The Sleeping Beauty Problem: Any halfers here?

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In summary, the Sleeping Beauty Problem is a thought experiment that challenges the concept of subjective probability. It poses the question of whether Sleeping Beauty, who is woken up multiple times during an experiment, should have the same belief about the outcome each time or if her belief should change based on the probability of the event. This problem has sparked debate among philosophers and has implications for understanding the nature of consciousness and the role of probability in decision-making.

What is Sleeping Beauty's credence now for the proposition that the coin landed heads?

  • 1/3

    Votes: 12 33.3%
  • 1/2

    Votes: 11 30.6%
  • It depends on the precise formulation of the problem

    Votes: 13 36.1%

  • Total voters
    36
  • #526
First let me apologize. I basically never leave a comment on a thread I haven't read all the way through. This time I'm going to make an exception to ask about my reasoning on the problem. I'm sorry, I read the first several pages thoroughly but 27 is just not in the cards for me right now, so please anyone feel free to ignore me if you want and I will not be offended.

Looking at the original formulation of the problem, as on the Wikipedia page. Say we have two princesses, one's a thirder and one's a halfer. Say both are put through the experiment many times (independently). Also say we award each princess a dollar for each time they correctly guess the coins toss result for that week.

The halfer thinks both heads and tails are equally likely, so she might as well guess heads every day, and it should make no difference from guessing tails, so that's what she does (always guess heads). The thirder, on the other hand, believes tails is twice as likely as heads, and so guesses tails every time. At the end of many iterations of the experiment, isn't it pretty clear that the thirder is going to have a lot more dollars than the halfer?

Is this a good argument that the thirders are right?
 
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  • #527
eloheim said:
At the end of many iterations of the experiment, isn't it pretty clear that the thirder is going to have a lot more dollars than the halfer?
Yes, but that's an expected value, not a probability, and the question is about a probability.

The situation you describe is one in which the probability of Heads is 0.5 and the Princess gets paid $1 over the course of the whole experiment if she correctly guesses Heads and $2 over the course of the whole experiment if she correctly guesses Tails. Hence the expected payoff for guessing heads is 50c and for guessing Tails it is $1. But those are not probabilities.
 
  • #528
Stephen Tashi said:
I think the answer is ( (1/2)(1) + (1/2)(1/99) , which may offend people's intuition, but it doesn't contradict any information given in the problem.
That would lead to some really bad betting
 
  • #529
Yes, it would offend my intuition to think I have a 50-50 chance of winning a poker hand with a pair of twos, but more to the point, I'd lose my shirt. So it's not about intuition, it's about payoff. There is no need to even invoke a concept of probability, one only needs to know the betting odds, which is about expected payoffs.
 
  • #530
andrewkirk said:
Yes, but that's an expected value, not a probability, and the question is about a probability.
Hmm..I'll have to do a little reading about definitions in probability theory I think to fully understand this. Thanks for the reply.
 
  • #531
Ken G said:
Yes, it would offend my intuition to think I have a 50-50 chance of winning a poker hand with a pair of twos, but more to the point, I'd lose my shirt. So it's not about intuition, it's about payoff. There is no need to even invoke a concept of probability, one only needs to know the betting odds, which is about expected payoffs.

You make an analogy to a different situation which does have as set of equilprobable outcomes. A person who has an intuitive objection the the "halfer"'s answers can reject the "halfer' model for the distribution of F and assume a different model. My only point is that assumptions are being made.

Beauty' can determine her betting strategy without making any assumption that the probability of heads changes to 1/3 after she awakens. She can compute her betting strategy using the assumption that the probabiiity of heads is 1/2 and using her knowledge of the conduct of the experiment. That computation does not imply that she should give even odds to (always) betting about what day of the week it is.
 
  • #532
Stephen Tashi said:
Beauty' can determine her betting strategy without making any assumption that the probability of heads changes to 1/3 after she awakens.
The betting strategy and the credence are linked by definition (specifically the definition of credence).
 
  • #533
eloheim said:
The halfer thinks both heads and tails are equally likely, so she might as well guess heads every day, and it should make no difference from guessing tails, so that's what she does (always guess heads).

That isn't true because a "betting strategy" is more than a simple estimate of a probability. Using the fact that the probability of heads is 1/2, Sleeping Beauty would arrive at a betting strategy that did not give even odds to the coin landing heads. This because if it lands tails, she is required to make her bet on Monday and Tuesday and is thus penalized twice, but if it lands heads, she only wins once.
 
  • #534
Dale said:
The betting strategy and the credence are linked by definition (specifically the definition of credence).

The definition of credence describes buying a single bet - see the definition of credence.
 
  • #535
Stephen Tashi said:
The definition of credence describes buying a single bet - see the definition of credence.
Yes. Exactly. It is a single bet purchased each time she determines her credence.
 
  • #536
Dale said:
Yes. Exactly. It is a single bet purchased each time she determines her credence.
And since she knows she might have to buy the bet twice, she is purchasing an agreement that includes the possibility of being forced to make two bets.
 
  • #537
Stephen Tashi said:
And since she knows she might have to buy the bet twice, she is purchasing an agreement that includes the possibility of being forced to make two bets.
More like she is rational so given the same information she will make the same bet multiple times. That is also stipulated in the problem setup.

Do you now agree that the halfer solution is incompatible with the problem as stated, in particular with the definition of credence and the rationality of Beauty.
 
  • #538
@Dale Is there a definition of credence that has been agreed somewhere in the 537 posts? With a quick skim, all I could see was a link to this Wikipedia article on credence, which does not give a formal definition, instead just offering a few vague sentences and a single example, none of which seem to help much with its interpretation in this context.

Thanks

Andrew
 
  • #539
andrewkirk said:
Is there a definition of credence that has been agreed somewhere in the 537 posts?
In post 384 I have several links I found useful. The second one has a brief definition that we have seemed to settle on.

The Sleeping Beauty Problem: Any halfers here?
 
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  • #540
Dale said:
Do you now agree that the halfer solution is incompatible with the problem as stated, in particular with the definition of credence and the rationality of Beauty.

No, why would I? (Goody, a chance to recapitulate.) So far nobody has shown the "halfer" solution contradicts information given in the problem. People have said they don't like the implications of the halfer solution - well, that's not a mathematical argument. Further, the halfer solution has nothing to do with a betting strategy. A betting strategy is computed using the value 1/2 for the probability of heads and the specification of the experiment - which says that a bet is offered each time Sleeping Beauty is awakened - not at randomly selected times that Sleeping Beauty awakens. A person's opinion about the probability distribution that describes the state of the experiment at a random awakening is irrelevant to computing a betting strategy. Both the "thirder" and "halfer" solutions for P(H|awake) are solutions. Which one you use, doesn't affect the betting strategy.

Since Sleeping Beauty can (rationally) determine a betting strategy by using the value P(Heads) = 1/2 and her knowledge of the conduct of the experiment, it is clear that her answer for "credence" is not a bet just on the value of P(Heads | Sleeping Beauty is awakened). She picks a strategy based on "what I ought say" given that she may have to say it twice - unless she takes the question "What is your credence that the coin landed heads?" literally. If she takes the question literally, she must be given a bet on that event alone, not a bet that involves that event plus some other consequences that affect the net payout.
 
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  • #541
Stephen Tashi said:
So far nobody has shown the "halfer" solution contradicts information given in the problem.
It contradicts the stipulation that Beauty is rational by requiring her to knowingly lose money on bets.

If Beauty's credence for the coin being heads is 1/2, then she will necessarily buy a $1.00 bet that it is heads at a price of $0.40. She would expect to lose money on this bet, but would buy it anyway. That is irrational.

Stephen Tashi said:
Which one you use, doesn't affect the betting strategy.
It had better affect it since that is the definition of credence.
 
  • #542
Dale said:
If Beauty's credence for the coin being heads is 1/2, then she will necessarily buy a $1.00 bet that it is heads at a price of $0.40.
Are you talking about a situation in the experiment? Rationally, she doesn't base her purchases of bets during the experiment on an estimate of P(Heads| awakened) (which has no unique answer that she can compute). She bases her answers in the experiment on a betting strategy, which is indeed computed using the fact that P(Heads) = 1/2.

She would expect to lose money on this bet, but would buy it anyway.
You need to define the bet you are talking about.
 
  • #543
Stephen Tashi said:
P(Heads| awakened) (which has no unique answer that she can compute).
This is wrong. I showed how to compute it uniquely in post 255.

Stephen Tashi said:
computed using the fact that P(Heads) = 1/2
Yes, P(H) =1/2 is given.

Stephen Tashi said:
You need to define the bet you are talking about.
This is completely standard usage (unlike you I am not trying to change the meaning of the terms here): A $1.00 bet that the coin landed heads, means that if the coin landed heads she gets a prize of $1.00 and if not she gets $0.00.

Such a bet, purchased at $0.40, would be expected to lose money. Yet she would necessarily (and irrationally) buy it if her credence that the coin landed heads is 1/2.
 
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  • #544
Dale said:
This is completely standard usage (unlike you I am not trying to change the meaning of the terms here): A $1.00 bet that the coin landed heads, means that if the coin landed heads she gets a prize of $1.00 and if not she gets $0.00.
Then , unlike the situation in the experiment, the price paid for the the bet cannot have the consequence that if the coin doesn't land heads, she will purchase the bet twice at that price and win nothing each time.

Such a bet, purchased at $0.40, would be expected to lose money. Yet she would necessarily (and irrationally) buy it if her credence that the coin landed heads is 1/2.

If you want to talk about Sleeping Beauty's credence that the coin lands heads (without any other consequences that affect the net payout) then please define a bet to offer Sleeping Beauty that has those properties.

A rational Sleeping Beauty is not a "thirder" or a "halfer" and knows she cannot compute P(Heads | | Aawkened) because finding that probability is an ill-posed problem unless she makes some assumptions. So when the experimenter proposes any bet about whether the coin landed heads" she bases her decisions on her betting strategy - assuming the experimenter will accept her stated credence as implying a price she will pay for the bet.

If Sleeping Beauty was rational and honest (or naive), she would point out to the experimenter that no bet has been offered to her that is simply "You get $1 if the coin landed heads" without any other strings attached. However if Sleeping Beauty is rational and devious, she will not point this out to the experiment and (if the experimenter is you) the experimenter will accept her answers as indicating her credence that the coin landed heads.
 
  • #545
Stephen Tashi said:
Then , unlike the situation in the experiment, the price paid for the the bet cannot have the consequence that if the coin doesn't land heads, she will purchase the bet twice at that price and win nothing each time.
Why not? If she expects to make money on the bet then she should want to purchase it as often as possible. Do you have a reference that says that a person cannot rationally purchase a good bet more than once?

Stephen Tashi said:
If you want to talk about Sleeping Beauty's credence that the coin lands heads (without any other consequences that affect the net payout) then please define a bet to offer Sleeping Beauty that has those properties.
The bet I described is exactly the one in the definition of credence. The price is below the "buy or sell" price. There is nothing in the definition restricting the number of times the bet is to be purchased or any of the other objections you raise or properties you require. You are adding those to bend the situation to your liking. Just apply the definition and you get the thirder result

Stephen Tashi said:
she cannot compute P(Heads | | Aawkened)
False

Stephen Tashi said:
no bet has been offered to her that is simply "You get $1 if the coin landed heads" without any other strings attached.
False
 
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  • #546
Stephen Tashi said:
Beauty' can determine her betting strategy without making any assumption that the probability of heads changes to 1/3 after she awakens. She can compute her betting strategy using the assumption that the probabiiity of heads is 1/2 and using her knowledge of the conduct of the experiment. That computation does not imply that she should give even odds to (always) betting about what day of the week it is.
The issue is only one thing-- what is her betting strategy. You said in the case of 99 days, she should bet at 50-50 odds that it is Monday, that this is a good bet for you. I say, she will lose her shirt.
 
  • #547
Dale said:
Why not? If she expects to make money on the bet then she should want to purchase it as often as possible. Do you have a reference that says that a person cannot rationally purchase a winning bet more than once?
Look at situation in the experiment. If she offers to pay X for the bet "You get $1 if the coin lands heads" every time it is offered then, assuming P(HEADS) = 1/2, her expected expenditure on bets is (1/2)X + (1/2)2X = (3/2)X. He expected gain is (1/2)(1). So the actual expected expenditure is not X, but (3/2)X and her expected gain is 1/2. She knows the price she offers for the bet will be paid once for an even bet (on Monday) and, with probability 1/2, paid again for a sure losing bet on Tuesday. She is not paying the same price twice for the same bet.

If Sleeping Beauty actually thought the probability that the coin landed heads was 1/3, she would do the above calculation with P(HEADS) = 1/3 and conclude her expected expenditure is (1/3)X + (2/3)2X = (5/3)X and her expected gain is 1/3. So when asked for her credence she would answer 1/5.

The bet I described is exactly the one in the definition of credence. There is nothing in the definition restricting the number of times the bet is to be purchased
As pointed out above, it isn't the same bet that is purchased.
 
  • #548
Stephen Tashi said:
She is not paying the same price twice for the same bet.
Yes, she is. It is exactly the same bet word for word. And it could be resolved identically and immediately each time. It is exactly the same bet as defined every time she is asked her credence.
 
  • #549
Ken G said:
Then why don't you answer my question-- what is her betting strategy, if she bets on it being Monday?

Asume she pays X for the bet "You get $1 if today is Monday" and must do this every time the bet is offered. The bet is offered each time she is awakened. Taking P(Heads) = 1/2, her expected net expenditure is (1/2)X + (1/2)2X = (3/2)X. Her expected gain is (1/2)1 + (1/2)(1) = 1. She should state her "credence" as 2/3 and hope the experimenter accepts that number as her credence for the event "Today is Monday".

Of course if she uses P(Heads) = 1/3, she computes a different fair price for the bet - a wrong one, I think.
 
  • #550
Dale said:
Yes, she is. It is exactly the same bet word for word. .
"Word for word" isn't sufficient to show two bets are the same. The expected payoff needs to be the same.
 
  • #551
Stephen Tashi said:
She should state her "credence" as 2/3 and hope the experimenter accepts that number as her credence for the event "Today is Monday".
As you have described it, that is in fact her credence since it gives the price she would buy or sell the defined bet.
 
  • #552
Stephen Tashi said:
"Word for word" isn't sufficient to show two bets are the same. The expected payoff needs to be the same.
Every bet has the same expected payoff. Procedurally they can all be done the same too. They are the same linguistically, financially, and procedurally.
 
  • #553
Dale said:
As you have described it, that is in fact her credence since it gives the price she would buy or sell the defined bet.
We're going round and round. The experimenter offers SB a bet stated as "If the coin landed heads you win $1". On Monday, this is fair bet. On Tuesday, it is a sure loser. SB knows that if she says she is wiling to pay X for the bet, she will do so on both bets whenever the second bet is offered. If you want to define a bet to measure SB's credence for the event "the coin landed heads" then please define a bet where she does not incur the possibility of having to buy a second bet with a different expected payoff.
 
  • #554
Stephen Tashi said:
The experimenter offers SB a bet stated as "If the coin landed heads you win $1". On Monday, this is fair bet. On Tuesday, it is a sure loser. SB knows that if she says she is wiling to pay X for the bet, she will do so on both bets whenever the second bet is offered.
Yes. So she had better factor all that into the price she is willing to pay for the bet.

Stephen Tashi said:
If you want to define a bet to measure SB's credence for the event "the coin landed heads" then please define a bet where she does not incur the possibility of having to buy a second bet with a different payoff
The bet to measure her credence is already defined. No exceptions are listed in the definition.

Stop trying to change the definition, just use it as stated.
 
  • #555
Dale said:
Every bet has the same expected payoff.
No.

If SB were awakened at random days in random experiments, the bet "You get $1 if the coin lands heads" could be assigned a single expected payoff because the 3 situations where the bet happens are random variables chosen by "random sampling with replacement"

However, in computing the betting strategy, the systematic plan of the experiment is considered. After the coin toss, the situations that arise are determined, not random.

The bet to measure her credence is already defined. No exceptions are listed in the definition.

Here's a bet: "You get 1$ if the coin lands heads - and we toss another coin and you lose $10 if it lands heads".

Do you really think the fair price for that bet is the same as the fair price for "You get $1 if the coin lands heads"?
 
  • #556
Dale said:
In post 384 I have several links I found useful. The second one has a brief definition that we have seemed to settle on.

The Sleeping Beauty Problem: Any halfers here?
thanks for that. From the Stanford article we get the following definition, which I'm reproducing here so everybody can see it without having to follow links:

SEP said:
Your degree of belief [credence] in E is p iff p units of utility is the price at which you would buy or sell a bet that pays 1 unit of utility if E, 0 if not E.

With that in place, we now need to know exactly what SB is betting on. In particular, we need to know whether she knows, before the experiment begins, that she will be offered a bet every day or, if not, on what basis it will be determined whether she will be offered a bet. If she does not know that, and remember it each time she wakes up, none of the analysis I've read so far about betting applies.

Let's assume that she does know, and always remembers, that each day she will be offered the opportunity to pay ##x## dollars for a chance to name Heads or Tails and then immediately win $1 if what she named was correct, and also that ##x## will be the same every day! Then it's a simple exercise in expected values of net payoffs, as per my previous post.

The expected payoff, over the course of the entire experiment, of a strategy that always guesses Heads (recalling that Heads is associated with only being woken on Monday, on the wiki presentation) is (giving the probability for the case where the coin was actually Heads in the first term):
$$0.5\times (1-x)+ 0.5\times 2 \times(0-x)=0.5-1.5x$$

So, for the bet to be worth taking, it must have an expected payoff greater than 0.

For Heads, this means that ##0.5-1.5x>0## so that ##x<1/3##

If SB is offered the Heads bet with ##x=0.33##, and remembers the full details of the betting program, including that she'll be offered the same bet, at the same price, every day, she will take the bet.

So on that definition, and with those very specific rules about betting, (which were not stated in the original problem statement), we might say that the credence in the proposition that the coin came up Heads is one third.

A different conclusion is reached if SB has not been told that she will be offered a bet every day. Then she cannot do the above calculation. Her expected payoff will be the expected value
$$0.5\times (1-x)\times P_1+ 0.5\times (0-x)\times P_2+ 0.5\times (0-x)\times P_2P_3+ 0.5\times (0-x)\times (1-P_2)P_4
$$

where ##P_1## is the probability of being offered a bet on Monday, given that the coin landed Heads,
##P_2## is the probability of being offered a bet on Monday, given that the coin landed Tails, and
##P_3## is the probability of being offered a bet on Tuesday given that a bet was offered on Monday, and that the coin landed Tails.
##P_4## is the probability of being offered a bet on Tuesday given that a bet was NOT offered on Monday, and that the coin landed Tails.

This is positive if ##x<\frac{P_1}{P_1+P_2+P_2P_3+(1-P_2)P_4}##. SB would make an assumption for these three probabilities in order to decide whether to bet. One possible assumption is that the probability of being offered a bet on any day is the same as any other day and independent of whether bets have been offered on other days. If so, the credence is 1/3 again.

However, an equally plausible is the assumption that ##P_1=P_2=P_4## but that only one bet will be offered, so that a bet will not be offered on both Monday and Tuesday. In that case ##P_3=0## and the credence is
$$\frac{P_1}{3P_1-P_1{}^2}=\frac1{3-P_1}$$
which will be somewhere between a third and a half.

So it depends on SB's knowledge of what betting opportunities will be made available to her.
 
  • #557
andrewkirk said:
we might say that the credence in the proposition that the coin came up Heads is one third.
That's a subject of debate in recent posts. SB can calculate her betting strategy using P= 1/2 and never worry about calculating P(heads | awakened). In fact P(heads|awakened) has no unique value and is irrelevant to planning a betting strategy. So SB can determine what she ought pay for the bet "The coin lands heads" without computing a posterior probability that the coin lands heads. Her fair price for the bet is dishonest as a report of her credence because, according to the definition of "credence", she should imagine a bet where "You get $1 if the coin lands heads" is bet with no strings attached. However, in computing her betting strategy, as you illustrated, she considers that it might obligate her pay price X for another bet on Tuesday that is worded the same, but is a sure loser.
 
  • #558
Stephen Tashi said:
Asume she pays X for the bet "You get $1 if today is Monday" and must do this every time the bet is offered. The bet is offered each time she is awakened. Taking P(Heads) = 1/2, her expected net expenditure is (1/2)X + (1/2)2X = (3/2)X. Her expected gain is (1/2)1 + (1/2)(1) = 1. She should state her "credence" as 2/3 and hope the experimenter accepts that number as her credence for the event "Today is Monday".

Of course if she uses P(Heads) = 1/3, she computes a different fair price for the bet - a wrong one, I think.
But you are mixing up the probability of heads in one way of framing the calculation, with her credence, which can be shown to be quite different. Most obvious is the 99 day version. When you say P(Heads)=1/2, you are talking about the probability that heads was flipped originally, which is of course indeed 1/2 (but that's not her credence, as we shall see). So in a single 99-day experiment, her expected expenditure is (1/2)X+ (1/2)99X = 50X. Her expected gain is 1, as you say. Ergo, she should state her credence that it is Monday as 1/50, as X=1/50 is the fair bet that it's Monday-- that is what you would get and that's correct. The problem is, the use of P=1/2 in that calculation is simply not her credence that heads was flipped, that's the probability of a heads flip taking the experiment from a kind of disinterested external perspective, analyzing the situation from the start (perhaps, from the perspective of the experimenters trying to figure out if they will bilk money out of SB). What you are not seeing yet is that if she has a credence of 1/50 that it is Monday (which we have agreed is true), and the only possible way it could be any other day is if the coin flip was tails, this clearly implies she strongly suspects the coin flip was tails, every time she is awakened. That simple truth is quite independent from the use of P(heads) = 1/2 in the above calculation.

To see the calculation from SB's perspective, it looks like this. Her credence that it is Monday is 1/50, and if it is Monday, her credence that the coin was heads is 1/2. That contributes 1/100 to her credence that the coin is heads. Her credence that it is any other day is 49/50, and if it is any other day, her credence that the coin was heads is 0. Hence her total credence that the coin was heads is 1/100, not 1/2. Furthermore, she can correctly determine her expected payoff on her bet that it is Monday by taking X = 1/50. So in the long run, for the 99 day version, she can break even if she always pays 1/50 for the bet "today is Monday" at $1 payoff, and she can also break even if she always pays 1/100 for the bet "the coin was heads" at $1 payoff. If she pays 1/2 for that latter bet, she is a chump, because her expected payoff is (1/2)*1, but her expected cost is 1/2*1 + 1/2*99 = 50. She's losing $49 every time she accepts that strategy, this is an error. So notice that even though the 1/2 appears in the calculation I just did, that's the probability of heads taken from the start of the experiment, it is not her credence on the bet "it was a heads." That's clear because if she thinks it's her credence, she gets bilked, and that proves it is not the correct credence. You should see the contradiction in thinking her credence that it is Monday is 1/50, but her credence that the coin was heads is 1/2.
 
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  • #559
andrewkirk said:
So it depends on SB's knowledge of what betting opportunities will be made available to her.
But that's obvious, it would be like playing poker and not being sure the dealer isn't dealing from the bottom of the deck. Of course that will mess up your chances, but that's just clearly not the way poker is played, it would be a cheat. SB only needs to know she is not being cheated, just as any poker player must assume, it goes without saying. This is also why the correct answer to the question is not "it depends on the precise formulation," because if you are playing bridge and you want to know the chances of dropping the queen if the defense holds 4 cards, you can look up that credence in a bridge book, and I guarantee you that you will never see "the answer depends on whether or not the dealer was cheating." SB must be told what the betting strategy is, and it must be the truth, and if that betting strategy is that she will be offered the bet every time she wakes, her credence is 1/3-- it's not an assumption, anything else would be a cheat.
 
  • #560
Ken G said:
SB only needs to know she is not being cheated, just as any poker player must assume.
only needs to know that, in order to reach what conclusion?

If the conclusion is to be that she is willing to pay 33c for a bet that pays $1 if the coin landed Heads and zero otherwise, she needs more information than that - specifically, what are the probabilities of me being offered the same bet on every other day.

I'm afraid I know nothing about poker, so I can't really assess the extent to which the rest of your post relates to the SB situation.
 

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