I took a look at your table, and I have to say, it was a great variant. It took me a while to de-confuse my thoughts about it, and made me realize I wasn’t as comfortable with the SB problem as I thought.
Initially I was a bit uncertain what you meant by “round”, whether it was a wake-up event of a full iteration of the experiment. After rereading a few times, I became more certain that it was the latter.
So, you gain $30 if the coin lands heads and $6 + $6 = $12 if it lands tails. So your expectation of gain per iteration of the experiment is 1⁄2 * $30 + 1⁄2 * $12 = $21
There are 3 equally likely wake-up events, so the expectation of gain per wake-up event is 1⁄3 * $30 + 1⁄3 * $6 + 1⁄3 * $6 = $14
We can notice that there are on average 1.5 wake-up events per iteration, and $14 * 1.5 = $21, so that checks out.
Now, let’s say I experience a wake-up event. I believe that there is 1⁄3 chance that the coin landed heads. You think that my calculation for the expected gain per iteration will be 1⁄3 * $30 + 2⁄3 * $6 = $14
However, I reason: if the coin landed tails, I won’t gain $6, I will gain $6 * 2, once for Monday and once for Tuesday. So my expected gain per iteration is 1⁄3 * $30 + 2⁄3 * $6 * 2 = $18
That’s the point where I was confused. What did that $18 mean? What actually helped me figure it out was the betting scheme that you added in your post.
Sleeping Beauty is put to sleep like normal with only the usual information supplied to her about the original SB problem. When she awakes, there is a surprise message left for her:
You may agree to play the following game: If the coin had flipped Heads, your bank account will receive $30 right now. If the coin had flipped Tails, your bank account will receive $6 right now. However, once this round of experiment is over and you awake, you will pay $18 for having played this game. (If you are inconsistent between wakings about whether you agree to play the game, the game is considered defunct and all bank transfers will be reverted.)
This betting scheme is a based on a version of the SB problem where the Monday guess and the Tuesday guess are counted as a single one. So, Tuesday is unnecessary, so (see previous comment) it boils down to the trivial prediction of the result of a fair coin toss. $21 - $18 = $3 gain expectation.
Next I tried a few variations:
if the bets are per wake-up (SB can bet once on Monday and another time on Tuesday), but SB is only paid for her current situation (if she bets on Monday she gains $6 and pay $18, same for Tuesday), then her gain expectation is 1/3($30 - $$6 - $18) = -$4. In other words, it’s $14 (average gain expectation per wake-up event) - $18 (cost to bet per wake-up event).
if the bets are per wake-up, but SB is paid for both Monday and Tuesday when she bets on either Monday or Tuesday, then her gain expectation is 1/3($30 - $$12 - $18) = $0. So that’s what the $18 found with the thirder calculation meant: it’s the gain expectation, over the whole iteration, but knowing that she’s in a wake-up event. In hindsight, I should have guessed, because it naturally corresponds to the formulations where the thirder answer is correct. It is not meaningless: if you repeat the experiment a large number of times, and then tag each wake-up event with the profit that SB made during the iteration that included that event, then the average of tags will be $18. In other words, when SB’s Monday and Tuesday answers are not conflated in a single one, it does answer the question (to SB during a wake-up event): how much money do you expect to gain during this iteration? It’s not intuitive, but it checks out.
So, cool variation, but ultimately each interpretation finds the result that it expected.
$14 was a mistake, that’s my bad and I’ve updated the post.
>> This betting scheme is a based on a version of the SB problem where the Monday guess and the Tuesday guess are counted as a single one
This is the piece I don’t understand about thirders (apologies if that sounds like aggressive wording): If you have a credence for the coin, why can’t you use it?
>> (your second variant) SB is paid for both Monday and Tuesday when she bets on either Monday or Tuesday… if you repeat the experiment a large number of times, and then tag each wake-up event with the profit that SB made during the iteration that included that event, then the average of tags will be $18
Not exactly. I laid out a separate $6 on Mon and $6 on Tues to make this easier to think about. Combining into $12, you end up with the same amount of money being equally shared between two days. I don’t think it’s accurate to say that SB made $12 on Mon and also say that SB made $12 on Tues. So in other words, I think this scheme is doing double counting. Which also means to me that the $18 value doesn’t really correspond to anything relevant and useful to the scenario?
This is the piece I don’t understand about thirders (apologies if that sounds like aggressive wording): If you have a credence for the coin, why can’t you use it?
No problem, I am often in that situation too of being puzzled by the other side of an argument :)
And, it’s not like I’m confident that I fully understand the SB problem.
I’m guessing that the crux is this: for me, if (let’s call it experiment 1) you interrogate SB once after the coin landed tails, she should say her credence (of heads) is 1⁄2. But if (experiment 2) you interrogate her twice (under amnesia), she should say her credence is 1⁄3. You seem to disagree with that, on the grounds that her answer means something else than what the “SB problem” is about? But if you run the experiments many times, in experiment 1 half of SB’s answers will happen after a heads outcome, while in experiment 2 it’s only a third, so it means something.
I don’t think it’s accurate to say that SB made $12 on Mon and also say that SB made $12 on Tues. So in other words, I think this scheme is doing double counting. Which also means to me that the $18 value doesn’t really correspond to anything relevant and useful to the scenario?
The $18 value corresponds to how much money SB expects to make during this iteration on average. I’m curious about your answer to the following experiment:
No coin. SB is put to sleep on Sunday, awoken on Monday, interrogated, and put to sleep+amnesia. Then the same happens on Tuesday: awoken, interrogated, and put to sleep+amnesia. When interrogated (on Monday and Tuesday), they tell her: “Here is $10, how much do you think you will have gained at the end of the experiment on Wednesday?” What should be her answer? If you answer $20, do you see how the same reasoning is part of my $18 answer on the previous problem?
I take back one piece, when I said $18 doesn’t correspond to anything. I believe that corresponds to how much Beauty should expect to be making per day on average: (1/2)(30) + (1/2)(6).
If we’re talking about expectation per iteration, though, then it should be (1/2)(30) + (1/2)(12) = $21. If you’re using the $18 value there, then you can get money pumped.
Let’s be sure we agree on the parameters of the bet. If SB agrees to bet, she pays a fee, then:
she gains $30 if the coin came up heads.
she gains $12 if the coin came up tails.
And the question is: How much should she be willing to pay to bet? (assuming she’s ok to break even)
Answer: depends on when you ask SB.
If you propose the bet to SB on Sunday or Wednesday, she should be willing to pay $21.
If you propose the bet during a wake-up event, she should be willing to pay only $18. That’s the meaning of the $18. The difference comes from the fact that when she wakes up, the coin has only 1⁄3 chance of having landed heads. That’s where the thirder position comes from. Test it if you don’t believe it checks out.
Over many repetitions (to balance out bad luck streaks), it’s rational to play the game for prices of $20 or $19. If you’re saying that SB should reject those prices when presented it while waking, that’s irrational, because an SB that accepts those prices will make money.
This is assuming that if asked twice, she’ll always make the same decision because her decision-making is deterministic. (We could assume there’s a chance she makes a different decision on Tuesday than on Monday but I think that’s an unnecessary and irrelevant complication?)
Just to be sure: in your original formulation, the bet counted only once for the two wake-up events in the Tails case, and it was canceled if she didn’t accept it on Monday and Tuesday.
That is not the scheme I described in previous comment. Each wake-up event is independent. She is awoken then asked to pay the fee now and gets the rewards now. If she pays twice she gets the rewards twice. It’s a bet that can be done, it is no less valid than yours. In this bet (when asked during a wake-up event), her gain expectation is $18. Accepting for $19 will lose money on average. Her decision-making process on the other day is irrelevant. Test it if you don’t believe it.
Thanks for taking a look! I’ll be sure to go over what you’ve written—I ran out of time to spend any more time on this stuff for this weekend / this week so I won’t be able to reply soon, just wanted to let you know that your response is appreciated
I took a look at your table, and I have to say, it was a great variant. It took me a while to de-confuse my thoughts about it, and made me realize I wasn’t as comfortable with the SB problem as I thought.
Initially I was a bit uncertain what you meant by “round”, whether it was a wake-up event of a full iteration of the experiment. After rereading a few times, I became more certain that it was the latter.
So, you gain $30 if the coin lands heads and $6 + $6 = $12 if it lands tails. So your expectation of gain per iteration of the experiment is 1⁄2 * $30 + 1⁄2 * $12 = $21
There are 3 equally likely wake-up events, so the expectation of gain per wake-up event is 1⁄3 * $30 + 1⁄3 * $6 + 1⁄3 * $6 = $14
We can notice that there are on average 1.5 wake-up events per iteration, and $14 * 1.5 = $21, so that checks out.
Now, let’s say I experience a wake-up event. I believe that there is 1⁄3 chance that the coin landed heads. You think that my calculation for the expected gain per iteration will be 1⁄3 * $30 + 2⁄3 * $6 = $14
However, I reason: if the coin landed tails, I won’t gain $6, I will gain $6 * 2, once for Monday and once for Tuesday. So my expected gain per iteration is 1⁄3 * $30 + 2⁄3 * $6 * 2 = $18
That’s the point where I was confused. What did that $18 mean? What actually helped me figure it out was the betting scheme that you added in your post.
This betting scheme is a based on a version of the SB problem where the Monday guess and the Tuesday guess are counted as a single one. So, Tuesday is unnecessary, so (see previous comment) it boils down to the trivial prediction of the result of a fair coin toss. $21 - $18 = $3 gain expectation.
Next I tried a few variations:
if the bets are per wake-up (SB can bet once on Monday and another time on Tuesday), but SB is only paid for her current situation (if she bets on Monday she gains $6 and pay $18, same for Tuesday), then her gain expectation is 1/3($30 - $
if the bets are per wake-up, but SB is paid for both Monday and Tuesday when she bets on either Monday or Tuesday, then her gain expectation is 1/3($30 - $
So, cool variation, but ultimately each interpretation finds the result that it expected.
Sorry for taking so long!
$14 was a mistake, that’s my bad and I’ve updated the post.
>> This betting scheme is a based on a version of the SB problem where the Monday guess and the Tuesday guess are counted as a single one
This is the piece I don’t understand about thirders (apologies if that sounds like aggressive wording): If you have a credence for the coin, why can’t you use it?
>> (your second variant) SB is paid for both Monday and Tuesday when she bets on either Monday or Tuesday… if you repeat the experiment a large number of times, and then tag each wake-up event with the profit that SB made during the iteration that included that event, then the average of tags will be $18
Not exactly. I laid out a separate $6 on Mon and $6 on Tues to make this easier to think about. Combining into $12, you end up with the same amount of money being equally shared between two days. I don’t think it’s accurate to say that SB made $12 on Mon and also say that SB made $12 on Tues. So in other words, I think this scheme is doing double counting. Which also means to me that the $18 value doesn’t really correspond to anything relevant and useful to the scenario?
No problem, I am often in that situation too of being puzzled by the other side of an argument :)
And, it’s not like I’m confident that I fully understand the SB problem.
I’m guessing that the crux is this: for me, if (let’s call it experiment 1) you interrogate SB once after the coin landed tails, she should say her credence (of heads) is 1⁄2. But if (experiment 2) you interrogate her twice (under amnesia), she should say her credence is 1⁄3. You seem to disagree with that, on the grounds that her answer means something else than what the “SB problem” is about? But if you run the experiments many times, in experiment 1 half of SB’s answers will happen after a heads outcome, while in experiment 2 it’s only a third, so it means something.
The $18 value corresponds to how much money SB expects to make during this iteration on average.
I’m curious about your answer to the following experiment:
No coin. SB is put to sleep on Sunday, awoken on Monday, interrogated, and put to sleep+amnesia. Then the same happens on Tuesday: awoken, interrogated, and put to sleep+amnesia.
When interrogated (on Monday and Tuesday), they tell her: “Here is $10, how much do you think you will have gained at the end of the experiment on Wednesday?”
What should be her answer?
If you answer $20, do you see how the same reasoning is part of my $18 answer on the previous problem?
I take back one piece, when I said $18 doesn’t correspond to anything. I believe that corresponds to how much Beauty should expect to be making per day on average: (1/2)(30) + (1/2)(6).
If we’re talking about expectation per iteration, though, then it should be (1/2)(30) + (1/2)(12) = $21. If you’re using the $18 value there, then you can get money pumped.
Let’s be sure we agree on the parameters of the bet. If SB agrees to bet, she pays a fee, then:
she gains $30 if the coin came up heads.
she gains $12 if the coin came up tails.
And the question is: How much should she be willing to pay to bet? (assuming she’s ok to break even)
Answer: depends on when you ask SB.
If you propose the bet to SB on Sunday or Wednesday, she should be willing to pay $21.
If you propose the bet during a wake-up event, she should be willing to pay only $18. That’s the meaning of the $18. The difference comes from the fact that when she wakes up, the coin has only 1⁄3 chance of having landed heads. That’s where the thirder position comes from. Test it if you don’t believe it checks out.
Over many repetitions (to balance out bad luck streaks), it’s rational to play the game for prices of $20 or $19. If you’re saying that SB should reject those prices when presented it while waking, that’s irrational, because an SB that accepts those prices will make money.
This is assuming that if asked twice, she’ll always make the same decision because her decision-making is deterministic. (We could assume there’s a chance she makes a different decision on Tuesday than on Monday but I think that’s an unnecessary and irrelevant complication?)
Just to be sure: in your original formulation, the bet counted only once for the two wake-up events in the Tails case, and it was canceled if she didn’t accept it on Monday and Tuesday.
That is not the scheme I described in previous comment. Each wake-up event is independent. She is awoken then asked to pay the fee now and gets the rewards now. If she pays twice she gets the rewards twice. It’s a bet that can be done, it is no less valid than yours.
In this bet (when asked during a wake-up event), her gain expectation is $18. Accepting for $19 will lose money on average. Her decision-making process on the other day is irrelevant. Test it if you don’t believe it.
Thanks for taking a look! I’ll be sure to go over what you’ve written—I ran out of time to spend any more time on this stuff for this weekend / this week so I won’t be able to reply soon, just wanted to let you know that your response is appreciated