Mathematical Conundrum or Not? Number Five

Srap Tasmaner

In the event of Tails, Beauty will be awakened on Monday and Tuesday, but due to the nature of the experiment she will not be able to tell the difference, either one is equally likely when interviewed — Jeremiah

So what? It's not a situation that arises. Neither she not the experimenters are ever in the position of knowing that the coin landed tails but wondering what day it is. Beauty only wonders what day it is to figure out how the coin landed.

Suppose there was another coin toss to determine whether heads was the single interview or the double this time around. Then half the time heads would be 1/3 of the interviews, and half the time 2/3, so heads would on average be half the interviews and same for tails.

But that is not the case here. The interviews are not randomly distributed.

Beauty knows that when she is asked for her credence, 1/3 of the time the coin has landed (or will land?) heads and 2/3 of the time the coin has landed (or will land?) tails.

Therefore her credence that the coin has landed (or will land?) heads must be 1/3.

Jeremiah

So what? It's not a situation that arises. Neither she not the experimenters are ever in the position of knowing that the coin landed tails but wondering what day it is. Beauty only wonders what day it is to figure out how the coin landed. — Srap Tasmaner

If my position was that she somehow magically knew it was tails, then why would I claim it has a 50% uncertainty? If she knew it was tails it would be a 100% certainty and M2 = 50% = Tu with temporal uncertainty. The 50% is the uncertainty of T or H and the 25% is the uncertainty of M2 or Tu.

The conditional probability of tails given that it is M2 or Tu is P(T|M2)= P(.5|.25) = .125/.25 = 1/2 = P(T|Tu) which is equal to P(H). Both days still have the same 50% uncertainty when considering H or T. In fact from that direction all three days have the same uncertainty when considering H or T, which is where we get the 1/3 argument.

The interviews are not randomly distributed. — Srap Tasmaner

Randomly just means equal probability. Figured I should clear that up. In the technical sense when talking about random, it means each element in the sample space has an equal chance of being selected. So in the event of tails, on any given consideration of the interview between M2 and Tu Beauty, has a 25% uncertainty of being in either of them. It is 25% and not 50% because of the uncertainty in it being Tails. We are stacking uncertainties.

How it comes out all depends on the considerations of the uncertainties.

Srap Tasmaner

The conditional probability of tails given that it is M2 or Tu is P(T|M2)= P(.5|.25) = .125/.25 = 1/2 = P(T|Tu) which is equal to P(H). — Jeremiah

This is still slightly puzzling to me.

P(H | M1) = 1, right? And this is the thing about the double interview track: both them happen if and only if the coin lands tails. From your calculation, P(T | M2 v Tu) = 1, yes? But it should be P(T | M2) = P(T | Tu) = 1, and P(T) = P(M2) = P(Tu) = 1/2. You always get both on tails. You get them one at a time, but we don't necessarily care.

That space of three possibilities, {M1, M2, Tu} has three elements each of which has an unconditional probability of 50%. Conditioned on the whole space, they'll each be 33%.

Jeremiah

P(H | M1) = 1, right? And this is the thing about the double interview track: both them happen if and only if the coin lands tails. From your calculation, P(T | M2 v Tu) = 1, yes? But it should be P(T | M2) = P(T | Tu) = 1, and P(T) = P(M2) = P(Tu) = 1/2. You always get both on tails. You get them one at a time, but we don't necessarily care.

That space of three possibilities, {M1, M2, Tu} has three elements each of which has an unconditional probability of 50%. Conditioned on the whole space, they'll each be 33%. — Srap Tasmaner

This is why I keep saying it depends on how Beauty decides to consider her uncertainties.

Remember conditional probability is the the probability of event K given that event L has already occurred. Order matters.

So what is T given M2 or Tu? 50%

Hence, .50 +.50 +.50 = 1.5 > 1. The sum of probability cannot be greater than 1 and since P(H|M1)=P(T|M2)=P(T|Tu) we reallocate the credibility to 1/3 each.

That is when Beauty is considering the uncertainty of her location in time.

Now what if Beauty considers instead the uncertainty of H or T?

Then, purely for demonstration, what is M2 given the event T?

P(M2|T) = P(.25|.5) = .125/.5 = 1/4.

However, if Beauty was considering the uncertainty of H or T, and not her location in time, the only reason to consider the conditional probability of M2 would be for completion; practically she could end at the uncertainty of H or T.

The real issue here is not that we get two different yet seemly reasonable answers; this is not about 1/2 vs 1/3. What the Sleeping Beauty Problem demonstrates is that decision affects the outcome of her solution.

What I find interesting is that this decision also is very likely an unconscious decision. Which may be why we get people who are convinced it is 1/3 and people who are convinced it is 1/2. The unconscious mind made a decision for them on how to consider the probability, a decision that consciously they were never aware of.

Srap Tasmaner

↪Jeremiah

Something I don't remember us talking about: should Beauty, knowing the rules of the experiment, subject her expectation of a tails interview to a discount? It occurs to me that this may be the regime Lewis is describing.

Here's a physical version. You decide to test if a coin is fair by throwing a red marble in an urn on heads, and a blue marble if tails. After a bunch of flips, you'll count the marbles, expecting them to be about equal. Drawing a marble randomly will have the same distribution as the coin itself.

Suppose instead on tails you throw in two blue marbles. Then you'd expect a 2:1 ratio if the coin is fair. A randomly selected marble is now twice as likely to be blue, but each blue is discounted, and is only half the total available evidence of a tails flip, unlike the reds each of which is all the evidence of a heads. Each blue does represent a tails, certainly, and only got in the urn because a tails was tossed, but there's another blue that's evidence of the same toss.

Now suppose the marbles are all white. Still true that you're twice as likely to draw a marble representing a tails toss, but you have to discount.

Srap Tasmaner

@Michael I think I'm a halfer now. (Still some things I'd like to be clearer on.)

@Andrew M, @andrewk, @JeffJo: do you find this argument as convincing as I do?

@Jeremiah

each blue is discounted, and is only half the total available evidence of a tails flip, unlike the reds each of which is all the evidence of a heads — Srap Tasmaner

Is this what the conditional probabilities we've been talking about are trying to express? I'm still not clear on how this idea is formalized.

Srap Tasmaner

@Jeremiah
It's as we were discussing: each marble represents an interview event. To count coin toss outcomes you only need one red marble, but two blues, to make up the entire double interview event. Each blue marble is one kind of event, but that event is half of the kind of event we want to count.

Andrew M

A randomly selected marble is now twice as likely to be blue — Srap Tasmaner

That's all we're asking Beauty about. That is, the probability that the next marble to be drawn (or the interview that is being conducted) will be associated with a heads outcome. Which is 1/3.

but each blue is discounted, and is only half the total available evidence of a tails flip, unlike the reds each of which is all the evidence of a heads. — Srap Tasmaner

Discounting doesn't help Beauty. Suppose she is a halfer. Should she discount the next drawn marble (or current interview) by 1/2 or not? Well, she should 2/3 of the time. So 2/3 * (1/2 * 1/2) + 1/3 * 1/2 = 1/3. The thirder says that is the real probability of each state for her.

Srap Tasmaner

Discounting doesn't help Beauty. Suppose she is a halfer. Should she discount the next drawn marble (or current interview) by 1/2 or not? Well, she should 2/3 of the time. So 2/3 * (1/2 * 1/2) + 1/3 * 1/2 = 1/3. The thirder says that is the real probability of each state for her. — Andrew M

This is the wrong model. This table

    Mon   Tue
H   1/2
T   1/4   1/4

is right, and here's why.

Suppose you have a machine set up like this: there's a hopper full of red marbles and a hopper with twice as many blue marbles; you push one button and it transfers a single red marble or two blue marbles to another hopper, one you can't see; you push a different button and it dispenses one of the marbles from the small hopper. What are your odds of getting a red marble? 1/2. Half the time only a single red marble goes into the small hopper and then gets dispensed in the second step. (Half the time, two blue marbles go in, and then one of those two is dispensed, so the chance of blue -- a blue, any blue, one of the two in the small hopper -- is also 1/2, despite the fact that twice as many marbles were dispensed at the stage you don't see.)

Now do it this way: you have a hopper full of white marbles; push one button and half the time a single marble is moved to the small hopper, half the time two; you push the second button and get a single marble. How many marbles are left in the small hopper? Dunno. Half the time there's still one there, and half the time there isn't.

You could randomize. Any number of marbles could be dispensed to the small hopper. Getting one tells you exactly nothing. You could have it transfer a random number of reds to the small hopper half the time and a random number of blues half the time. When you push the second button to get your marble, the chances will still be 50:50 of getting a red or a blue.

Andrew M

What are your odds of getting a red marble? 1/2. — Srap Tasmaner

Agreed. But that scenario is equivalent to randomly waking Beauty on either Monday or Tuesday if tails, but not both days. To be analogous to the Sleeping Beauty scenario, the second blue marble has to be dispensed in a separate event (with amnesia in between). That is an additional possible state that Beauty could be in which, for the thirder, decreases the probability for the Heads and Monday (or red marble) state to 1/3.

All that means is that if Beauty is asked to guess which state she is in and she guesses Heads and Monday, then she will be correct 1/3 of the time. Similarly for Tails and Monday and Tails and Tuesday. On the thirder view, probability is about the state she is in, not the coin toss (or day) outcome itself.

The halfer view, while seemingly just representing a fair coin toss as coming up heads half the time, has the consequence that P(Heads|Monday) = 2/3 instead of 1/2. I think that is a reductio of the halfer view.

Srap Tasmaner

↪Andrew M

If you toss a fair coin 100 times and throw one red marble into an urn on heads and two blues on tails, you'll end up with (roughly) 50 reds and 100 blues. If you count each of the marbles as an outcome of the coin toss, without discounting the blues, you'll end up with 100 tosses having 150 outcomes, which is absurd. It's an attempt at alchemy.

One of the side effects of this attempted alchemy is that each red represents twice as much of an outcome as each blue. Yes, there is something absurd about the 2/3, but it's a result of putting in twice as many blues per toss but then taking them out one at a time, as if they were the same as the reds. You can pair off each red with two blues -- that is, taking the marbles back out of the urn the same way you put them in -- without absurdity. If you insist on taking the blues back out singly, the absurdity of the result (a marble representing half an outcome) is on you.

As for Beauty's state, try thinking of the Tuesday interview this way: I am (still) being interviewed about a tails outcome (the same one as yesterday). There was just the one coin toss, with just the one outcome. Smearing the interview across two days doesn't change that. Beauty does not know which interview this is, but she knows there will be two interviews for each tails outcome and she discounts accordingly.

Think about what's going on if she wagers. She can make a Dutch book on tosses of a fair coin at even money. That should not be possible. That's just as strong a principle as the business about no updating without new information. If it is possible, someone's performing alchemy or cheating.

Srap Tasmaner

I keep saying Beauty can make a Dutch book, which is wrong of course and I'll quit it: the odds are coherent but they don't match the odds of the event being given odds on, so Beauty can expect a profit just by betting tails.

Srap Tasmaner

↪Andrew M

Here's another way to look at it.

You could say it depends on what we take to be the event that must be predicted -- I've made that argument myself, and not long ago. Should Beauty predict the outcome of the coin toss or being asked about the coin toss?

I used to think that being asked was itself evidence, but that's only true on two conditions: (a) Beauty doesn't know the rules and is unable to figure out the true chances of heads and tails; (b) she is willing to accept the absurdity that a single coin toss has, on average, 1.5 outcomes.

It is certainly true that there are on average 1.5 interviews, but some of those interviews (the extra .5 on average) are about the same outcome. That extra blue marble left in the hopper is not another outcome; it's just the rest of the outcome you already know about from the first blue marble.

Srap Tasmaner

Still another point:

I have argued, as thirder, that there are three outcomes each of which has a 50% chance of occurring (the one heads interview and the two for tails). Taken conditional to the whole set {H1, T1, T2}, each would shift to a 33% chance.

The thing is, this 2:1 proportion of interviews is right, but remember that SB does not payout like a wager on a 2:1 biased coin. It pays out like a 3:1 coin.

It's not just the ratio of wins to losses, but the actual payouts that bothers me. If it were a genuine 2:1 deal, we'd expect a $1 wager on tails to pay out 2/3(1) - 1/3(1) = 1/3. It doesn't. It pays out 2*1/2(1) - 1/2(1) = 1/2.

That means that scaling 50:50:50 to 33:33:33 never happens. The actual payouts represent a coin that has a 50% chance of heads and a 100% chance of tails, not a coin that is 33:67. That's pretty weird.

Andrew M

Yes, there is something absurd about the 2/3, but it's a result of putting in twice as many blues per toss but then taking them out one at a time, as if they were the same as the reds. — Srap Tasmaner

I'd like to analyze the halfer's P(Heads|Monday) = 2/3 consequence further because I think it is key to how we see the Sleeping Beauty scenario. Just to make the consequence more stark, suppose that Beauty knows she will undergo 1000 memory-erased interviews on tails (subject to ethics approval). The halfer probability distribution is:

       Mon     Tue     Wed     ...
Heads  1/2
Tails  1/2000  1/2000  1/2000  ...

P(Heads and Monday) = 1/2
P(Monday) = 1/2 + 1/2000 = 1001/2000
P(Heads|Monday) = 1/2 / 1001/2000 = 1000/1001 (approx. 99.9%)

Now suppose that Beauty knows before the experiment begins that the coin will be flipped Monday evening after she goes to sleep. The experiment proceeds and, during her Monday interview, she is told that it is Monday. As a halfer, she concludes that it is a virtual certainty (99.9%) that the outcome of the yet-to-be-tossed coin will be heads!

Is that how she should understand P(Heads|Monday)? If she should discount the possible tails interviews to get the more sensible result of 1/2, then where does that show up in the halfer's math?

The thing is, this 2:1 proportion of interviews is right, but remember that SB does not payout like a wager on a 2:1 biased coin. It pays out like a 3:1 coin. — Srap Tasmaner

Yes, this is an important point. I think the intuitive comparison with a weighted coin is misleading since SB is just structured differently. Adding more interviews (and thus bets) on tails is not like increasing coin bias.

Michael

↪Andrew M

I don’t think any of these probability tables are relevant if you’re told what day it is. If you’re told it’s Monday then the probability is 1/2 that it’s heads and if you’re told it’s Tuesday then the probability is 0. This is worked out not by some equation but just by knowing the rules and how coin flips work.

The thirder reasoning works against intuition as well, especially in the extreme version. It suggests that if we’re to be woken a thousand times in the case of tails then we should be almost certain that it’s tails upon waking, despite the fact that it’s an unbiased coin flip. And all because if we’re right then we’re right more often? That shouldn’t be the measure.

I think the sensible position is to just accept that a coin toss is 50:50 and so we shouldn’t have a preference, no matter how many more times we are woken if it’s tails. The probability tables are only useful if we have no further information and want to work out the probability that today is Tuesday, in which case I think the halfer’s table is more intuitive.

Srap Tasmaner

Yes, this is an important point. I think the intuitive comparison with a weighted coin is misleading since SB is just structured differently. Adding more interviews (and thus bets) on tails is not like increasing coin bias. — Andrew M

Indeed, I think it means that the odds here are not truly 2:1 at all.

I can't figure out how to make this into a normal wager of any kind. The closest I can come looks very similar to the problem itself: the bet is at even money, but on tails the wager is doubled. Now this is very strange. Conditional wagers are fine, but the one thing they would generally not condition on is the event whose outcome determines the outcome of the wager! Even your stake is unknown until the bet itself is resolved?! I don't know why people who find the wagering argument so attractive, as I once did, aren't more troubled by this.

If you’re told it’s Monday then the probability is 1/2 that it’s heads and if you’re told it’s Tuesday then the probability is 0. — Michael

This position is attractive, but I just don't understand how it works.

Your expectation of heads is 1/2 before you know what day it is, right? So that includes the possibility that it's Tuesday. If you go with the way Lewis splits up the space, that's fine:

$\small P(H) = \cfrac{3}{4}(67\%) + \cfrac{1}{4}(0\%) = 50\%$ .

Learning that it's not Tuesday should increase your expectation of heads if you were taking the possibility of it being Tuesday into account, and that's how we end up with the 2/3.

Is there a principled way to just ignore Tuesday, and any subsequent Tuesdays?

I'd like to analyze the halfer's P(Heads|Monday) = 2/3 consequence further — Andrew M

Michael is certainly right that this is the mirror image of the thirder problem -- we each seem committed to unreasonable confidence about heads or about tails as you increase the number of Tuesdays.

I'm not convinced this is what the halfer position really implies, but the two are clearly related.

That extra blue marble left in the hopper is not another outcome; it's just the rest of the outcome you already know about from the first blue marble. — Srap Tasmaner

This is what I keep thinking about. You can add as many Tuesdays as you like to represent the tails outcome, but Beauty would only need to know about one of them to know the toss landed tails. All of the Tuesdays are clones of each other -- if you know the fact of the matter about one, you know them all.

Still working on it. Maybe in another day or so I'll have something new to say.

Michael

This position is attractive, but I just don't understand how it works. — Srap Tasmaner

As Andrew M said the coin isn’t flipped until Monday evening, so if you know it’s Monday then you’re being asked about your credence that a future coin flip will land heads, and of course that’s going to be 1/2. And as you’re only woken on Tuesday if it’s tails then if you’re told it’s Tuesday then you know it didn’t land heads.

Andrew M

This is worked out not by some equation but just by knowing the rules and how coin flips work. — Michael

I see those equations as a formalization of the rules that we learn in familiar settings. In unfamiliar settings, the question would be whether it is the equations that are at fault (or inapplicable) or our intuitions.

The thirder reasoning works against intuition as well, especially in the extreme version. It suggests that if we’re to be woken a thousand times in the case of tails then we should be almost certain that it’s tails upon waking, despite the fact that it’s an unbiased coin flip. And all because if we’re right then we’re right more often? That shouldn’t be the measure. — Michael

The thirder is not measuring the outcome of the coin toss as an isolated event. She is instead measuring uncertainty about her location in a state space.

In a familiar setting, there is no difference between those two ways of thinking about probabilities. It does not matter whether you think of the coin as being in one of two possible states or, alternatively, you as being located in one of two possible states with respect to the coin. There are only two states in both views, so 50:50 is the agreed answer.

In the Sleeping Beauty scenario, those two methods give different answers. The first method is the halfer view which understands the coin to be in one of two states. But that leads to absurdity when conditioning on Monday. The thirder doesn't have that problem, because conditioning on Monday eliminates all but two states that she could possibly be in. So 50:50 is the natural answer as with a coin toss in an everyday setting.

In the initial Sleeping Beauty scenario, Beauty is located in one of three possible states, she just doesn't know which one. Since they are indistinguishable to her, she assigns each of them an equal probability. That assignment would make no sense if it meant that the coin's intrinsically equally likely states had somehow become weighted. But it does make sense if it means that there are three indistinguishable states that she could be located in and two of them are associated with a tails outcome.

Indeed, I think it means that the odds here are not truly 2:1 at all. — Srap Tasmaner

I think they are, but not as a measure of the coin toss outcome. I think probability is a measure of self-locating uncertainty in a state space. It's not directly about coin outcomes, days, or even interviews at all, except in so far as they contribute to the construction of the state space.

I can't figure out how to make this into a normal wager of any kind. — Srap Tasmaner

The simplest wager is just to bet on being located in a single state. If you bet on being located in the Tails and Monday state, then you'll win once and lose twice (on average), just as the thirder probabilities imply. Similarly for each of the other states. Whereas if you bet on being located in a Tails state, you'll win twice and lose once. And sure enough, there are two tails states that add to 2/3. Similarly if you bet on being located in a Monday state, you'll win twice and lose once and those states indeed add to 2/3.

That makes sense to me.

Michael

In the Sleeping Beauty scenario, those two methods give different answers. The first method is the halfer view which understands the coin to be in one of two states. But that leads to absurdity when conditioning on Monday. — Andrew M

I think the halfer reasoning should just be that it’s a 50:50 chance that it’s heads, whether unconditioned or conditioned to Monday. We shouldn’t be applying some formula and should just consider what we know about coin flips.

Srap Tasmaner

You are both ignoring part of the problem.

I think the halfer reasoning should just be that it’s a 50:50 chance that it’s heads, whether unconditioned or conditioned to Monday. — Michael

That's saying P(A | B) = P(A), and therefore A and B are independent events. Of course in one sense the coin toss and the day of the week are independent, but whether Beauty is interviewed on that day is clearly not independent of the coin toss: you yourself noted that if she's told it's Tuesday then her credence should be 0. Is that right? You have to decide whether you're using this table

   Mon  Tue
H  1/4  1/4
T  1/4  1/4

or this table

   Mon  Tue
H  1/2  0
T  1/4  1/4

Switching between them as needed is just equivocation.

I think probability is a measure of self-locating uncertainty in a state space. It's not directly about coin outcomes, days, or even interviews at all, except in so far as they contribute to the construction of the state space. — Andrew M

But Beauty is asked about the coin toss itself: it has one outcome which gives rise on average to 1.5 states for Beauty. Not discriminating between Beauty's states and the contribution of the coin toss to those states is another sort of equivocation. Beauty can make this distinction, just as you could in counting 1 red marble on heads and 2 blue marbles on tails, expecting a 2:1 ratio if the coin is fair.

The thirder model relies explicitly on there being a single toss of a coin with heads and tails distributed 50:50. (And we've agreed you cannot construct an alternate model with a weighted coin.) How can Beauty take that as a premise and then be unable to reach the conclusion that the chances of heads were 1/2?

Michael

That's saying P(A | B) = P(A), and therefore A and B are independent events. Of course in one sense the coin toss and the day of the week are independent, but whether Beauty is interviewed on that day is clearly not independent of the coin toss — Srap Tasmaner

It is independent. She’s interviewed on Monday regardless, and in Andrew’s example the coin isn’t flipped until after the Monday interview.

Srap Tasmaner

↪Michael

So Monday is independent of the coin toss but ¬Monday isn't? Aren't we just using ambiguous vocabulary here?

Speaking of ambiguity, does when the coin toss happens affect Beauty's credences? Don't we want a solution that applies to a toss before the Monday interview as well as after?

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 OP uses the toss-before model.

Michael

So Monday is independent of the coin toss but ¬Monday isn't? — Srap Tasmaner

Yes, because the Monday interview happens regardless, and in this example before the toss, whereas a Tuesday interview only happens if there’s a toss of tails.

Speaking of ambiguity, does when the coin toss happens affect Beauty's credences? Don't we want a solution that applies to a toss before the Monday interview as well as after? — Srap Tasmaner

Perhaps. And as it’s obvious that if you know it’s Monday and the coin toss hasn’t happened that the probability that it will be heads is 1/2 then it should be obvious that if you know it’s Monday and the coin toss has happened that the probability that it was heads is 1/2.

Srap Tasmaner

I think I've finally got the right model now, and it's nuts.

Two urns, one with a one red marble, and one with one blue marble. One blue marble.

You toss a fair coin to determine which urn you'll be taking a marble from. One coin toss with one outcome. The unchosen urn is removed.

If we stop here, it's obvious that the chance of getting a red marble is 1/2, and the chance of getting a blue marble is the same. (In each case, there's a single marble with a 100% chance of getting chosen from the urn.) This is Monday.

Now here's the wonky part: selection from the red urn is without replacement; selection from the blue urn is with replacement. You can go just twice, as in standard SB, or you can go a thousand times. Whenever you want, there's a single blue marble available.

Looked at this way, you can see why there doesn't ever seem to be any discounting or conditioning -- not in the chances of a tails interview on Monday, not in the payoffs from wagering, not in the chances of a second interview, nowhere. Each marble inherits the full 1/2 probability of its urn because it is the only marble there. Tails doesn't show up as an event chopped up into parts, but as an event that repeats as often as you like.

The coin toss determines whether you draw the one red marble or some number of blue marbles. The number of marbles drawn has nothing to do with the chances that you're drawing red or blue though.

I'm not sure how to finish formalizing this, but I think it represents SB pretty well.

Andrew M

I think the halfer reasoning should just be that it’s a 50:50 chance that it’s heads, whether unconditioned or conditioned to Monday. We shouldn’t be applying some formula and should just consider what we know about coin flips. — Michael

OK, that's the double-halfer view. When Beauty is told it is Monday, all the Tuesday and Tails probability is reallocated to Monday and Tails which violates conditionalization. That is:

P(Heads) = 1/2          P(Heads|Monday) = 1/2

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

The thirder model relies explicitly on there being a single toss of a coin with heads and tails distributed 50:50. (And we've agreed you cannot construct an alternate model with a weighted coin.) How can Beauty take that as a premise and then be unable to reach the conclusion that the chances of heads were 1/2? — Srap Tasmaner

Because of the initial conditioning on being awake. You also accept that conditionalization changes Beauty's probability of heads when she is told it is Monday, so probability isn't simply about the nature of fair coins. It's also about the information that an agent has that she can update on.

Srap Tasmaner

You also accept that conditionalization changes Beauty's probability of heads when she is told it is Monday — Andrew M

No, I'm in the double halfer camp now. The post right above explains my current thinking.

((This is, I don't know, maybe the third time I've argued with @Michael about something and then concluded he was right all along.))

Srap Tasmaner

↪Andrew M

Just to clarify my position a little:

According to the design of the experiment, there are two protocols:

a single interview on heads;
multiple interviews on tails.

Beauty is asked for her degree of belief that the coin landed heads, that is, that the experiment is following the heads protocol. Chances of that are 1/2. As it happens, being told it's Monday provides no additional information because both protocols include a Monday interview. (If she were told it's Tuesday, she'd know it was the tails protocol.)

Back when @Michael said it's just the difference between being asked once and being asked repeatedly, and that this makes no difference, he was absolutely right.

Andrew M

↪Michael

↪Srap Tasmaner

I think you are both looking at the experiment from an independent observer's perspective (or Beauty's Sunday perspective) and not from Beauty's perspective when she is awakened and interviewed in the experiment.

Here's a slight variation. Beauty will receive payments at the end of the experiment for each state that she was either awake or asleep in, as follows:

       Mon       Tue        Total
Heads  Awake:$1  Asleep:$2  $3
Tails  Awake:$4  Awake: $8  $12

When Beauty is awakened and interviewed, she is asked how likely it is that she is in the state that pays $1.

My answer is that she should condition on being awake and so her answer should be 1/3.

Srap Tasmaner

I think you are both looking at the experiment from an independent observer's perspective (or Beauty's Sunday perspective) and not from Beauty's perspective when she is awakened and interviewed in the experiment. — Andrew M

I'm just following the principal principle. If I can figure out what the objective chances are, so can Beauty, and she can set her credences accordingly.

Mathematical Conundrum or Not? Number Five

Welcome to The Philosophy Forum!

Categories

More Discussions