Mathematical Conundrum or Not? Number Five

Michael

But this probability space is the epistemological probability space of Beauty, and we have no axioms that justify our assuming that the probabilities will be the same in that probability space as in the independent observer's one. — andrewk

But Beauty is aware of how the experiment is to be conducted.

If I know that you flipped a coin and that if it's heads you'll give me £1 and if it's tails you'll give me £2 then I know that there's a 50% chance that I will get £2.

If Beauty knows that you flipped a coin and that if it's heads you'll wake her once and if it's tails you'll wake her twice then she knows that there's a 50% chance that she'll be woken twice.

BlueBanana

There's nothing to eliminate. This is it. — Michael

Because that's where we get to after we eliminate the one possibility from the aforementioned case.

We flip a coin. If it's heads then we wake Mary once. If it's tails then we wake her twice.

What is the probability that it's her first awakening? 2/3, because two of the three outcomes are first awakenings? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)? — Michael

You forgot to weigh the calculations for the potential two wakings. Run the test 100 times, she'll wake up the first time 100 times out of 150.

Michael

Because that's where we get to after we eliminate the one possibility from the aforementioned case. — BlueBanana

What possibility have we eliminated?

BlueBanana

↪Michael

As I said, Tuesday and heads. From the trivial scenario where she's woken up twice regardless of the coin flip.

Michael

As I said, Tuesday and heads. — BlueBanana

Where has Tuesday come from? It wasn't mentioned at all in the experiment I described here.

From the trivial scenario where she's woken up twice regardless of the coin flip.

Then you might as well say that we've eliminated the trivial scenario of heads + heads in this experiment. Heads + heads is utterly irrelevant and doesn't change the probability at all. The probability of heads is 0.5 + (0.5 * 0.5) = 0.75.

Heads + two awakenings is utterly irrelevant and doesn't change the probability at all. The probability of first awakening is 0.5 + (0.5 * 0.5) = 0.75.

BlueBanana

Where has Tuesday come from? It wasn't mentioned at all in the experiment I described here. — Michael

Because that's the original scenario where it's already eliminated.

Then you might as well say that we've eliminated the trivial scenario of heads + heads in this experiment. — Michael

Completely possible to do so.

Heads + heads utterly irrelevant and doesn't change the probability at all. — Michael

You mean modifying that scenario so that it includes heads+heads doesn't change the probabilities in it?

Michael

Because that's the original scenario where it's already eliminated. — BlueBanana

I have a new original scenario.

You mean modifying that scenario so that it includes heads+heads doesn't change the probabilities in it? — BlueBanana

Here you said "Completely different situation. There's no 'eliminated' outcome." as if to explain why the probability is 0.75 and not 2/3.

I'm saying that there is an eliminated outcome – heads + heads – and yet the probability is still 0.75. So it doesn't matter if you talk about an eliminated outcome or not. The probability is 0.75 whether you consider it or don't.

BlueBanana

I'm saying that there is an eliminated outcome – heads + heads – and yet the probability is still 0.75. — Michael

With the outcome added there are certain odds, and when that is modified by eliminating that one possibility the odds are changed in a way that somehow makes sense. It's just another way to look at the problem, like how in the Sleeping Beauty problem the chances are 2/3 whether one thinks about the eliminated extra possibility or not. Thinking about it doesn't change the situation but coming to the same conclusion through eliminating it can be use to confirm the solution.

But as we disagree about what conclusion is reached by looking at the problem that way I don't think we can achieve an agreement through it.

Michael

With the outcome added there are certain odds, and when that is modified by eliminating that one possibility the odds are changed in a way that somehow makes sense. — BlueBanana

No, it doesn't. If you toss a coin once and then again if it's tails then there's a 0.75 chance that you'll finish with a heads. It doesn't matter if you consider heads + heads or heads + tails or tails + tails + heads or any number of trivially eliminated outcomes. It's always just going to be 0.75.

BlueBanana

↪Michael

No, if you consider heads + tails the chances go to 0,5. Eliminating that possibility results in 0,75.

Michael

No, if you consider heads + tails the chances go to 0,5. Eliminating that possibility results in 0,75. — BlueBanana

You're really confusing me now. Perhaps you could explain what you meant by this objection:

We flip a coin. If it's heads then the result stands. If it's tails then we flip again and the new result stands.

What is the probability that it's heads? 2/3, because two of the three outcomes are heads? Or 3/4 because the probability is 0.5 + (0.5 * 0.5)? — Michael

Completely different situation. There's no "eliminated" outcome. — BlueBanana

BlueBanana

↪Michael

In the original problem adding any outcomes, like heads and Tuesday, doesn't affect the other odds in respect to each other. If some specific interview is as likely as another one (like tails and Monday and heads and Monday) we can stuff in new outcomes and those relations don't change.

In the coin flip example to get into any outcome you need to take away some outcome from which to get to the new outcome. Like heads is twice as likely as tails + heads, but heads + tails removes heads because you have to proceed with a new flip.

Srap Tasmaner

The probability of first awakening is 0.5 + (0.5 * 0.5) = 0.75. — Michael

There is a 50% chance that Beauty will be interviewed once, and a 50% chance that she will be interviewed twice, determined by the toss of a fair coin.

Because the second interview, if it happens, takes place on Tuesday, the chance of being interviewed on Tuesday is also 50%.

There are two orthogonal partitions of the space, one by Heads/Tails, and one by Monday/Tuesday. Each divides the space into subspaces of equal probability. Thus

$\small H \cap M = H \cap M^C = H^C \cap M = H^C \cap M^C = \cfrac{1}{4}$ .

If you now restrict the total space to occasions when Beauty might be awakened, according to the rules of the experiment, you lose $\small H \cap M^C$ , and now you have

$\small H \cap M = H^C \cap M = H^C \cap M^C = \cfrac{1}{3}$ .

It is this identity that Beauty will rely on, given that she has been awakened.

Andrew M

Then there's 6 states, not 7. You're counting the tails state twice, which you shouldn't do. The two tails days need to share the probability that it's the tails state (1/6) giving each 1/12 which is the correct figure you get when you apply the probability rule:

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

P(Tails and Tuesday) = P(Tails) * P(Tuesday|Tails)

P(Tails and Tuesday) = 1/6 * 1/2 = 1/12 — Michael

I'm not sure if you're just disputing the thirder position, or disputing my characterization of the thirder position for the 5/6 heads-weighted coin. Assuming the latter, there are twelve states, including awake and asleep over the two days, each with 1/12 probability. Seven of those states are awake states.

Conditionalizing on being awake, the probabilities for the halfer and thirder positions are:

Halfer (5/6 heads-weighted coin):

       Mon   Tue
Heads  5/6   0
Tails  1/12  1/12

Thirder (5/6 heads-weighted coin):

       Mon   Tue
Heads  5/7   0
Tails  1/7   1/7

The halfer distributes all the Tue/Heads state probability to Mon/Heads. That is, 5/12 + 5/12 = 5/6. The thirder distributes the Tue/Heads state probability evenly to each awake state. That is, (5/12 / 7) + 1/12 = 1/7.

For the thirder:

P(Tails and Tuesday) = P(Tails) * P(Tuesday|Tails) = 2/7 * 1/2 = 1/7

Edit:

Oops, the above result is conditionalized on being awake, per the above thirder table. So actually P(Tails and Tuesday) = 1/6 * 1/2 = 1/12 as Michael correctly noted. The unconditional probabilities are:

       Mon   Tue
Heads  5/12   5/12
Tails  1/12   1/12

P(Heads|Awake) = P(Heads and Awake) / P(Awake) = 5/12 / 7/12 = 5/7.

Andrew M

Beauty doesn't gain relevant new information when awakening, she knew all this before hand. If we do the experiment on you, then you have a prior belief that it is 1/3, what new information would then update that? Priors need relevant new information which would allow us to update it, not just any old information that you think happened. — Jeremiah

Beauty (as a thirder) knew at the time before the experiment that P(Heads) = 1/2. She also knew at the time before the experiment that P(Heads|Awake) = 1/3. When she does awake in the experiment and learns this, that self-locating information enables her to update P(Heads) to be 1/3.

Note that Beauty (as a halfer) does the same thing when she is told that it is Monday. For a halfer, P(Heads|Monday) = 2/3 so, on attaining that self-locating information, she updates P(Heads) to be 2/3.

Andrew M

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.
— Andrew M

The host does not, that's the trick. — tom

How so? Isn't the host telling you a specific door number (that doesn't contain the prize) information?

Not as the problem was described at the top of the thread. No information is given to Sleeping Beauty beyond what she was told would happen. To her each awakening is identical, and there are three of them. — tom

The event of waking provides information about which states one can eliminate (not merely the conditional probabilities, which were previously known). BTW, do you assign 1/2 or 1/3 to P(Heads|Awake)?

Srap Tasmaner

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. — Andrew M

I didn't follow this part of the thread, so sorry for the late reply.

Beauty isn't assigning 1/3 from a principle of indifference. (I made the same mischaracterization earlier,)

Monday splits 50:50 because the coin is fair; Tuesday happens or not on a toss of the same fair coin, so each quadrant is 25%, by stipulation.

Beauty conditionalizes on being awakened, so the values change to

$\small \cfrac{\cfrac{1}{4}}{\cfrac{3}{4}}=\cfrac{1}{3}$ .

andrewk

↪Michael

First let me say that my inclination is to being a halfer, and I argued for that 'cause' on physicsforums last year, so here I'm arguing not against halfer being a preferred position (IMHO) but rather against the idea that it can be proven that it is preferable.

With that preface, perhaps I misunderstood your post here. In the fifth line of the reasoning, you assert, seemingly as an axiom, that P(Heads)=0.5. While - as indicated above - I sympathise with that position, it does look like assuming one's conclusion, if the object of the exercise is to determine what probability Beauty assigns to the coin having landed Heads.

If we do not assume our conclusion, we need to make P(Heads) an unknown, which I called p, and see if we can assert other, less controversial relationships that enable us to deduce that. But I can't see how we can do that without assuming something that is essentially equivalent to P(Heads)=0.5.

To me it looks irresolvable, which is why I like to turn to betting games to make the question concrete.

Jeremiah

Note that Beauty (as a halfer) does the same thing when she is told that it is Monday. For a halfer, P(Heads|Monday) = 2/3 so, on attaining that self-locating information, she updates P(Heads) to be 2/3. — Andrew M

She is never told it is Monday, each awaking is the same, there is no hint as to which day it is; temporally she is uncertain of her location.

So what "self-locating information" allows her to reallocate credibility? You need to be specific. If I am working on a Bayesian analysis, and I get up to go to the bathroom, siting on the toilet doesn't count as new relevant information. Just her waking up is not good enough, she has to actually gain new information.

tom

How so? Isn't the host telling you a specific door number (that doesn't contain the prize) information? — Andrew M

Well, it certainly seems like you are being given information, but you're not. The host, because he has total knowledge of the situation, can always open an empty door. There will always be an empty door. So, the problem is identical to choosing 1 door or 2 doors.

You, as the contestant, know for certain that one of the other two doors is empty, once the door is opened, you still know for certain that one of the doors is empty. OK, you now now which one is empty, and that IS information of sorts, but is it relevant information?

Anyway, I've never encountered anyone who agrees with me on this.

The event of waking provides information about which states one can eliminate (not merely the conditional probabilities, which were previously known). BTW, do you assign 1/2 or 1/3 to P(Heads|Awake)? — Andrew M

In the original description of the problem, Sleeping Beauty wasn't told what day it is, so all waking events are identical. She knows there is a probability space of 3 independent events, and that 2 of them are associated with tails.

All she knows is that she is awake, and that is twice as likely to be associated with tails.

Srap Tasmaner

↪andrewk

I think there are two issues.

First, getting conditionalizing on being awake right. It's clearest perhaps to imagine the coin being tossed after the first interview. Could that coin toss have come up heads? Did it have a 50% chance of doing so? Yes and yes, but Beauty will not be asked about it if it did.

Second, should Beauty conditionalize on being awakened, since she already knew she would be and cannot know how many times she is awakened. I think it only makes sense to consider the odds, in the absence of knowledge. What's more, if you don't, you're in the position of saying, "I'm right about the odds and can prove it by consistently losing money on tosses of this fair coin."

Michael

With that preface, perhaps I misunderstood your post here. In the fifth line of the reasoning, you assert, seemingly as an axiom, that P(Heads)=0.5. While - as indicated above - I sympathise with that position, it does look like assuming one's conclusion, if the object of the exercise is to determine what probability Beauty assigns to the coin having landed Heads. — andrewk

I think it's clearer with this:

$P(Heads|Awake) = \frac{P(Heads ∩ Awake)}{P(Awake)}$

Our objective is to determine P(Heads|Awake), and so it's not necessarily begging the question to assert that P(Heads) = 0.5.

It really depends on what we understand by the (unconditional) probability that it's heads. I understand it as being the probability that a fair coin toss lands heads.

Jeremiah

Anyway, I've never encountered anyone who agrees with me on this. — tom

That is because you are wrong. Knowing which door is empty is the new information which calls for a reallocation of credibility. You know now one of the the possible empty doors. Yes there is still an empty door but before there were two empty doors and now you know at least one of them. That is a real reason to reallocate credibility.

Srap Tasmaner

P(Awake) = 1 (as the sample space of interviews is always when awake) — Michael

That is not how conditionals work. If it were, there would never be a point to conditional probability. You'd just always use 1, and P(A|B) would just be a funny way of writing P(A). In how much of the total space is she awakened? Your way is just wrong.

Michael

Beauty (as a thirder) knew at the time before the experiment that P(Heads) = 1/2. She also knew at the time before the experiment that P(Heads|Awake) = 1/3. — Andrew M

P(Heads|Awake) = P(Heads) * P(Awake|Heads) / P(Awake)

If she applies this before the experiment then she knows that P(Heads|Awake) = 0.5 * 1 / 1 = 0.5.

Michael

That is not how conditionals work. If it were, there would never be a point to conditional probability. You'd just always use 1, and P(A|B) would just be a funny way of writing P(A). In how much of the total space is she awakened? Your way is just wrong. — Srap Tasmaner

So what is P(Awake)?

Srap Tasmaner

↪Michael

3/4.

It is possible to have a heads on Tuesday, same as always. It's within the total space, just not the conditional space.

Michael

3/4.

It is possible to have a heads on Tuesday, same as always. It's within the total space, just not the conditional space. — Srap Tasmaner

Why count heads on Tuesday?

What if I were to change the experiment so that it ends on Tuesday if it's heads rather than on Wednesday? In the original we don't count Heads + Wednesday (or Tails + Wednesday) so in my variation we don't count Heads + Tuesday.

Or what if we wake her on Monday if it's heads and wake her on Tuesday and Wednesday if it's tails, ending the experiment on Thursday? Is P(Awake) = 3/6?

This is why I suggested the alternative here where there's no mention of days at all; just one awakening if heads and two awakenings if tails. The sleep day is a red herring.

tom

That is because you are wrong. Knowing which door is empty is the new information which calls for a reallocation of credibility. — Jeremiah

How does this new information alter the fact that one door has a probability of 1/3, and 2 doors has a probability of 2/3?

Michael

How does this new information alter the fact that one door has a probability of 1/3, and 2 doors has a probability of 2/3? — tom

If you have to pick one of three doors then the probability of being right is 1/3.
If you have to pick one of two doors then the probability of being right is 1/2.

Later I might write a script to test the Monty Hall problem. I believe prior experiments have supported the hypothesis.

Mathematical Conundrum or Not? Number Five

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