For my own benefit, i’ll try to explain my thinking on this problem, in my own words, because the discussions here are making my head spin. Then the rest of you can tell me whether i understand. The following is what i reasoned out before looking at neq1′s explanations.
Firstly, before the experiment begins, i’d expect a 50% chance of heads and a 50% chance of tails. Simple enough.
If it lands on heads, then i wake up only once, on Monday. If it lands on tails, then i wake up once on Monday, and a second time on Tuesday.
So, upon waking with amnesia, i’d expect a 50% chance of it being my first-and-only interview on Monday. I’d expect a 25% chance of it being my first-of-two interviews on Monday, and a 25% chance of it being my second-of-two interviews on Tuesday.
And due to the amnesia, and my having no indication of what day it is, i’d basically have no new information to act on after i wake up. So my probability estimates would remain the same after waking as they were before.
So, upon waking, i’d say:
50% chance that the coin landed on heads, and it’s currently Monday.
25% chance that the coin landed on tails, and it’s currently Monday.
25% chance that the coin landed on tails, and it’s currently Tuesday.
In other words, neq1′s probability-tree picture turned out to most clearly match my own reasoning on the problem. Does this make sense?
* 33% chance that the coin landed on heads, and it's currently Monday.
* 33% chance that the coin landed on tails, and it's currently Monday.
* 33% chance that the coin landed on tails, and it's currently Tuesday.
p(heads) and p(tails) on Monday should be equal (a fair coin was flipped). p(tails) on Monday and p(tails) on Tuesday should also be equal (nothing important changes in the interim).
Even though you knew ahead of time that there was a 50% chance you’d be on the heads path, and a 50% chance you’d be on the tails path, you’d shift those around without probability law justification?
I also think you are not careful with your wording. What does p(heads) on Monday mean? Is it a joint or conditional probability? p(heads | monday) = p(tails | monday), yes, but Beauty can’t condition on Monday since she doesn’t know the day. If you are talking about joint probabilities, p(heads and monday) does not equal p(tails and monday).
Re: a 50% chance you’d be on the heads path, and a 50% chance you’d be on the tails path.
Those are not the probabilities in advance of the experiment being perfomed. Once the experimental procedure is known the subjective probabilites for Beauty on awakening are 33% for heads and 67% for tails. These probabilities do not change during the experiment—since Beauty learns nothing.
“Once the experimental procedure is known the subjective probabilites for Beauty on awakening are 33% for heads and 67% for tails.”
Suppose 50% of the population has some asymptomatic form of cancer. We randomly select someone and do a diagnostic test. If they have cancer (we don’t tell them), we wake them up 9 times and ask their credence for cancer (administering amnesia-inducing drug each time). If they don’t have cancer, we wake them up once.
The person selected for this experiment knows there is a 50% chance they have cancer. And they decide ahead of time that, upon awakening, they’ll be 90% sure they have cancer. And this makes sense to you.
“What is your credence now for the proposition that our coin landed heads?”
...as being equivalent a bet along these lines:
“the scenario where at each awakening we offer a bet where she’d lose $1.50 if heads and win $1 if tails, and we tell her that we will only accept whichever bet she made on the final interview.”
...which is a tortured interpretation.
The question says “now”. I think the correct corresponding wager is for Beauty to make a bet which is judged according to its truth value there and then—not for it to be interpreted later and the payout modified or cancelled as a result of other subsequent events.
Prior to the experiment, the probability of heads was 50%, tails 50%. Upon waking.. she learns no new information. She knew in advance she was going to wake up, and they tell her nothing.
One of the major take-aways I got from actually reading Jaynes was how he is always careful to write probabilities as conditioned on all prior knowledge: P(A|X) where X is our “background knowledge”.
This is useful in the present case since we can distinguish X, Beauty’s background knowledge about which way a given coin might land, and X’, which represents X plus the description of the experimental setup, including the number of awakenings in each case.
That—the difference between X and X’ - is the new information that Beauty learns and which might make P(heads|X’) different from P(heads|X).
She knew from the start that she is twice as likely to be asked when it is tails. So, her estimate of the chances of her being awakened facing tails should be bigger from the beginning.
Thank you, your explanation for the 1⁄3 answer makes sense to me. I’m still a bit confused about it, but i think i feel like i might be changing my mind.
I’ll try to figure out what would happen if SB makes a bet on the coin flip at each interview. Suppose she guesses heads each time, then:
Given that the result was heads, then she is interviewed once, and she is right once.
Given that the result was tails, then she is interviewed twice, and she is wrong twice.
… meaning that if the experiment is repeated several times, the guess “heads” will be correct for one out of three guesses. Just like you said.
(Perhaps it’s important to realize that, if the coin lands on tails, then she’s guaranteed to wake up once on Monday, and also guaranteed to wake up once on Tuesday. Now that i read your other comment again, i see your meaning when you say that p(heads) and p(tails) for each day is the same.)
I couldn’t decide exactly you meant by “twice as likely to be asked (woken) when it’s tails” either. I’m going to guess that you’re averaging evenly over Monday and Tuesday, in which case I agree. After marginalizing over M/T, P(wake|heads)=1/2 and P(wake|tails)=1.
“She knew from the start that she is twice as likely to be asked when it is tails. ”
The probability that she would be asked is 1, regardless of the outcome of the coin. Her estimate of the chances of her being awakened should have been 1.
For my own benefit, i’ll try to explain my thinking on this problem, in my own words, because the discussions here are making my head spin. Then the rest of you can tell me whether i understand. The following is what i reasoned out before looking at neq1′s explanations.
Firstly, before the experiment begins, i’d expect a 50% chance of heads and a 50% chance of tails. Simple enough.
If it lands on heads, then i wake up only once, on Monday. If it lands on tails, then i wake up once on Monday, and a second time on Tuesday.
So, upon waking with amnesia, i’d expect a 50% chance of it being my first-and-only interview on Monday. I’d expect a 25% chance of it being my first-of-two interviews on Monday, and a 25% chance of it being my second-of-two interviews on Tuesday.
And due to the amnesia, and my having no indication of what day it is, i’d basically have no new information to act on after i wake up. So my probability estimates would remain the same after waking as they were before.
So, upon waking, i’d say:
50% chance that the coin landed on heads, and it’s currently Monday.
25% chance that the coin landed on tails, and it’s currently Monday.
25% chance that the coin landed on tails, and it’s currently Tuesday.
In other words, neq1′s probability-tree picture turned out to most clearly match my own reasoning on the problem. Does this make sense?
This was also my understanding of the problem. Are we missing something?
On awakening, I would give:
p(heads) and p(tails) on Monday should be equal (a fair coin was flipped). p(tails) on Monday and p(tails) on Tuesday should also be equal (nothing important changes in the interim).
Even though you knew ahead of time that there was a 50% chance you’d be on the heads path, and a 50% chance you’d be on the tails path, you’d shift those around without probability law justification?
I also think you are not careful with your wording. What does p(heads) on Monday mean? Is it a joint or conditional probability? p(heads | monday) = p(tails | monday), yes, but Beauty can’t condition on Monday since she doesn’t know the day. If you are talking about joint probabilities, p(heads and monday) does not equal p(tails and monday).
Re: a 50% chance you’d be on the heads path, and a 50% chance you’d be on the tails path.
Those are not the probabilities in advance of the experiment being perfomed. Once the experimental procedure is known the subjective probabilites for Beauty on awakening are 33% for heads and 67% for tails. These probabilities do not change during the experiment—since Beauty learns nothing.
“Once the experimental procedure is known the subjective probabilites for Beauty on awakening are 33% for heads and 67% for tails.”
Suppose 50% of the population has some asymptomatic form of cancer. We randomly select someone and do a diagnostic test. If they have cancer (we don’t tell them), we wake them up 9 times and ask their credence for cancer (administering amnesia-inducing drug each time). If they don’t have cancer, we wake them up once.
The person selected for this experiment knows there is a 50% chance they have cancer. And they decide ahead of time that, upon awakening, they’ll be 90% sure they have cancer. And this makes sense to you.
Re: “but Beauty can’t condition on Monday since she doesn’t know the day.”
She could make a bet. You do not have to know what day of the week it is in order to make a bet that it is Monday.
Re: “If you are talking about joint probabilities, p(heads and monday) does not equal p(tails and monday).”
Sure it does—if a fair coin was flipped!
Maybe instead of just saying it’s true, you could look at my proof and show me where I made a mistake. I’ve done that with yours.
I think you already clarified that here.
You interpreted:
“What is your credence now for the proposition that our coin landed heads?”
...as being equivalent a bet along these lines:
“the scenario where at each awakening we offer a bet where she’d lose $1.50 if heads and win $1 if tails, and we tell her that we will only accept whichever bet she made on the final interview.”
...which is a tortured interpretation.
The question says “now”. I think the correct corresponding wager is for Beauty to make a bet which is judged according to its truth value there and then—not for it to be interpreted later and the payout modified or cancelled as a result of other subsequent events.
Yes, this is correct.
My reasoning was a bit simpler.
Prior to the experiment, the probability of heads was 50%, tails 50%. Upon waking.. she learns no new information. She knew in advance she was going to wake up, and they tell her nothing.
So how could her beliefs possibly change?
One of the major take-aways I got from actually reading Jaynes was how he is always careful to write probabilities as conditioned on all prior knowledge: P(A|X) where X is our “background knowledge”.
This is useful in the present case since we can distinguish X, Beauty’s background knowledge about which way a given coin might land, and X’, which represents X plus the description of the experimental setup, including the number of awakenings in each case.
That—the difference between X and X’ - is the new information that Beauty learns and which might make P(heads|X’) different from P(heads|X).
She knew from the start that she is twice as likely to be asked when it is tails. So, her estimate of the chances of her being awakened facing tails should be bigger from the beginning.
Thank you, your explanation for the 1⁄3 answer makes sense to me. I’m still a bit confused about it, but i think i feel like i might be changing my mind.
I’ll try to figure out what would happen if SB makes a bet on the coin flip at each interview. Suppose she guesses heads each time, then:
Given that the result was heads, then she is interviewed once, and she is right once.
Given that the result was tails, then she is interviewed twice, and she is wrong twice.
… meaning that if the experiment is repeated several times, the guess “heads” will be correct for one out of three guesses. Just like you said.
(Perhaps it’s important to realize that, if the coin lands on tails, then she’s guaranteed to wake up once on Monday, and also guaranteed to wake up once on Tuesday. Now that i read your other comment again, i see your meaning when you say that p(heads) and p(tails) for each day is the same.)
I couldn’t decide exactly you meant by “twice as likely to be asked (woken) when it’s tails” either. I’m going to guess that you’re averaging evenly over Monday and Tuesday, in which case I agree. After marginalizing over M/T, P(wake|heads)=1/2 and P(wake|tails)=1.
“She knew from the start that she is twice as likely to be asked when it is tails. ”
The probability that she would be asked is 1, regardless of the outcome of the coin. Her estimate of the chances of her being awakened should have been 1.
Yes: her estimate of the chances of her being awakened is indeed 1.