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. — Michael

If you were given the choice of 1 or 2 doors, which would you choose? I hope you would choose 2. What about the extra information that bears sh*t in the woods? I suspect you would still choose 2 doors. What about if you were told that if you choose 2 doors, the host will open an empty door as an irrelevant dramatic flourish? It makes no difference.

The host can always open an empty door, an act as relevant to the choice as the behaviour of bears in the woods. It's a choice between 1 or 2 doors, that's all there is to it.

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? — Michael

If whether each happens is determined by the toss of a fair coin, they're all equal and 3/6 is right. (If the subspaces had unequal probabilities, you might still be picking half of them numerically but they wouldn't add up to half the space.)

If whether each happens is determined by the toss of a fair coin, they're all equal and 3/6 is right. (If the subspaces had unequal probabilities, you might still be picking half of them numerically but they wouldn't add up to half the space.) — Srap Tasmaner

So we do my variation where there are no sleep days. The experiment is just; we ask her once if it's heads and twice if it's tails.

What do you think this changes?

Or, better still, can we just stick to the problem at hand?

Here's the result of 100,000 games where you don't re-guess. 1/3 games are winners

Here's the result of 100,000 games where you re-guess. 1/2 games are winners

Go down towards the bottom and hit "Execute".

What do you think this changes? — Srap Tasmaner

It changes P(Awake) to 1. Which means:

P(Heads|Awake) = 0.5 * 1 / 1 = 0.5

Dude, if she's not sleeping we wouldn't call the condition "awake". What matters is when she's asked.

Dude, if she's not sleeping we wouldn't call the condition "awake". What matters is when she's asked. — Srap Tasmaner

Then:

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

P(Heads|Asked) = 0.5 * 1 / 1

P(Heads|Asked) = 0.5

You really just need to look up conditional probability again.

Are you suggesting that, in my amended case, being awake/asked shouldn't be treated as a conditional?

You scaled by 1/1. That's not the idea.

Seriously, read a tutorial and then we can talk about the philosophy. I don't know what else to say.

Here's the result of 100,000 games where you don't re-guess. 1/3 games are winners

Here's the result of 100,000 games where you re-guess. 1/2 games are winners

Go down towards the bottom and hit "Execute". — Michael

The probability of winning, when you choose 2 doors is 2/3. That is all there is to it. The fact that one of these doors is open is irrelevant. One door can always be opened, because one of them is certain to be empty.

A related question: if we have two theories – one which posits a single world and one which posits multiple worlds – should our credence favour the theory which posits multiple worlds?

The probability of winning, when you choose 2 doors is 2/3. That is all there is to it. The fact that one of these doors is open is irrelevant. One door can always be opened, because one of them is certain to be empty. — tom

Having two opportunities to choose one door (i.e. change your selection) isn't the same as having one opportunity to choose two doors (i.e. select both).

If I choose door A then I have 1/3 chance of winning. If I change my mind and choose door B then I still have 1/3 chance of winning (for the moment let's forget about Monty opening a door).

The option of selecting both A and B isn't an option in the Monty Hall problem, though, so I'm not sure why you'd bring it up.

If two doors had a 2/3 chance, and you eliminate one of them, then the 2/3 is reallocated to the remaining door and now that one door has a 2/3 chance. That is the Monty Hall problem and why you should change doors.

Four Beauties volunteer to undergo the following experiment and are told all of the following details: Four cards have been prepared; one says "You will sleep through Tuesday if the coin lands Heads." Another says "You will sleep through Tuesday if the coin lands Tails." The other two change "Tuesday" to "Monday." The cards are dealt to the four volunteers. Each can look at hers, but cannot share its information with the others.

On Sunday all four she will be put to sleep, and a fair coin will be flipped. Three of the Beauties will be wakened on Monday, according to the coin result and their cards.They will be interviewed, and put back to sleep with an amnesia-inducing drug that makes them forget that awakening.

This will be repeated on Tuesday.

In the interview, each awake volunteer will be asked for her belief that the coin result is the one named on her card.

Each volunteer knows that three volunteers are awake. Each knows that exactly one of these three has a card that names the coin result. None have any information that could make her believe that she is more,or less, likely to be the one. So what should her belief be?

One of these volunteers is experiencing the exact same problem that started this thread. The others are experiencing equivalent problems that must have the same answer.

re the Monty Hall problem

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? — tom

If the door you had selected has the car behind it (call this event C) then it is not relevant information.

But if your selected door does not have the car behind it then it is very relevant information, because it means the car is behind the remaining closed door that you did not select.

Since P(C)=1/3 there is a 1/3 probability of the information being irrrelevant and a 2/3 probability of it being relevant. So the probability of a win is increased by assuming the information is relevant, and using it, which means switching. The calcs are:

We have P(C) = 1/3. If you retain your selection then the probability of a win is:

1/3 x P(win | C & not switch) + 2/3 x P(Win | ~C & not switch) = 1/3 x 1 + 2/3 x 0 = 1/3.

On the other hand if you switch the probability of a win is:

1/3 x P(win | C & switch) + 2/3 x P(Win | ~C & switch) = 1/3 x 0 + 2/3 x 1 = 2/3.

We could call the new information 'conditionally relevant information', since its helpfulness is conditional on an unknown, which is whether C is the case.

Another approach, for those that don't find the above persuasive, is to say that there is no new information, but reject the dictum that one should not change one's guess in the absence of new information, replacing it with a dictum that one should not change one's strategy unless one receives new information. The best strategy is to change one's guess, after Monty has opened a door. That strategy does not change when Monty opens the door.

Consider a slightly different game in which Monty doesn't open a door but you can still switch your guess and, if you do so, you will win if the car is behind either of the two doors you did not originally select. The probabilities in this game are identical to those in the original Monty Hall problem, but in this case you clearly get no new info, since no door is opened until the final reveal-all. But it is in your interests to switch. You will have changed your guess, but not your strategy.

Imagine a stock market strategy of buying stocks of companies that have had a certain event, say a change of CEO. Say our research has showed that on average there is an upwards surge in stock price on appointment of a new CEO, that peaks on average three days after the announcement, and then often subsides. Then we could set up a hedge fund that buys stocks that have had such announcements, and sells them after three days. In most cases there will be no important new info about the company in those three days, but we will still sell the stock. We've had no new info. We've changed our stock-pick (our guess) but not our strategy.

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. — Jeremiah

I'm saying that for a halfer, P(Heads|Monday) = 2/3. As a separate hypothetical, if Beauty is told that it is Monday during the experiment, then she will have received self-locating information. So she would update P(Heads) to 2/3.

The thirder is making a parallel argument. Before the experiment, there are four states of 1/4 probability, where one is a sleep state. If she is in a state where she awakes, that is information that would rule out the sleep state. So she would update P(Heads) to 1/3.

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

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? — tom

Yes. Your initially chosen door has 1/3 probability of containing the prize. The other two doors have a total of 2/3 probability. When the host shows you that one of those doors is empty, its probability goes to 0. So the remaining door now has 2/3 probability. So you should switch to that door and you will win the prize 2/3 of the time.

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

Agreed.

She is never told it is Monday, there is no relevant self-locating information, and she knew there were only three possible awake periods before the experiment. Everything we know is everything she knows before the experiment therefore 1/3 is a prior. We don't have any privy information here.

If you can argue the 1/3 without doing the experiment, then you are arguing a prior.

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

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

There are four equally probable states in the experiment, three awake states and one sleep state, so P(Awake) = 3/4 and P(Awake|Heads) = 1/2.

↪Andrew M She is never told it is Monday, there is no relevant self-locating information, and she knew there were only three possible awake periods before the experiment. Everything we know is everything she knows before the experiment therefore 1/3 is a prior. We don't have any privy information here. — Jeremiah

I'm saying that P(Heads|Monday) = 2/3 is a consequence of the halfer position. Do you agree?

BTW, that is David Lewis' (L6) which contradicts Elga's (E1) that P(Heads|Monday) = 1/2. See http://fitelson.org/probability/lewis_sb.pdf

She is never told it is Monday, there is no relevant self-locating information, and she knew there were only three possible awake periods before the experiment. Everything we know is everything she knows before the experiment therefore 1/3 is a prior. We don't have any privy information here. — Jeremiah

Oh good. I thought the OP didn't make this clear. The halvers would have it right if she went into this thing blind.

Michael, I see you're still going on about this. Spent a bit skimming the posts since I dropped off.

I’ve showed the reasoning multiple times. It’s the Kolmogorov definition of conditional probability. — Michael

You're doing it wrong. You are mixing probabilities from different times, different points of view, or from positions with different information.

Choose a point and stick with it. P(Heads) of 50% is true only before the coin is tossed but you seem to assign that probability to Beauty's POV when in a waking period. That is question begging (as I pointed out early in the thread) since that's the answer we're looking for, not the premise. At that time, it is not 50% from her POV nor the administrators, where it is 0% or 100%.

Do you agree? — Andrew M

No, and I already commented on Lewis and Elga.

Also Lewis is arguing that 2/3 is a consequence of Elga's 1/3 position. Lewis is arguing for the 1/2. Elga is the one arguing the 1/3, and both of their responses have already been posted, by myself.

Lewis argues two cases, L2, which is P(Heads)=1/2=P(Tails) and then he notes quod erat demonstrandum. Then moves on to L6 which is P with the plus. Read the article to find out why it has a plus.

Beauty conditionalizes on being awakened, so the values change to 1/4 / 3/4 = 1/3. — Srap Tasmaner

Yes, to expand, P(Heads|Awake) = P(Heads and Awake) / P(Awake) = 1/4 / 3/4 = 1/3.

What that equation is doing is redistributing probability from the sleep states to the awake states in proportion to each awake state's probability. In the case of Sleeping Beauty, that is equivalent to being indifferent about which awake state she is currently in, since each awake state has the same 1/4 probability. That is, when conditionalizing on being awake, the sleep state probability of 1/4 is evenly distributed to each awake state (1/4 + (1/4 / 3) = 1/3).

In other scenarios where the state probabilities are unequal (such as the 5/6 heads-weighted coin scenario), it is first necessary to transform the original state space to states of equal probability. Then the indifference principle similarly applies. So for the 5/6 heads-weighted scenario, we get:

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

Lewis agrees with me that there is no new relevant information which would allow reallocation of credibility. I just disagree with Lewis that that condition excludes the 1/3 position.

Oh good. I thought the OP didn't make this clear. The halvers would have it right if she went into this thing blind. — noAxioms

It says right in the OP, " Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details." That is the first line of the problem.

Read the article to find out why it has a plus. — Jeremiah

I am aware why it has the plus. As the paper says, "Let P+ be her credence function just after she's told that it's Monday". From Lewis' halfer argument:

(L6) P+(Heads) = 2/3

(L6) is a consequence of the halfer position. Do you agree?

Do you agree? — Andrew M

What on Earth do you think that tag question does?

I just gave you my response and I am not going to wade through your clear misunderstanding of Lewis's argument.