Mathematical Conundrum or Not? Number Five

Srap Tasmaner

1 point for each would also be a Dutch book wouldn't it? — Michael

Missed this.

Yes. That was exactly my argument: if you offer even money, I am guaranteed a profit.

Michael

And I seriously don't understand how you can't see the 50% argument; disagreeing with it is one thing but it has been explained to you at this point numerous different ways. Either you see it and are too stubborn to let on that you see it, or it is going right over your head.

Michael

She knows that P(Heads) = 0.5 and that P(Monday|Heads) = 1. So she knows that P(Heads ∧ Monday) = 0.5. — Michael

This right here is the most succinct explanation.

Michael

↪Jeremiah

Then I think you're wrong in suggesting that both are valid.

The general multiplication rule, from here is:

P(A and B) = P(A) * P(B|A)

Where A is heads and B is Monday:

P(heads and Monday) = P(heads) * P(Monday|heads)

P(heads and Monday) = 0.5 * 1 = 0.5

What rule allows you to get P(heads and Monday) = 1/3? If there were one wouldn't that mean that we're using an inconsistent system?

Srap Tasmaner

I understand how it works, now that I finally got a moment to do the wagering argument.

By betting tails, I get to double what I risk only when my profit is guaranteed to double. I should not be allowed to do that.

I also understand now why @andrewk remembered that it all turns on what you mean by "credence". The weirdness here is that even though my own description has the two scenarios arising half the time each -- fair coin, after all -- I should bet "as if" the odds are really 2-1 against heads.

This is annoying, because we (Bayesians, subjective probabilists of all stripes, and the Bayes-curious like me) want to use wagering to measure degree of belief. If you can show that I must wager in a way that systematically deviates from my degree of belief, that's trouble.

So there's a good meaty philosophical issue waiting at the end of this one, which I will now ponder (and probably read Lewis). Again, nice thread @Jeremiah.

Michael

Adam Elga's reasoning from the Wikipedia article is this:

Suppose Sleeping Beauty is told and she comes to fully believe that the coin landed tails. By even a highly restricted principle of indifference, her credence that it is Monday should equal her credence that it is Tuesday since being in one situation would be subjectively indistinguishable from the other. In other words, P(Monday | Tails) = P(Tuesday | Tails), and thus

P(Tails and Tuesday) = P(Tails and Monday).
Consider now that Sleeping Beauty is told upon awakening and comes to fully believe that it is Monday. She knows the experimental procedure doesn't require the coin to actually be tossed until Tuesday morning, as the result only affects what happens after the Monday interview. Guided by the objective chance of heads landing being equal to the chance of tails landing, it should therefore hold that P(Tails | Monday) = P(Heads | Monday), and thus

P(Tails and Tuesday) = P(Tails and Monday) = P(Heads and Monday).
Since these three outcomes are exhaustive and exclusive for one trial, the probability of each is one-third by the previous two steps in the argument.

But let's apply the general multiplication rule from above:

P(Tails and Monday) = P(Tails) * P(Monday|Tails) = 0.5 * 0.5 = 0.25
P(Tails and Tuesday) = P(Tails) * P(Tuesday|Tails) = 0.5 * 0.5 = 0.25
P(Heads and Monday) = P(Heads) * P(Monday|Heads) = 0.5 * 1 = 0.5

But he says that they're all equal. So where has he gone wrong? It's in saying that P(Tails | Monday) = P(Heads | Monday).

P(Heads and Monday) = P(Monday) * P(Heads|Monday)

From the above, we know that P(Heads and Monday) = 0.5, so:

0.5 = P(Monday) * P(Heads|Monday)

What's the probability that it's Monday? It's tempting to say 2/3, but that would be wrong. From the above, we know that P(Tails and Monday) = 0.25 and that P(Heads and Monday) = 0.5. So it must be that P(Monday) = 0.75.

This then gives us:

0.5 = 0.75 * P(Heads|Monday)

Meaning that P(Heads|Monday) = 2/3. And the same reasoning gives us P(Tails|Monday) = 1/3. This is Lewis' conclusion in response to Elga.

Jeremiah

↪Michael

I already posted both of their responses.

Michael

What's the probability that it's Monday? It's tempting to say 2/3, but that would be wrong. From the above, we know that P(Tails and Monday) = 0.25 and that P(Heads and Monday) = 0.5. So it must be that P(Monday) = 0.75. — Michael

To better explain this, imagine I toss a coin. If it's heads then I give you a red ball, and if it's tails I toss again. If it's heads then I give you a red ball and if it's tails then I give you a blue ball. What's the chance that I'm given a red ball? 0.75.

In our case, the red ball is Monday and the blue ball is Tuesday.

andrewk

My policy is to wager $1 on tails whenever I'm asked. You're paying even money.
If the toss is heads, I lose $1; this happens half the time, so my expected loss is $0.50.
If the toss is tails, I make $1 each time I'm asked; this scenario happens half the time, so I have an expected profit of $1.
So I make at least $0.50 on average each time I play, no matter how the toss goes.
I have made a Dutch book against you. — Srap Tasmaner

This is Betting Game 2 from this post.

My calculation is that expecting that profit is the same as if there were a single $1 bet at even odds that the result of a coin toss will be tails and the coin had a 3/4 probability of coming up tails. To see this, note that in that case the expected profit is

$1 x 3/4 + (-$1) x 1/4 = $0.50

Hence, in interpreting the term 'degree of belief' it seems reasonable to make it the probability of tails when betting on a single flip of an unfair coin, at which one would have the same expected profit. So under this interpretation of 'degree of belief', the answer is 3/4 for tails, and hence 1/4 for heads.

Examiner tosses a fair coin, and then tosses another. If the first toss was tails, she asks Beauty her credence that the first toss came up heads; if the first toss was heads she only asks for Beauty's credence if the second toss was heads as well, otherwise the round is over.

Done this way, Beauty will know that when she was not asked the first toss was heads, but she can do nothing with that knowledge. She's not asked and the round is over. What matters is that she's always asked when it was tails and asked half the time when it was heads. So her credence that it was heads should be 1/3. — Srap Tasmaner

I can't quite follow this. I think there are a couple of grammar glitches in it that make it hard to understand what it is saying. Can you please expand on it, maybe using bullet points for clarity, and explain in what way you find it equivalent to the original question?

thanks

andrewk

Any takers on the challenge to define a probability space in which the statement 'today is Monday' corresponds to a well-defined Event?

noAxioms

So, I ran 100,000 games and gave 1 point for successfully guessing heads and 0.5 points for successfully guessing tails (because you get two opportunities). It doesn't matter if you always select heads, always select tails, select tails 1/2 the time, or select tails 2/3 of the time. The average score is 0.5 in every case. — Michael

Ouch... Does miss Beauty know that she's getting fewer points on some of the bets. In that case the outcome is certain and she can win every bet. If not, she's not the one doing the gambling.

This whole betting thing demonstrates the correct answer. It only needs to be done twice, not 100000 times, because there are only two unique cases occurring in equal probability.
I wake up and am expecting to bet a coin with even odds. I bet tails and win 2 coins (one each day) if it is tails, and lose one coin on Monday if it is heads. Sounds like a winning bet to me.

If the stakes depend on the coin toss, then I bet tails if the stakes are 0.5 coins per bet, and heads if the stakes are 1 coin in the bet. I win every time in that case.

Michael

I already posted both of their responses. — Jeremiah

Lewis has the same answer to me, but it looks like I got there a different way. His depends on a premise that Elga rejects whereas mine doesn't. I only apply the general multiplication rule of probability.

Michael

This whole betting thing demonstrates the correct answer. It only needs to be done twice, not 100000 times, because there are only two unique cases occurring in equal probability.
I wake up and am expected to bet a coin with even odds. I bet tails and win 2 coins if it is tails, and lose one coin if it is heads. Sounds like a winning bet to me. — noAxioms

Sure, but it doesn't mean that tails is twice as likely to occur as heads, which is why these betting examples miss the point. All the betting examples show is that it's better to bet on whichever outcome provides more payouts, which is obvious.

Jeremiah

↪Michael

This was my prior response, that I already posted before a few pages back.

I am not convinced it contradicts the 1/3, but instead just leads back to the initial problem of the inherited subjectivity in the notion that you should start with a prior belief and update that prior into a posterior belief when you get new information. This carries that subjectivity into any posterior, and as we see here those priors can look very different based on the viewer. Not really objective science. — Jeremiah

Michael

Any takers on how to define a probability space in which the statement 'today is Monday' corresponds to a well-defined Event? — andrewk

Is that what I've done above?

P(Tails and Monday) = P(Tails) * P(Monday|Tails) = 0.5 * 0.5 = 0.25
P(Heads and Monday) = P(Heads) * P(Monday|Heads) = 0.5 * 1 = 0.5

Therefore, P(Monday) = 0.75

noAxioms

Sure, but it doesn't mean that tails is twice as likely to occur as heads, — Michael

It is twice as likely to occur to miss Beauty. Tails happens twice, and heads only once.

Michael

It is twice as likely to occur to miss Beauty. Tails happens twice, and heads only once. — noAxioms

It doesn't happen twice to her. It only happens once to her, given that it was only tossed once. She just wakes up to it twice.

noAxioms

It doesn't happen twice to her. It only happens once to her, given that it was only tossed once. She just wakes up to it twice. — Michael

And in what way is that not happening twice to her? She gets to bet twice in that case, despite the fact that she is unaware of which times she's betting twice.

Srap Tasmaner

This is Betting Game 2 from this post. — andrewk

Nearly.

Game 2: At each interview, Beauty bets $1 to guess what coin came up, and loses that dollar if wrong or wins $2 if right. — andrewk

This is why I said I couldn't understand how you were setting the odds. Doing it this way is paying off 2-1 on both heads and tails, which is incoherent. (Unless you meant both tails interviews would taken together pay off $2, but that's still incoherent.)

My calculation is that expecting that profit is the same as if there were a single $1 bet at even odds that the result of a coin toss will be tails and the coin had a 3/4 probability of coming up tails. To see this, note that in that case the expected profit is

$1 x 3/4 + (-$1) x 1/4 = $0.50

Hence, in interpreting the term 'degree of belief' it seems reasonable to make it the probability of tails when betting on a single flip of an unfair coin, at which one would have the same expected profit. So under this interpretation of 'degree of belief', the answer is 3/4 for tails, and hence 1/4 for heads. — andrewk

Yeah that's really interesting. On the one hand, my pass at making a fair book says the odds should be 2-1 against heads. But the expected profit on betting tails clearly matches the 3-1 version.

Did I get the fair book wrong, or is there some other explanation for why they're coming out the same? (The payoff system here is so wonky, I think this could be another side effect of that, but I can't see how.)

Either way -- 2-1 or 3-1 -- offering even odds in this game loses money. We agree on that, right?

How to interpret this as credence or degree of belief, I'm not sure yet.

Can you please expand on it — andrewk

I guess I could do this again -- I wouldn't mind, and will if you're really interested, but to me it's no longer relevant (if it ever was). I was still trying to figure out how the thing works when I wrote that.

Srap Tasmaner

↪Michael

I don't understand at all what you mean by P(Monday|Heads) and P(Monday|Tails).

andrewk

Game 2: At each interview, Beauty bets $1 to guess what coin came up, and loses that dollar if wrong or wins $2 if right — andrewk

This is why I said I couldn't understand how you were setting the odds. Doing it this way is paying off 2-1 on both heads and tails, which is incoherent. (Unless you meant both tails interviews would taken together pay off $2, but that's still incoherent.) — Srap Tasmaner

I don't know betting terminology, so I may have used the wrong words. What I meant by 'wins $2' is that she gets her own dollar back, plus another dollar. Perhaps the correct betting terminology for that is 'wins $1'. My ignorance of gambling terminology is gargantuan.

Either way -- 2-1 or 3-1 -- offering even odds in this game loses money. We agree on that, right?

Yes I think we can agree that if a bookmaker offers Beauty Game 2, which has even odds, Beauty can select a strategy under which the expected value of the bookmaker's profit is negative.

Srap Tasmaner

↪andrewk

Yeah, that's right. When you win, you get your stake back and the odds represent your profit. So 2-1 against heads gives you back the $1 and pays you $2 more, total of $3. On tails, you'd get your dollar plus $0.50, total of $1.50. Losing, you lose your dollar. (I only know as much about gambling as I need to to understand philosophers who talk about probability this way!)

Srap Tasmaner

So put your money where your mouth is. What should you bet? I say bet £1 on heads. — Michael

Sure, but it doesn't mean that tails is twice as likely to occur as heads, which is why these betting examples miss the point. All the betting examples show is that it's better to bet on whichever outcome provides more payouts, which is obvious. — Michael

The betting matters, and be honest: you were happy enough to use wagering arguments when it suited you. (What's more the examples don't show what you say is obvious; they assume it and it's irrelevant here anyway.)

Why does the betting matter?

What I'm saying is that there is no reason for her to have a greater belief that it was tails than heads. When she's asked what her belief is that it was heads the rational answer is 50:50. — Michael

Because of this point of @andrewk's:

in interpreting the term 'degree of belief' it seems reasonable to make it the probability of tails when betting on a single flip of an unfair coin — andrewk

Suppose Beauty is not told all the rules, the amnesia pills, etc. She's told she'll get even money -- she has to be told the odds to bet. Given enough trials -- more than is reasonable, I know, it's a story -- she'd conclude from her accumulated payoffs that the coin is unfair.

Now that's quite curious. If betting is a stand-in for hypothesis testing, it'll support a theory that we, in our omniscience, are inclined to call "false" but it will serve Beauty perfectly. Beauty will have plenty of evidence to support her theory.

It'd be interesting to flesh out competition between the two theories. If Beauty doesn't learn what's really going on, she would be incredulous that we would claim the coin is fair. Even if we explained, she might see that our theory is as adequate to the evidence gathered so far as hers, but she'd have no reason to switch to our view without new evidence, something like witnessing new trials showing the coin to be much closer to fair than she thought. That whole process is really cool.

PS: Our explanation is spectacularly more outlandish than a coin being unfair.

Andrew M

Assume she's woken on Monday if it's heads or Tuesday and Wednesday if it's tails.

Do you agree that P(Monday|Awake) = 1/2? — Michael

No, 1/3.

Both the halfer and the thirder positions are consistent. The difference stems from how the probabilities are distributed when conditioning on being awake.

Halfer:

         Mon  Tue
  Heads  1/2  0
  Tails  1/4  1/4

P(Heads|Awake) = 1/2
P(Heads|Monday) = P(Heads and Monday) / P(Monday) = 1/2 / 3/4 = 2/3

Thirder:

         Mon  Tue
  Heads  1/3  0
  Tails  1/3  1/3

P(Heads|Awake) = 1/3
P(Heads|Monday) = P(Heads and Monday) / P(Monday) = 1/3 / 2/3 = 1/2

One characteristic of the thirder view is that it doesn't imply an external perspective that "knows" what the probability is for Beauty's current awake state. Since Beauty has no available information distinguishing the three states from her point-of-view, she is simply indifferent about which state she is currently in, and so assigns a probability of 1/3 for each awake state.

So the thirder view can be preferred on pragmatic grounds.

Andrew M

Except in this, and in the Monty Hall problem, there is no new information. — tom

In the Monty Hall problem, the host gives you information that changes the probabilities that you assign to each door. That information is new to you.

Similarly, in the Sleeping Beauty problem, awakening provides information that enables you to rule out one of the four states. However since you have no information distinguishing the remaining states, you should be indifferent about which state you are currently in.

Michael

I don't understand at all what you mean by P(Monday|Heads) and P(Monday|Tails). — Srap Tasmaner

The probability that it's Monday given the fact that it's heads/tails.

Michael

She gets to bet twice in that case, despite the fact that she is unaware of which times she's betting twice. — noAxioms

Getting to bet twice on tails doesn't mean that it's more likely to be tails. It just means she gets to bet twice on tails.

Michael

And in what way is that not happening twice to her? She gets to bet twice in that case, despite the fact that she is unaware of which times she's betting twice. — noAxioms

You seem to be thinking of it as:

1. Sleeping Beauty picks tails.
2. We flip a coin.
3. If it's tails, we flip again.

In that case; yes, it's more likely to be tails. This was the reasoning in my first response. But that's not happening in this case. Instead it's:

1. Sleeping Beauty picks tails.
2. We flip a coin.
3. If it's tails, we payout twice.

That doesn't mean it's more likely to be tails. It just means that tails is the better bet.

Michael

Thirder:
Mon Tue
Heads 1/3 0
Tails 1/3 1/3

P(Heads|Awake) = 1/3
P(Heads|Monday) = P(Heads and Monday) / P(Monday) = 1/3 / 2/3 = 1/2

One characteristic of the thirder view is that it doesn't imply an external perspective that "knows" what the probability is for Beauty's current awake state. Since Beauty has no available information distinguishing the three states from her point-of-view, she is simply indifferent about which state she is currently in, and so assigns a probability of 1/3 for each awake state.

So the thirder view can be preferred on pragmatic grounds. — Andrew M

I'm questioning the validity of those probability choices. Imagine instead if it were:

            Mon  Tue
  Roll 1-5  1/3  0
  Roll 6    1/3  1/3

Is it right to distribute the probabilities that way? I say no. We have to apply the general multiplication rule:

P(Roll 1-5 and Monday) = P(Roll 1-5) * P(Monday|Roll 1-5) = 5/6 * 1 = 5/6

            Mon   Tue
  Roll 1-5  5/6   0
  Roll 6    1/12  1/12

Benkei

I liked the solution I read on the physics forum from PeroK. Let's change the story shall we?

Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice during the experiment, Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening.

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. The interview consists of her being asked whether the coin toss was heads or tails, if she guesses wrong, she'll be executed. If she still lives, she will be awakened on Wednesday without interview and the experiment ends.

Given these facts, what answer will Sleeping Beauty give?

If the toss is heads, she will awaken Monday. If she'd then say heads, she lives. If she'd say tails she'd be dead.

If the toss is tails, she will awaken first on Monday. If she then says heads, she dies. If she says tails, she'll live.

If she said tails, she will awaken again on Tuesday. If she then says heads, she dies. If she says tails, she'll live.

In the above we see there's only one event where saying tails gets you killed and two events where saying heads gets you killed. Sleeping Beauty would be smartest to state tails.

Mathematical Conundrum or Not? Number Five

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