10 ms·

The sleeping beauty paradox

77 points by xtacy 11y ago

baddox 11y ago
A similar problem is God's Coin Toss as described in Scott Aaronson's excellent lecture series (and book) "Quantum Computing Since Democritus": http://www.scottaaronson.com/democritus/lec17.html http://www.scottaaronson.com/democritus/lec17.html
Gravityloss 11y ago
Maybe it's really an argument about this: Upon awakening: Halfer: P(monday,heads)=0,5 P(monday,tails)=0,25 P(tuesday,tails)=0,25 Thirder: P(monday,heads)=0,33 P(monday,tails)=0,33 P(tuesday,tails)=0,33
- Gravityloss 11y ago
  I think it boils down to the thirder position. P(monday,heads)=0,33 since P(heads)=0,5 and P(monday)=0,67. This is because we are sampling awakenings, not coin tosses or beauties. It's like betting on a coin toss, but with a side twist that the bet is evaluated twice (no new toss) when the coin is tails. Hence a majority of evaluations will be on tails: Guess, Coin, Profit H, H, +1 H, T, -2 T, H, -1 T, T, +2 Analysis for betting heads: average is +1 + -2 = -1 Betting tails: average is -1 + +2 = +1. I'd bet tails!
acchow 11y ago
This reminds me of the Monty Hall problem, but in reverse - you were playing the Monty Hall game and just won the car! You forgot whether or not you switched doors during the second step - what is the probability you switched doors?
- deleted 11y ago
  [deleted]
  - Nadya 11y ago
    I also related it to Monty Hall in that one option is two options bundled together. The options are not [Heads||Tails] but rather [Heads||Tails+Tails]. The Tails+Tails is like having the Car+Goat bundled as a single option. Sleeping Beauty knows the coin is 50/50 but that if she is woken there is more opportunity to be woken during a Tails flip. To show that it is biased in favor of Tails - the experiment must be repeated. It shows more readily if you increase the number of times she is woken if the coin lands on Tails, similar to how the Monty Hall problem becomes more intuitive when you increase the number of doors. Instead of being woken 2 times - let's have her woken 1,000 times. Every time she is woken she is asked if the coin was Heads or Tails, with no recollection of her previous answers. She is given a dollar for every time she is correct. Should she guess Heads or Tails? If she guesses Heads every time - she can only win $1. If she guesses Tails every time - she can win $1,000. If she alternates answers she can only win $500.
- Bartweiss 11y ago
  There's actually a strong analogy to an extended Monty Hall here. Specifically: 1. Monty offers the challenge, and I pick door B. 2. Monty opens door C, showing a goat. 3. Monty asks if I want to switch. My odds of winning are now P(A) = 2/3, P(B) = 1/3. 4. My mom flips on the TV to see me play, but has missed my answers and only sees the state of the doors. What are her odds of guessing right? Result: my mom's odds of winning are 1/2, even though the odds of a given door winning are not. There's a bias between the two doors, but her perspective is neutral - she's guessing which door has better odds, and that guess is unbiased. As you say, this question is the reverse: the odds on the coin flip are unbiased, but my odds of being asked the question are biased towards one of the two outcomes.
  - CamperBob2 11y ago
    See my other post; I actually think that a Monty Hall comparison argues in favor of an unbiased 1:2 guess, because of how the terms of the two problems differ. In the SB question, your odds of being asked the question are 1:1. You just don't know how many times you'll be asked. You didn't know the night before (and neither did the researchers), and you still don't know, even though the researchers now do. In the MH game, you give the host some new information when you make your initial choice. He already knew what door not to open, but now he knows what door he must open, and the rules require him to communicate that to you. That's when you receive the new information (as you point out with the example of your mom walking into the room). In the SB problem, you don't get any new information before the question is asked, including whether or not you're going to be put back to sleep. So the only answer you can rationally give is 1:2. If the researchers used a d20 instead of a coin to determine how many times to wake you up, you could safely guess that any given awakening wasn't your first or your last. But you still can't give any answer about the number on the die, other than a random guess from 1-20. You need to store some information for later recall, and they're not letting you do that.
  - darkmighty 11y ago
    I quite like your phrasing because it highlights that probability is a model of uncertainty for an observer. So the mother has no information on the door, so she is 'neutral' (p=1/2 of guessing right); you have a small amount of information (p=2/3); you can even include Monty which of course has total information (p=1). You can't ask what are the probabilities of finding goats behind each door without specifying an observer and the available information.
- sidcypher 11y ago
  The Monty Hall problem is actually much better for educating people, as it highlights the bias hidden behind the "50/50" oversimplification. This Sleeping Beauty problem, on the other hand, is phrased to confuse you, so that arguing between the "halfers" and "thirders" boils down to their different assumptions on what is asked, I think. Similarly, the question "if a tree falls in a forest and nobody's there, does it make a sound?" has two valid answers for two meanings of "sound", an objectivistic "pressure wave" vs. a (somewhat?)solipsistic "consciousness' hearing".
jamesrom 11y ago
Can someone explain the half position clearly? I don't understand how anyone could think half is the correct answer. Let me frame the question a different way. You are one of three volunteers in separate rooms. I flip a coin and if it's heads I ask one volunteer (at random) to guess the outcome. If I flip tails I ask two of the volunteers (again, at random) to guess the outcome. You know the rules I will follow, but you cannot tell if anyone else has been asked before you. I open the door and ask you to guess the outcome of the flip. What do you guess?
- deleted 11y ago
  [deleted]
  - tantalor 11y ago
    > the question asked about the odds of the coin toss No, the question asked about the outcome of the coin toss, which is much less likely to be Heads (1 wake) than Tails (2 wakes).
  - Bartweiss 11y ago
    I think the issue here is that the two of you are answering subtly different questions - which statement of the problem encourages to arrive at a 'paradox'. In particular, we need to agree that Sleeping Beauty is questioned both times on a result of tails to make progress. The coin is fair, so the chance of getting heads is obviously 1/2. That's the answer to "What are the odds that a given coin flip came up heads?", but it's not what Sleeping Beauty is being asked when she wakes up. What SB is being asked is "Given that you have just been awoken, what are the odds that you were awoken as a result of the coin landing heads?" There's an implied conditional present. The coin flip was fair, but the questioning is biased by result - if SB is awake and being asked, the odds are 2/3 that she's in a world where the coin came up tails. To demonstrate, resort to extreme cases, as you did. If SB isn't woken at all for heads, then the answer to the question "you're awake, how did the coin land?" is tails with 100% certainty. If she's woken up once for heads and 99 times for tails, then the answer to "you're awake, how did the coin land?" is tails with 99% certainty, and so on. This all hinges on the assertion that SB is questioned on every awakening. If she's only asked on the first awakening (or is asked about the coin flip as a general concept) then the answer is of course 1/2, as you assert.
Bartweiss 11y ago
I'm having trouble finding the paradox here. As with many probability puzzles, the problem seems to be a hidden conditional in the probabilities. There's no change in belief happening, only two different beliefs. First is "1/2 of the times it is flipped, the coin will be heads". Seconds is "1/3 of the times I am awakened, the coin will be heads". The coin flip is fair, but the decision to ask the question is biased. This is extremely simple to demonstrate by resorting to the absurd case. If we change the awakening ratio away from 1-2, the absurdity of saying "1/2" becomes increasingly clear. At 1-9, tails will be the correct guess 90% of the time the question is asked. At 0-1, heads isn't even a possible outcome on awakening.
- leohutson 11y ago
  For 1-9, tails being the correct guess 90% of the time still makes sense to me. If you run the experiment twice, typically the first time you get heads, you are woken once, and so heads is correct answer once, the second time you get tails, but you are woken 9 times, so tails is the correct answer 9 times. Even from the frame of reference of the experimenter, heads and tails are equally likely, but it if you guess heads incorrectly, you will be wrong 9 times, or 9 times as wrong ;). Basically it depends on what you consider to be a "trial" of the experiment, if a coin toss is one trial, the wakings are a red herring, the probability is the same as the coin toss. If one trial is one waking, 90% of wakings are going to be tails.
  - timv 11y ago
    The phrasing of the question is When you are awakened, to what degree should you believe that the outcome of the coin toss was Heads? It's not actually asking for a strategy about guessing the correct answer, or how many times you'll be right or wrong. It think whether you are a halfer or a thirder is going to depend on how you interpret the question. If you see it as asking about the frequency at which "Heads" will the right answer, then it's clearly 1/3. But that's not what the question actually asks (probably - all language is interpreted).
- tempestn 11y ago
  Exactly this. I came to make the same comment, but unsurprisingly someone has beat me to it. In particular, the answer to the "paradox" in a nutshell is this: > There's no change in belief happening, only two different beliefs. First is "1/2 of the times it is flipped, the coin will be heads". Seconds is "1/3 of the times I am awakened, the coin will be heads".
deleted 11y ago
[deleted]
deleted 11y ago
[deleted]
personjerry 11y ago
From my understanding, the answer is 1/2. Let us assert that the probability of the coin flip being heads is 1/2. Now, you have awoken. Regardless of if you've awoken to the Heads flip, or the first time to the Tails flip, or the second time to the Tails flip, the original flip's chance was still 1/2. The possible misunderstanding comes from the fact that you will awake TWICE to Tails, which is more than to Heads! But the thing is this is irrelevant to the question, because you don't know whether this is the first or the second time, and you only need to wonder whether the original flip (recall it is probability 1/2) was heads or not. Potentially you could think that you'd be "wrong" more often, but we are only looking at one specific instance of you waking up in isolation, for which you have no additional information. For example, consider waking up 1000 times if the flip is tails. Upon waking up, do you think the probability heads becomes 1/1001? I think regardless of waking up the first time or the 1000th time, you have no more information so it might as well have been the first time, and hence the probability is 1/2.
- swsieber 11y ago
  You are not being asked the probability of a coin flip. You are a being asked the probability of a coin flip given that it woke you up. Simple Bayesian statistics. You don't know which of the three scenarios your in (Heads: 1st wake up, Tails: 1st Wake up or Tails: 2nd wake up). Given that there are three possible scenarios you are in, when finding myself in a scenario, I would give each scenario equal weight. There's my devil's advocate view. I can see what you're saying though. Edit: Additionally, even though the awakenings are not in the same 'stream of consciousness' if one tails waking happen, they both happen. They are linked even if you're trying to view them in isolation.
- URSpider94 11y ago
  Let me try to explain why this is the wrong answer. Let's assume there are 100 people undergoing this experiment. Half of them will flip heads, half tails. The half that flip heads will be woken up a total of 50 times. The half that flip tails will be woken up a total of 100 times. So, we have a total of 150 wake-ups. 100 of those came from tails, and 50 from heads. So, if you're woken up and have no prior knowledge, you have a 100/150 = 2/3 chance that you flipped tails, and a 50/150 = 1/3 chance that you flipped heads. Put another way, when you are woken up, it could be one of three cases: -- Awakened on the only time for heads -- Awakened on the first time for tails -- Awakened on the second time for tails Two of those cases correspond to tails, one to heads. So, it's twice as likely that you are being awakened due to a tails flip.
staz 11y ago
I'm a layman in probability but isn't it the same thing as the Inspection paradox? http://allendowney.blogspot.be/2015/08/the-inspection-paradox-is-everywhere.html http://allendowney.blogspot.be/2015/08/the-inspection-parado...
ThugNasty 11y ago
I think I'm a 3/8-er. When SB is woken up, the probability it's monday is: P(Monday) = P(Monday|Heads)P(Heads) + P(Monday|Tails)P(Tails) = 1(.5) + .5(.5) = 3/4 So then: P(Heads) = P(Heads|Monday)P(Monday) + P(Heads|Tuesday)P(Tuesday) = .5(.75) + 0(.25) = 3/8
deleted 11y ago
[deleted]
jsnell 11y ago
I remember reading rec.puzzles occasionally in the day when the Sleeping Beauty question was first posed. Coming back after not reading the group was for a couple of weeks was incredibly confused because it was full of minor variants of this single question constructed by people to support their position. I'd not only missed the original question and was thus lacking the context completely, but also had missed the megathread that followed and didn't understand the political undercurrents which would have been obvious if I'd only known who was halfer and who a thirder. It was just crazy. Here's a great account of how it unfolded: http://www.maproom.co.uk/sb.html http://www.maproom.co.uk/sb.html
nicholas73 11y ago
Even if you get awakened 99 times with Tails, you have equal probability going down the Heads (1 awakening) or Tails (1/99 awakenings) paths. So if you awaken with no knowledge of other awakenings, you are equally likely to be on either path, with the coin being Heads or Tails. You are NOT equally likely to guess that your awakening was due to Heads or Tails however, and that's the paradox.
- deleted 11y ago
  [deleted]
- ikeboy 11y ago
  This is basically just asserting SSA. One problem with the halfer claim is that if you learn it is the first awakening, you then place much more probability on it landing heads (basic Bayes theorem application), even when the coin is yet to be flipped. How would you resolve that?
colanderman 11y ago
"[…] to what degree should you believe that the outcome of the coin toss was Heads?" is a terribly unrigorous way to phrase this question. I suspect it's where all the confusion sets in. Are we adjusting the natural prior of a coin toss based on information we now have? We have no information; 50%. Are you placing a bet each time you are awake? You get to bet twice if the coin comes up Tails; 2:1 (33%) odds against Heads balances the tables. Are we judging how often we'd be correct if we guessed Heads every time we were awoken? Depends on the number of trials. With only 1 trial; 50%. With 2 trials; 42%. With infinite trials; 33%. Nothing begets a paradox like an ill-posed question.
- ikeboy 11y ago
  >We have no information We do, we know that we are now in the experiment. In the case where you don't wake up on heads at all, have you learned new information upon waking? If yes, why don't you learn something anytime the relative probabilities differ?
  - colanderman 11y ago
    That's not new information. You know that you will be in the experiment before the coin toss. Information does not flow acausally. To contrast with Monty Hall, after the MC opens the first door, you do have new information: you glean information from the MC's choice, which he made based on his knowledge of where the prize is. In this problem, the researchers take no action visible to you after the coin flip, and your memory is wiped before each new observation you can make. So your knowledge about the flip's outcome is exactly the same after it occurs as before: 50% chance.
    - ikeboy 11y ago
      If that doesn't count as new information, does waking up if you were to only wake on tails count as new information? If yes, what's the justification for distinguishing the two? What happens if you are told it's your first awakening as soon as you get up? Bayes theorem would imply you can't think the chances of heads stay the same before and after hearing this, so at least one must not be 50%.
millstone 11y ago
Here is a surefire way to win the lottery. Start by picking numbers. Your chances of winning naturally are very small, so I will make an arrangement with you. You go to sleep before the winning numbers are read. Afterwards, if you have won, I will wake you up N times, administering our trusty forget potion. But if you have not won, I will wake you up only once. As we have established in this thread, by increasing N, your chances of having guessed the lotto numbers correctly upon awakening approach 100%. I wake you up, you apply Bayes' Theorem, and then rejoice, for you almost certainly have won the lottery!
stephengillie 11y ago
How do the researchers get you back to sleep for the second (tails) awakening? You're woken once or twice and given the potion. Or you're woken and given the potion each time. This riddle wants it both ways, and that is part of the problem with it.
- ikeboy 11y ago
  You're only given the potion right before going to sleep, after you need to state your probability estimates.
blahblah3 11y ago
Not a paradox...here is a proof that it is 1/3 Let P(T) = probability that coin landed on tails, P(H) = probability that coin landed on heads. Let "1st" denote the event that it is the first time you are awaken, "2nd" the second time. Note that P(T|1st) = 1/2 (if you were told that this is the first time you were woken up, it's equally likely that the coin landed heads or tails). And of course, P(T|2nd) = 1 By the definition of conditional probability, 1/2 = P(T|1st) = P(1st|T)P(T)/P(1st) = (1/2)P(T)/P(1st) Hence P(T) = P(1st) = P(T)1/2 + P(H) = P(T)1/2+(1-P(T)) = 1 - P(T)/2 -> P(T) = 2/3
mcphage 11y ago
I don't see the problem in their problem: > Whenever SB awakens, she has learned absolutely nothing she did not know Sunday night. What rational argument can she give, then, for stating that her belief in heads is now one-third and not one-half? The argument that she can give is clear: she might have learned something, but the memory was taken from her. And she knew that they would take the memory from her. So yes, she hasn't learned anything new, but that's a cop-out—the problem explicitly prevents it. I mean, from that perspective, she doesn't even need to go to sleep! They could ask her, "when you wake up, how likely will you think that the coin was heads?" and get the same (correct) response of 1/3. That result is not based on her "learning new information" (since the situation forbids it), it's based purely on the situation as described.
- IkmoIkmo 11y ago
  > They could ask her, "when you wake up, how likely will you think that the coin was heads?" and get the same (correct) response of 1/3. Exactly. Let's stretch the example to the extreme, if you get heads you get woken once, if you get tails you get woken 1 million times. Now say you run this experiment 10 times and you happen to get 5 heads and 5 tails. You'd be woken up 5 million and 5 times. And with each wakening, the chance of the coin having been heads isn't 50%, if you'd say and guessed it was heads every time you wouldn't get half of them right, you'd be statistically wrong about 5 million times out of the 5m and 5 times. The correct answer is 1 in a million, and if you'd guessed tails on every awakening instead you'd be wrong on average only once in a million. It's this chance to be correct or incorrect in your answer when waking that would inform the question that was posed 'When you are awakened, to what degree should you believe that the outcome of the coin toss was Heads', which then would be 1/3.
  - daxfohl 11y ago
    First of all, how often you're correct doesn't matter; say the tails option wasn't "you'll be woken up a million times", but instead "you'll be asked the same stupid question a million times" (and have to answer as if you'd forgotten the previous answer). Of course you'll be wrong more often in the tails situation. Second, the experiment isn't run 10 times. It's run once. A single coin flip. With a multitude of flips, you've got no idea where you are in the sequence so it pushes your averages toward the 1:1000000. But a single coin flip is a single coin flip, 50/50.
drblast 11y ago
I'm not sure I understand this given the way the question is posed. We're doing a single experiment, and I am put to sleep without remembering either once or twice, and then awaken (potentially a third time) after that? Or could this keep going indefinitely until heads comes up? And am I guessing each time I'm woken, or only after the final time?
ucaetano 11y ago
You're flipping a fair coin again and again. Every time you flip heads, you add 1 black ball to a (initially empty) bag. Every time you flip tails, you add 2 red balls to the same bag bag. After a large number of flips, you pick a ball randomly from the bag. What are the odds that the ball was added when a heads was flipped? What are the odds that the ball is black?
- millstone 11y ago
  The question specified one coin flip. In your thought experiment, after one flip, the answer to both your questions is 50%.
daxfohl 11y ago
Both are right: the two camps posit completely different things. Halvers stipulate a single experiment. Thirders stipulate infinite experiments. It is pretty straightforward math to show these are not inconsistent, and there are even options in between. https://stats.stackexchange.com/questions/41208/the-sleeping-beauty-paradox/169582#169582 https://stats.stackexchange.com/questions/41208/the-sleeping...
fenomas 11y ago
It seems to me that the paradox/confusion here comes from asking someone to consider a distribution over a bound variable - i.e. you are asked to examine the odds of X happening in a situation where there are already side effects of whether or not X happened. In this sense, it reminds me of the puzzle where a man gives you a choice between two envelopes, one of which is specified to contain twice as much money as the other, with the paradox centering on a bystander's argument that you should then switch envelopes, since the other one must have either half or twice your value, giving you a 1.25x higher expected value for switching. As I understand it, in both cases the "traditional" solution to the problem is to recognize that probability doesn't work that way, and you can't consider distributions over bound variables, but the more interesting solution is to rephrase things in Bayesian terms, in which case the analysis is reasonably straightforward. I'm a dabbler though; experts please tell me if I'm spouting gibberish.
JohnLeTigre 11y ago
I'm not sure this is a paradox. There are 2 expected outcomes. All the added outcomes rely on the experimenter to disrespect the set rules. This is a conditional probability with a hard to predict condition (ie. human factor) portrayed as a non-conditional probability. This looks more like a bad representation of a problem rather than a paradox. This reminds me of my youth. Regardless of the project I was trying to accomplish, as soon as my little brother got involved, all bets where off. He was a hard to predict little bugger.
deleted 11y ago
[deleted]

77 points by xtacy 11y ago

A similar problem is God's Coin Toss as described in Scott Aaronson's excellent lecture series (and book) "Quantum Computing Since Democritus": http://www.scottaaronson.com/democritus/lec17.html http://www.scottaaronson.com/democritus/lec17.html

Gravityloss 11y ago

Maybe it's really an argument about this: Upon awakening: Halfer: P(monday,heads)=0,5 P(monday,tails)=0,25 P(tuesday,tails)=0,25 Thirder: P(monday,heads)=0,33 P(monday,tails)=0,33 P(tuesday,tails)=0,33

Gravityloss 11y ago
I think it boils down to the thirder position. P(monday,heads)=0,33 since P(heads)=0,5 and P(monday)=0,67. This is because we are sampling awakenings, not coin tosses or beauties. It's like betting on a coin toss, but with a side twist that the bet is evaluated twice (no new toss) when the coin is tails. Hence a majority of evaluations will be on tails: Guess, Coin, Profit H, H, +1 H, T, -2 T, H, -1 T, T, +2 Analysis for betting heads: average is +1 + -2 = -1 Betting tails: average is -1 + +2 = +1. I'd bet tails!

acchow 11y ago

This reminds me of the Monty Hall problem, but in reverse - you were playing the Monty Hall game and just won the car! You forgot whether or not you switched doors during the second step - what is the probability you switched doors?

deleted 11y ago
[deleted]
- Nadya 11y ago
  I also related it to Monty Hall in that one option is two options bundled together. The options are not [Heads||Tails] but rather [Heads||Tails+Tails]. The Tails+Tails is like having the Car+Goat bundled as a single option. Sleeping Beauty knows the coin is 50/50 but that if she is woken there is more opportunity to be woken during a Tails flip. To show that it is biased in favor of Tails - the experiment must be repeated. It shows more readily if you increase the number of times she is woken if the coin lands on Tails, similar to how the Monty Hall problem becomes more intuitive when you increase the number of doors. Instead of being woken 2 times - let's have her woken 1,000 times. Every time she is woken she is asked if the coin was Heads or Tails, with no recollection of her previous answers. She is given a dollar for every time she is correct. Should she guess Heads or Tails? If she guesses Heads every time - she can only win $1. If she guesses Tails every time - she can win $1,000. If she alternates answers she can only win $500.
Bartweiss 11y ago
There's actually a strong analogy to an extended Monty Hall here. Specifically: 1. Monty offers the challenge, and I pick door B. 2. Monty opens door C, showing a goat. 3. Monty asks if I want to switch. My odds of winning are now P(A) = 2/3, P(B) = 1/3. 4. My mom flips on the TV to see me play, but has missed my answers and only sees the state of the doors. What are her odds of guessing right? Result: my mom's odds of winning are 1/2, even though the odds of a given door winning are not. There's a bias between the two doors, but her perspective is neutral - she's guessing which door has better odds, and that guess is unbiased. As you say, this question is the reverse: the odds on the coin flip are unbiased, but my odds of being asked the question are biased towards one of the two outcomes.
- CamperBob2 11y ago
  See my other post; I actually think that a Monty Hall comparison argues in favor of an unbiased 1:2 guess, because of how the terms of the two problems differ. In the SB question, your odds of being asked the question are 1:1. You just don't know how many times you'll be asked. You didn't know the night before (and neither did the researchers), and you still don't know, even though the researchers now do. In the MH game, you give the host some new information when you make your initial choice. He already knew what door not to open, but now he knows what door he must open, and the rules require him to communicate that to you. That's when you receive the new information (as you point out with the example of your mom walking into the room). In the SB problem, you don't get any new information before the question is asked, including whether or not you're going to be put back to sleep. So the only answer you can rationally give is 1:2. If the researchers used a d20 instead of a coin to determine how many times to wake you up, you could safely guess that any given awakening wasn't your first or your last. But you still can't give any answer about the number on the die, other than a random guess from 1-20. You need to store some information for later recall, and they're not letting you do that.
- darkmighty 11y ago
  I quite like your phrasing because it highlights that probability is a model of uncertainty for an observer. So the mother has no information on the door, so she is 'neutral' (p=1/2 of guessing right); you have a small amount of information (p=2/3); you can even include Monty which of course has total information (p=1). You can't ask what are the probabilities of finding goats behind each door without specifying an observer and the available information.
sidcypher 11y ago
The Monty Hall problem is actually much better for educating people, as it highlights the bias hidden behind the "50/50" oversimplification. This Sleeping Beauty problem, on the other hand, is phrased to confuse you, so that arguing between the "halfers" and "thirders" boils down to their different assumptions on what is asked, I think. Similarly, the question "if a tree falls in a forest and nobody's there, does it make a sound?" has two valid answers for two meanings of "sound", an objectivistic "pressure wave" vs. a (somewhat?)solipsistic "consciousness' hearing".

jamesrom 11y ago

Can someone explain the half position clearly? I don't understand how anyone could think half is the correct answer. Let me frame the question a different way. You are one of three volunteers in separate rooms. I flip a coin and if it's heads I ask one volunteer (at random) to guess the outcome. If I flip tails I ask two of the volunteers (again, at random) to guess the outcome. You know the rules I will follow, but you cannot tell if anyone else has been asked before you. I open the door and ask you to guess the outcome of the flip. What do you guess?

deleted 11y ago
[deleted]
- tantalor 11y ago
  > the question asked about the odds of the coin toss No, the question asked about the outcome of the coin toss, which is much less likely to be Heads (1 wake) than Tails (2 wakes).
- Bartweiss 11y ago
  I think the issue here is that the two of you are answering subtly different questions - which statement of the problem encourages to arrive at a 'paradox'. In particular, we need to agree that Sleeping Beauty is questioned both times on a result of tails to make progress. The coin is fair, so the chance of getting heads is obviously 1/2. That's the answer to "What are the odds that a given coin flip came up heads?", but it's not what Sleeping Beauty is being asked when she wakes up. What SB is being asked is "Given that you have just been awoken, what are the odds that you were awoken as a result of the coin landing heads?" There's an implied conditional present. The coin flip was fair, but the questioning is biased by result - if SB is awake and being asked, the odds are 2/3 that she's in a world where the coin came up tails. To demonstrate, resort to extreme cases, as you did. If SB isn't woken at all for heads, then the answer to the question "you're awake, how did the coin land?" is tails with 100% certainty. If she's woken up once for heads and 99 times for tails, then the answer to "you're awake, how did the coin land?" is tails with 99% certainty, and so on. This all hinges on the assertion that SB is questioned on every awakening. If she's only asked on the first awakening (or is asked about the coin flip as a general concept) then the answer is of course 1/2, as you assert.

Bartweiss 11y ago

I'm having trouble finding the paradox here. As with many probability puzzles, the problem seems to be a hidden conditional in the probabilities. There's no change in belief happening, only two different beliefs. First is "1/2 of the times it is flipped, the coin will be heads". Seconds is "1/3 of the times I am awakened, the coin will be heads". The coin flip is fair, but the decision to ask the question is biased. This is extremely simple to demonstrate by resorting to the absurd case. If we change the awakening ratio away from 1-2, the absurdity of saying "1/2" becomes increasingly clear. At 1-9, tails will be the correct guess 90% of the time the question is asked. At 0-1, heads isn't even a possible outcome on awakening.

leohutson 11y ago
For 1-9, tails being the correct guess 90% of the time still makes sense to me. If you run the experiment twice, typically the first time you get heads, you are woken once, and so heads is correct answer once, the second time you get tails, but you are woken 9 times, so tails is the correct answer 9 times. Even from the frame of reference of the experimenter, heads and tails are equally likely, but it if you guess heads incorrectly, you will be wrong 9 times, or 9 times as wrong ;). Basically it depends on what you consider to be a "trial" of the experiment, if a coin toss is one trial, the wakings are a red herring, the probability is the same as the coin toss. If one trial is one waking, 90% of wakings are going to be tails.
- timv 11y ago
  The phrasing of the question is When you are awakened, to what degree should you believe that the outcome of the coin toss was Heads? It's not actually asking for a strategy about guessing the correct answer, or how many times you'll be right or wrong. It think whether you are a halfer or a thirder is going to depend on how you interpret the question. If you see it as asking about the frequency at which "Heads" will the right answer, then it's clearly 1/3. But that's not what the question actually asks (probably - all language is interpreted).
tempestn 11y ago
Exactly this. I came to make the same comment, but unsurprisingly someone has beat me to it. In particular, the answer to the "paradox" in a nutshell is this: > There's no change in belief happening, only two different beliefs. First is "1/2 of the times it is flipped, the coin will be heads". Seconds is "1/3 of the times I am awakened, the coin will be heads".

deleted 11y ago

[deleted]

deleted 11y ago

[deleted]

personjerry 11y ago

From my understanding, the answer is 1/2. Let us assert that the probability of the coin flip being heads is 1/2. Now, you have awoken. Regardless of if you've awoken to the Heads flip, or the first time to the Tails flip, or the second time to the Tails flip, the original flip's chance was still 1/2. The possible misunderstanding comes from the fact that you will awake TWICE to Tails, which is more than to Heads! But the thing is this is irrelevant to the question, because you don't know whether this is the first or the second time, and you only need to wonder whether the original flip (recall it is probability 1/2) was heads or not. Potentially you could think that you'd be "wrong" more often, but we are only looking at one specific instance of you waking up in isolation, for which you have no additional information. For example, consider waking up 1000 times if the flip is tails. Upon waking up, do you think the probability heads becomes 1/1001? I think regardless of waking up the first time or the 1000th time, you have no more information so it might as well have been the first time, and hence the probability is 1/2.

swsieber 11y ago
You are not being asked the probability of a coin flip. You are a being asked the probability of a coin flip given that it woke you up. Simple Bayesian statistics. You don't know which of the three scenarios your in (Heads: 1st wake up, Tails: 1st Wake up or Tails: 2nd wake up). Given that there are three possible scenarios you are in, when finding myself in a scenario, I would give each scenario equal weight. There's my devil's advocate view. I can see what you're saying though. Edit: Additionally, even though the awakenings are not in the same 'stream of consciousness' if one tails waking happen, they both happen. They are linked even if you're trying to view them in isolation.
URSpider94 11y ago
Let me try to explain why this is the wrong answer. Let's assume there are 100 people undergoing this experiment. Half of them will flip heads, half tails. The half that flip heads will be woken up a total of 50 times. The half that flip tails will be woken up a total of 100 times. So, we have a total of 150 wake-ups. 100 of those came from tails, and 50 from heads. So, if you're woken up and have no prior knowledge, you have a 100/150 = 2/3 chance that you flipped tails, and a 50/150 = 1/3 chance that you flipped heads. Put another way, when you are woken up, it could be one of three cases: -- Awakened on the only time for heads -- Awakened on the first time for tails -- Awakened on the second time for tails Two of those cases correspond to tails, one to heads. So, it's twice as likely that you are being awakened due to a tails flip.

staz 11y ago

I'm a layman in probability but isn't it the same thing as the Inspection paradox? http://allendowney.blogspot.be/2015/08/the-inspection-paradox-is-everywhere.html http://allendowney.blogspot.be/2015/08/the-inspection-parado...

ThugNasty 11y ago

I think I'm a 3/8-er. When SB is woken up, the probability it's monday is: P(Monday) = P(Monday|Heads)P(Heads) + P(Monday|Tails)P(Tails) = 1(.5) + .5(.5) = 3/4 So then: P(Heads) = P(Heads|Monday)P(Monday) + P(Heads|Tuesday)P(Tuesday) = .5(.75) + 0(.25) = 3/8

deleted 11y ago

[deleted]

jsnell 11y ago

I remember reading rec.puzzles occasionally in the day when the Sleeping Beauty question was first posed. Coming back after not reading the group was for a couple of weeks was incredibly confused because it was full of minor variants of this single question constructed by people to support their position. I'd not only missed the original question and was thus lacking the context completely, but also had missed the megathread that followed and didn't understand the political undercurrents which would have been obvious if I'd only known who was halfer and who a thirder. It was just crazy. Here's a great account of how it unfolded: http://www.maproom.co.uk/sb.html http://www.maproom.co.uk/sb.html

nicholas73 11y ago

Even if you get awakened 99 times with Tails, you have equal probability going down the Heads (1 awakening) or Tails (1/99 awakenings) paths. So if you awaken with no knowledge of other awakenings, you are equally likely to be on either path, with the coin being Heads or Tails. You are NOT equally likely to guess that your awakening was due to Heads or Tails however, and that's the paradox.

deleted 11y ago
[deleted]
ikeboy 11y ago
This is basically just asserting SSA. One problem with the halfer claim is that if you learn it is the first awakening, you then place much more probability on it landing heads (basic Bayes theorem application), even when the coin is yet to be flipped. How would you resolve that?

colanderman 11y ago

"[…] to what degree should you believe that the outcome of the coin toss was Heads?" is a terribly unrigorous way to phrase this question. I suspect it's where all the confusion sets in. Are we adjusting the natural prior of a coin toss based on information we now have? We have no information; 50%. Are you placing a bet each time you are awake? You get to bet twice if the coin comes up Tails; 2:1 (33%) odds against Heads balances the tables. Are we judging how often we'd be correct if we guessed Heads every time we were awoken? Depends on the number of trials. With only 1 trial; 50%. With 2 trials; 42%. With infinite trials; 33%. Nothing begets a paradox like an ill-posed question.

ikeboy 11y ago
>We have no information We do, we know that we are now in the experiment. In the case where you don't wake up on heads at all, have you learned new information upon waking? If yes, why don't you learn something anytime the relative probabilities differ?
- colanderman 11y ago
  That's not new information. You know that you will be in the experiment before the coin toss. Information does not flow acausally. To contrast with Monty Hall, after the MC opens the first door, you do have new information: you glean information from the MC's choice, which he made based on his knowledge of where the prize is. In this problem, the researchers take no action visible to you after the coin flip, and your memory is wiped before each new observation you can make. So your knowledge about the flip's outcome is exactly the same after it occurs as before: 50% chance.
  - ikeboy 11y ago
    If that doesn't count as new information, does waking up if you were to only wake on tails count as new information? If yes, what's the justification for distinguishing the two? What happens if you are told it's your first awakening as soon as you get up? Bayes theorem would imply you can't think the chances of heads stay the same before and after hearing this, so at least one must not be 50%.

millstone 11y ago

Here is a surefire way to win the lottery. Start by picking numbers. Your chances of winning naturally are very small, so I will make an arrangement with you. You go to sleep before the winning numbers are read. Afterwards, if you have won, I will wake you up N times, administering our trusty forget potion. But if you have not won, I will wake you up only once. As we have established in this thread, by increasing N, your chances of having guessed the lotto numbers correctly upon awakening approach 100%. I wake you up, you apply Bayes' Theorem, and then rejoice, for you almost certainly have won the lottery!

stephengillie 11y ago

How do the researchers get you back to sleep for the second (tails) awakening? You're woken once or twice and given the potion. Or you're woken and given the potion each time. This riddle wants it both ways, and that is part of the problem with it.

ikeboy 11y ago
You're only given the potion right before going to sleep, after you need to state your probability estimates.

blahblah3 11y ago

Not a paradox...here is a proof that it is 1/3 Let P(T) = probability that coin landed on tails, P(H) = probability that coin landed on heads. Let "1st" denote the event that it is the first time you are awaken, "2nd" the second time. Note that P(T|1st) = 1/2 (if you were told that this is the first time you were woken up, it's equally likely that the coin landed heads or tails). And of course, P(T|2nd) = 1 By the definition of conditional probability, 1/2 = P(T|1st) = P(1st|T)P(T)/P(1st) = (1/2)P(T)/P(1st) Hence P(T) = P(1st) = P(T)1/2 + P(H) = P(T)1/2+(1-P(T)) = 1 - P(T)/2 -> P(T) = 2/3

mcphage 11y ago

I don't see the problem in their problem: > Whenever SB awakens, she has learned absolutely nothing she did not know Sunday night. What rational argument can she give, then, for stating that her belief in heads is now one-third and not one-half? The argument that she can give is clear: she might have learned something, but the memory was taken from her. And she knew that they would take the memory from her. So yes, she hasn't learned anything new, but that's a cop-out—the problem explicitly prevents it. I mean, from that perspective, she doesn't even need to go to sleep! They could ask her, "when you wake up, how likely will you think that the coin was heads?" and get the same (correct) response of 1/3. That result is not based on her "learning new information" (since the situation forbids it), it's based purely on the situation as described.

IkmoIkmo 11y ago
> They could ask her, "when you wake up, how likely will you think that the coin was heads?" and get the same (correct) response of 1/3. Exactly. Let's stretch the example to the extreme, if you get heads you get woken once, if you get tails you get woken 1 million times. Now say you run this experiment 10 times and you happen to get 5 heads and 5 tails. You'd be woken up 5 million and 5 times. And with each wakening, the chance of the coin having been heads isn't 50%, if you'd say and guessed it was heads every time you wouldn't get half of them right, you'd be statistically wrong about 5 million times out of the 5m and 5 times. The correct answer is 1 in a million, and if you'd guessed tails on every awakening instead you'd be wrong on average only once in a million. It's this chance to be correct or incorrect in your answer when waking that would inform the question that was posed 'When you are awakened, to what degree should you believe that the outcome of the coin toss was Heads', which then would be 1/3.
- daxfohl 11y ago
  First of all, how often you're correct doesn't matter; say the tails option wasn't "you'll be woken up a million times", but instead "you'll be asked the same stupid question a million times" (and have to answer as if you'd forgotten the previous answer). Of course you'll be wrong more often in the tails situation. Second, the experiment isn't run 10 times. It's run once. A single coin flip. With a multitude of flips, you've got no idea where you are in the sequence so it pushes your averages toward the 1:1000000. But a single coin flip is a single coin flip, 50/50.

drblast 11y ago

I'm not sure I understand this given the way the question is posed. We're doing a single experiment, and I am put to sleep without remembering either once or twice, and then awaken (potentially a third time) after that? Or could this keep going indefinitely until heads comes up? And am I guessing each time I'm woken, or only after the final time?

ucaetano 11y ago

You're flipping a fair coin again and again. Every time you flip heads, you add 1 black ball to a (initially empty) bag. Every time you flip tails, you add 2 red balls to the same bag bag. After a large number of flips, you pick a ball randomly from the bag. What are the odds that the ball was added when a heads was flipped? What are the odds that the ball is black?

millstone 11y ago
The question specified one coin flip. In your thought experiment, after one flip, the answer to both your questions is 50%.

daxfohl 11y ago

Both are right: the two camps posit completely different things. Halvers stipulate a single experiment. Thirders stipulate infinite experiments. It is pretty straightforward math to show these are not inconsistent, and there are even options in between. https://stats.stackexchange.com/questions/41208/the-sleeping-beauty-paradox/169582#169582 https://stats.stackexchange.com/questions/41208/the-sleeping...

fenomas 11y ago

It seems to me that the paradox/confusion here comes from asking someone to consider a distribution over a bound variable - i.e. you are asked to examine the odds of X happening in a situation where there are already side effects of whether or not X happened. In this sense, it reminds me of the puzzle where a man gives you a choice between two envelopes, one of which is specified to contain twice as much money as the other, with the paradox centering on a bystander's argument that you should then switch envelopes, since the other one must have either half or twice your value, giving you a 1.25x higher expected value for switching. As I understand it, in both cases the "traditional" solution to the problem is to recognize that probability doesn't work that way, and you can't consider distributions over bound variables, but the more interesting solution is to rephrase things in Bayesian terms, in which case the analysis is reasonably straightforward. I'm a dabbler though; experts please tell me if I'm spouting gibberish.

JohnLeTigre 11y ago

I'm not sure this is a paradox. There are 2 expected outcomes. All the added outcomes rely on the experimenter to disrespect the set rules. This is a conditional probability with a hard to predict condition (ie. human factor) portrayed as a non-conditional probability. This looks more like a bad representation of a problem rather than a paradox. This reminds me of my youth. Regardless of the project I was trying to accomplish, as soon as my little brother got involved, all bets where off. He was a hard to predict little bugger.

deleted 11y ago

[deleted]