem, and try to understand it, your formula does not apply to it, because the problem is NOT defined enough for that (also read post#67)

If P(heads | Monday & awake) = 1/2 and P(Monday | awake) < 1, then it follows that P(heads | awake) < 1/2

And again, P(Monday | awake) = 1, because of the drug.

 

[PeterDonis]

Beauty is asked her credence about the coin being heads. So the clear appropriate prior would be the 0.5 prior probability that a fair coin toss is heads. The prior probability to which day it is is not as clear and not necessary for the problem.

I'm not sure I understand. If the only prior that is relevant is P(Heads) = 1/2, and no probabilities related to which day it is are relevant, then you are basically agreeing with the halfer argument. Is that your intent? Or are you just saying that P(Monday|Awake) = 2/3, which is the crucial factor needed for the thirder's argument, is not a "prior" but something else?

 

[PeterDonis]

P(Monday | awake) = 1, because of the drug

This is not correct. P(Monday|Awake) is the conditional probability that it is Monday, given that Beauty is awake. It is not the conditional probability that Beauty is awake, given that it is Monday; that would be P(Awake|Monday). The latter conditional probability is indeed 1, but it is not the one being used in the argument.

 

[Dale]

A big part of solving any problem is recognizing the best formulation of a problem. In this case, the best formulation would be Bayes rule stated in odds form:
$$O(H:T|A) = O(H:T) \frac{P(A|H)}{P(A|T)}$$
Where H is the coin landed heads, T is the coin landed tails, A is Beauty is awakened during the experiment (i.e. with amnesia, being interviewed, and being asked her credence that it is heads).

Since the coin is fair O(H:T) is 1 (1:1 odds). The conditional probability P(A|H) is strange, but it is actually not important. What is important is the ratio of P(A|H)/P(A|T), which is clearly 1/2. So then O(H:T|A) = 1/2 (2:1 odds against H), which is a conditional probability of 1/3 for H.

The information is clear, the prior is clear, and the day of the week doesn't even need to enter into the calculation.

 

[PeterDonis]

What is important is the ratio of P(A|H)/P(A|T), which is clearly 1/2.

Ah, I see. The day of the week only plays an indirect role, in deriving this ratio, and the conditional probability for it being a particular day of the week does not enter into it. I agree this is a much cleaner formulation for the odds. However, I'm wondering how you would compute the expected payoff for bets in this way in a scenario like the second one in post #67 (where only one bet is paid off even if Beauty is awakened twice).

 

[Boing3000]

This is not correct. P(Monday|Awake) is the conditional probability that it is Monday, given that Beauty is awake. It is not the conditional probability that Beauty is awake, given that it is Monday; that would be P(Awake|Monday). The latter conditional probability is indeed 1, but it is not the one being used in the argument.

I agree, but my statement is : the drug only use, in the context in this exact scenario, is to make P(Monday) = 1 => P(Monday|Anything) = 1 (anything including no coin flipped at all)
If not, there is no reason to put beauty to sleep and ask her anything. She can answer everything Sunday evening, and do something better with her live, because no information appears past Sunday evening. I cannot get Dale or Steven to acknowledge that.

 

[Dale]

But just knowing the probabilities isn't enough. You also have to know the payoffs attached to each possible outcome. In other words, you are betting on the weighted expectation of the payoff.

Yes, this is true. What I mean is that a rational assessment of the probability of an event must be the same probability as used to calculate the payoff odds at which the rational assessor would be indifferent to a wager on that event occurring.

 

[PeterDonis]

the drug only use, in the context in this exact scenario, is to make P(Monday) = 1 => P(Monday|Anything) = 1 (anything including no coin flipped at all)

I don't understand. What do you mean by P(Monday) = 1 => P(Monday|Anything) = 1?

 

[stevendaryl]

You could simple have answered the question "explain your statement that she can test her amnesia or "update her information", instead of running in circle a get back to this formula which does not apply to the Sleeping Beauty problem.

So you are rejecting a standard formula of the theory of probability. That's what I thought, but you earlier denied rejecting probability theory.

 

[PeterDonis]

What I mean is that a rational assessment of the probability must be the same probability as used to calculate the payoff odds at which the rational assessor would be indifferent.

Yes, but which odds are those? That will depend on how the payoffs are structured. In the two scenarios described in post #67, the payoffs are structured such that different odds are relevant:

In the first scenario, where all bets made are paid off at the end, the relevant odds are O(H:T|A).

In the second scenario, where only the last bet made is paid off at the end, and any other bets made are discarded, the relevant odds are just O(H:T).

 

[Dale]

However, I'm wondering how you would compute the expected payoff for bets in this way in a scenario like the second one in post #67 (where only one bet is paid off even if Beauty is awakened twice).

That is actually a malicious wager, and I don't know how to compute odds with malicious agents who refuse to honor certain bets. I would suspect that such a computation should come out to 1/2 in this case, but that is a gut feeling with no analysis behind it.

But that is not the scenario considered in the Wikipedia description where Beauty is asked her credence "now".

 

[Boing3000]

I don't understand. What do you mean by P(Monday) = 1 => P(Monday|Anything) = 1?

I mean: no recollection of monday while having recollection of sunday make you certain that your are monday.
That's the world Beauty is in. The opinions of outsider are useless to her.

 

[stevendaryl]

I mean: no recollection of monday while having recollection of sunday make you certain that your are monday.
That's the world Beauty is in. The opinions of outsider are useless to her.

That's clearly wrong. The fact that she doesn't know whether it is Monday or Tuesday does not imply that it is Monday.

 

[stevendaryl]

And again, P(Monday | awake) = 1, because of the drug.

Do you really believe that? Sleeping Beauty knows the rules of the experiment. So she knows upon wakening that there are three possibilities:

1. It is monday, and the coin flip result was heads
2. It is monday, and the coin flip result was tails
3. It is tuesday, and the coin flip result was tails.

You're saying that her conclusion is: Today must be Monday?

 

[PeterDonis]

no recollection of monday while having recollection of sunday make you certain that your are monday

No, it doesn't, because she is in exactly the same position on Tuesday, since the drug erases her recollection of Monday.

 

[Boing3000]

However, I'm wondering how you would compute the expected payoff for bets in this way in a scenario like the second one in post #67 (where only one bet is paid off even if Beauty is awakened twice).

I have modified the code to simulate post#67 setup. Second parameter is Beauty strategy likelihood to bet on tail

(function BeautyBets67(run, beautyGuess, betAmount, rule) {
    var counts = { Wins: 0 }
    var experiments = [
            function headBet(counts) {
                var headBet = Math.random() >= beautyGuess;
                if (headBet)
                    counts.Wins += betAmount;
                else
                    counts.Wins -= betAmount;
            },
            function Tail(counts) {
                var tailBet = Math.random() < beautyGuess;
                if (tailBet)
                    counts.Wins += betAmount;
                else
                    counts.Wins -= betAmount;

                if (rule !== "B") {
                    if (tailBet) {
                        counts.Wins += betAmount;
                    }
                    else {
                        counts.Wins -= betAmount;
                    }
                }
            }
    ];
    while (--run >= 0) {
        experiments[Math.floor(Math.random() * experiments.length)](counts);
    }
    alert("Money " + counts.Wins);
})(10000, 0.5, 100, "A");

 

[Boing3000]

No, it doesn't, because she is in exactly the same position on Tuesday, since the drug erases her recollection of Monday.

Which is exactly the reason why she is drugged, to make her certain that she is Monday, which is the exact same position (your words)

 

[stevendaryl]

Ok, but we should really start by defining the "probability spaces" that are involved.

That's the challenge: to figure that out.

The original problem mixes up two different notions of possibility (which maybe is the source of the confusion?):

1. There are alternate "possible worlds": One where the coin flip result is heads, and one where the coin flip result is tails.
2. There are alternate days within one world.

(awake versus asleep is a function of the other two variables). So there are 4 possibilities:

1. (tails, monday, awake)
2. (tails, tuesday, awake)
3. (heads, monday, awake)
4. (heads, tuesday, asleep)

From Sleeping Beauty's point of view, she wakes up and knows that she is in situations 1-3 (she can't actually experience #4). She is uncertain about what day it is, and she is uncertain about what the coin flip result was. The point of the problem is to come up with a sensible way to quantify these uncertainties. So from a Bayesian reasoning point of view, she's being asked to come up with sensible prior probabilities.

 

[stevendaryl]

Which is exactly the reason why she is drugged, to make her certain that she is Monday, which is the exact same position (your words)

The point of the drug is so that she can't tell the difference between Monday and Tuesday. It's not to fool her into thinking it's Monday when it isn't. She knows that it's possible that she's been drugged.

There is a difference between not knowing what day it is and falsely believing that it is Monday when it's not.