tag:blogger.com,1999:blog-6894866515532737257.post2991331003880341971..comments2024-04-22T21:33:32.590-07:00Comments on Probably Overthinking It: The Sleeping Beauty ProblemAllen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comBlogger83125tag:blogger.com,1999:blog-6894866515532737257.post-73832185952477179532017-01-02T09:20:23.283-08:002017-01-02T09:20:23.283-08:00Have you seen my four-volunteer version? Each will...Have you seen my four-volunteer version? Each will be wakened at least once, and maybe twice, based on the same coin flip. Each will be left asleep in only one set of conditions, different for each, as defined by the cross product {Monday,Tuesday}x{Heads,Tails}. Each will be asked for her credence that the coin landed on the side that would let her sleep through one day.<br /><br />One of these volunteers is undergoing the identical problem as in the original problem. Three are undergoing a functionally equivalent one, that must have the same answer.<br /><br />On any day, exactly three volunteers will be wakened. On any day, exactly one of the three awake volunteers is in the set of conditions she is asked about. On any day, each of the three awake volunteers has the same information upon which to base her credence.<br /><br />Yet that credence is found by a "bunch of probability calculations." It is 1/3.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-14019635407645045782016-12-20T14:23:45.564-08:002016-12-20T14:23:45.564-08:00Christopher> Am I missing something here?
No-o...Christopher> Am I missing something here?<br /><br />No-one's disputing the maths:<br /> * [A] One half of experiments in which Beauty wakes are those in which the coin is Heads.<br /> * [B] One third of Beauty's experimental wake-ups are those in which the coin is Heads.<br /><br />Beauty knows the facts, but she is obliged to pick just one of them as the basis of her "credence for Heads". Halfers insist she pick [A], Thirders insist on [B]. She's not allowed to state the reference class (w.r.t. experiments/wakeups) in her answer: her "credence" must be unqualified and absolute.<br /><br />So there's an interesting philosophical question lurking at the heart of this: when Beauty can see both perspectives, [A] and [B], what then do we mean, fundamentally, by her "credence"? If there is a correct answer to the Sleeping Beauty problem then it's a philosophical one, not bunch of probability calculations.Creosotehttps://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-72168284068907806242016-12-17T21:28:06.322-08:002016-12-17T21:28:06.322-08:00Alright, let's assume Beauty believes the thir...Alright, let's assume Beauty believes the thirder perspective and decides that she will answer "I think it landed on Tails, the odds are 2 in 3" every time she is questioned because she thinks she'll be correct more often. She gets asked twice if it comes up Tails after all, and all these mathematicians can't be wrong.<br /><br />As soon as she enters sleep/stasis and the coin is flipped there emerge two timelines, T1 where the fair coin came up Heads and T2 where the fair coin came up Tails. Since it's a fair coin T1 happens 50% of the time and T2 happens 50% of the time. This means that in T1 Beauty says Tails and is incorrect, whereas in T2 Beauty says Tails and is correct.<br /><br />Doesn't this mean that she's correct in 50% of timelines and incorrect in 50% of timelines? There are only two possible timelines and they happen 50% of the time each since it's a fair coin. In fact she could be asked a thousand times in T2 and since her memories are wiped and reset she'll always answer Tails. Furthermore, doesn't this mean that if you repeat the experiment 1,000 times, you get 500 Heads and 500 Tails, and she answers Tails every time... that she's correct in 500 experiments and incorrect in 500 experiments?<br /><br />The fact that she answers incorrectly once in T1 and correctly twice in T2 doesn't make her answer twice as correct. You're only tracking whether she answered correctly, not how many times she gets it right. I mean, you're not giving her 100$ every time she gets the answer right. If you did then she'd always say Tails because she has a 50% of getting 200$ and a 50% of getting 0$, whereas if she said Heads she'd have a 50% chance of getting 100$ and a 50% chance of getting 0$. What you're looking at is her certainty that the coin came up Heads or Tails. And if my previous logic is correct, that means that guessing Tails makes her correct in 50% of timelines... so she'd break even instead of being ultimately correct approximately 666-667 times out of 1,000 experiments.<br /><br />I can't believe that the process of being interviewed would lead her to believe that the odds of it being Heads or Tails had changed. This isn't a Monty Haul problem, she's not getting any new information. She knows that she's going to be woken up at least once and has no capacity to distinguish between Heads(Monday), Tails(Monday), and Tails(Tuesday)... or the 997th Tails day for that matter.<br /><br />Am I missing something here?Christopherhttps://www.blogger.com/profile/16259364826167235453noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-90890487169420147272016-11-02T16:10:40.004-07:002016-11-02T16:10:40.004-07:00TheRingshifter> Does the "amnesia" el...TheRingshifter> Does the "amnesia" element really change it that much?<br /><br />The amnesia is necessary in order to argue about Beauty's credence within the experiment itself, i.e. to prevent Tuesday Beauty saying "oh, hang on, you woke me yesterday ... this must be a Tails time-line".<br /><br />But should Beauty feel any differently about the coin flip within the experiment than she did prior to it? That's a key sticking-point of the Sleeping Beauty problem.Creosotehttps://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-74342049259542223102016-11-01T09:04:43.904-07:002016-11-01T09:04:43.904-07:00Am I wrong saying this...
Probabilistically, the ...Am I wrong saying this...<br /><br />Probabilistically, the question is quite simple. Every time the coin is flipped, it IS 1/2 odds (it's a coin). But every time she is WOKEN UP it's 2/3 tails and 1/3 heads. Is it really that different to just saying something like, "I'm going to slap you twice on a tails, and once on a heads. How likely is it, at each point I've slapped you, that you got a heads or a tails?". Does the "amnesia" element really change it that much? It's the same here... if you are betting on the coin flip, it's still 1/2, but if you are being slapped, then it's more likely it's because you got a tails since 2/3rds of the slaps will end up being attributable to tails and only 1/3rd of them to heads. <br /><br />So to me... it seems like although the "wager" problem is interesting, you are indeed "overthinking" it. Surely the thing that makes the difference is that she is possibly being asked twice about what the coin landed on when she wakes up, but when the bet is resolved, she is only being asked once. <br /><br />Simply, if you do this, say, 6 times, and get 3 heads and 3 tails, and she bets on heads, she will earn money by being correct about the wager, but will also be MOSTLY correct about saying tails is more likely - because she has been asked about the coin NINE times (3 on heads, 6 on tails) and been correct SIX times.TheRingshifterhttps://www.blogger.com/profile/18363611127790007978noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-27465393678760252022016-09-02T10:50:31.434-07:002016-09-02T10:50:31.434-07:00Thanks for asking, and for letting me know about t...Thanks for asking, and for letting me know about this paper. I posted my response in a followup article: http://allendowney.blogspot.com/2016/09/sleeping-beauty-and-red-dice.htmlAllen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-73123764827715619832016-09-01T15:47:10.699-07:002016-09-01T15:47:10.699-07:00The late great philosopher David Lewis was a halfe...The late great philosopher David Lewis was a halfer. I'd be interested in any reactions to his paper on it. http://fitelson.org/probability/lewis_sb.pdfElendilhttps://www.blogger.com/profile/03952102910141707647noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-30909504693229373082015-11-12T08:26:15.076-08:002015-11-12T08:26:15.076-08:00There are three events of interest to a "thir...There are three events of interest to a "thirder" SB when she is awake: H1=Today is Monday and Heads flipped, T1=Today is Monday and Tails flipped, and T2=Today is Tuesday and Tails flipped.<br /><br />White called his event H, but he really was treating it as H1. His expression for P_(H|W) is more accurately called P_(H1|W) even though they represent the same thing. That's why the term (c/2) is used: the prior probability of being awake on Monday after Heads is c, and the prior probability of Heads is 1/2. He multiplied them to get his (c/2).<br /><br />Note that the denominator of his equation [P_(H) P_(W|H) + P_(~H) P_(W|~H)] has nothing to do with the fact that he is calculating P(H), it is merely an expression of the Law of Total Probability using the partition {H,~H}. So the same denominator can he used in a similar calculation for P_(T1|W) and P_(T2|W). Since the numerators for these calculations are P(H1)*P_(H1|W) = P_(T1)*P_(W|T1) = P_(T2)*P_(W|T2) = (c/2), you get the same result for each calculation. Like I said.<br /><br />I know you disagree with this approach. But it is what a thirder would use. What White accused a thirder of using is not.<br /><br />+++++<br />What you, and White, ignore is that H2=Today is Tuesday and Heads flipped *is* *also* *an* *event* *in* *the* *experiment*. SB can't observe it, but it is still an event. On Sunday night, it has the same status, and prior probability, as H1, T1, and T2.<br /><br />Her "new information" when she is awake is that H2 is not her current state.<br /><br />The partition of SB's posterior event space contains four independent events, not two. Since she samples twice, with amnesia in between, the sum pf the prior probabilities for these four events is 200%, not 100%. The laws of probability work just fine on these events.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-84587916463279975952015-11-11T11:04:59.703-08:002015-11-11T11:04:59.703-08:00No. His result is for P(H | SB didn't sleep th...No. His result is for P(H | SB didn't sleep through the experiment).<br /><br />Monday and Tuesday have nothing to do with it. SB doesn't know whether it is Monday or Tuesday, because it could be either day. All she knows is that she didn't sleep through the entire experiment.<br /><br />The credence for heads actually becomes P(H)=1/3 (the Thirder's answer), but only in the limit that it is extremely unlikely for her to have ever wake at all -- i.e., c->0.Brian Mayshttps://www.blogger.com/profile/13962229896535398120noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-48221413887639888972015-11-11T09:16:19.329-08:002015-11-11T09:16:19.329-08:00There are two "definitions of the problem.&qu...There are two "definitions of the problem." One is what the experiment director sees (and SB sees on Sunday), and a different one is what SB sees when she is awake. That one exactly matches the situation that H2 finds herself in, where the answer is 1/3. Which should she base her confidence on? I've shown you several ways why it must be 1/3, and all you do is repeat the other solution.<br /><br />So again, I am less interested in seeing your solution that gets 1/2, then knowing why you think she can treat Monday and Tuesday as the same day. Which is what White did.<br /><br />Digging out my notes from years ago (and I'm not re-deriving it), I see that White stated that P(H) = (c/2)/[(c/2+(1-(1-c)^2)/2] = 1/(3-c). But his event "H" really is "H&Mon." You can use his exact same logic to get P(T&Mon)=P(T&Tue)=1/(3-c). These should add up to 1, and they don't.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-80838081445465780952015-11-11T08:03:10.196-08:002015-11-11T08:03:10.196-08:00"The question Brian refuses to address is, wh..."The question Brian refuses to address is, which is more appropriate to the original question?"<br /><br />Nonsense! I have said over and over that the one that is more "appropriate" is the one that corresponds to the definition of the problem. X2 depends on the outcome of X1. That is explicitly stated without any ambiguities.<br /><br />By the way, you still haven't addressed White's "generalized Sleeping Beauty Problem." Google it if you want to find the original article published in <em>Analysis</em> in 2006. Or you can click on the link that Creosote provided above. It probably has a link that will take you there.Brian Mayshttps://www.blogger.com/profile/13962229896535398120noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-79190030919130049072015-11-11T07:34:38.971-08:002015-11-11T07:34:38.971-08:00Okey-dokey, I'm done. I look forward to other...Okey-dokey, I'm done. I look forward to other Thirders attempting the "beans experiment" in future, and will be interested to see what conclusion (if any) is eventually reached at www.sleepingbeautyproblem.org .Creosotehttps://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-89515331165509920462015-11-11T07:14:43.702-08:002015-11-11T07:14:43.702-08:00@JeffJo @Creosote @BrianMays My general policy is ...@JeffJo @Creosote @BrianMays My general policy is that I'll publish a comment as long as it is on point and not abusive. But I encourage you to keep the tone civil. And if it looks like you can't reach consensus, you might have to agree to disagree.Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-88338670347070456842015-11-11T06:40:33.401-08:002015-11-11T06:40:33.401-08:00In the original problem, say a six-sided die is al...In the original problem, say a six-sided die is also rolled on Sunday Night after SB is asleep. If it comes up even, an alarm is programed to go off at noon on Monday. If odd, it will go off at noon on Tuesday, but she may not be awake to hear it. SB is instructed to enter her confidence in Heads into a computer at precisely 12:30PM.<br /> <br />On Sunday Night, she recognizes that the coin and the die represent independent events, and the alarm is equally likely to go off under any of the four possible situations {H,Mon}, {T,Mon}, {H,Tue}, and {T,Tue}. Of course, she will not be awake to hear it under {H,Tue}. Since she is not assured of hearing it, if she does it constitutes “new information” and allows her to update her confidence in Heads to 1/3. But not hearing it is also “new information,” and also allows her to update her confidence in Heads to 1/3.<br /><br />What if SB is, unknown to the experiment director, unable to hear certain frequencies and so can’t hear the alarm? She can still update her confidence to 1/3 because the alarm was in only one state of going off, or not going off, and both allow the same update. Essentially, this is now a form of Bertrand’s Box Paradox. She can update her confidence to 1/3 under every possible state of the alarm, so her confidence must be 1/3 without considering the alarm. But please, please, please note that this answer only applies when the status of one day is all that is known.<br /><br />The point I didn’t want to get into when Brian mentioned independence, is that independence changes as a result of the amnesia drug. On Sunday Night, being awake under T1 and T2 represent the same future, so they are dependent events. Brian’s solution treats them as dependent and gets 1/2. As such, it is addressing the problem from Sunday’s (or Wednesday’s) viewpoint only; that is, a state where all possible awake days are considered. But when SB is awake, she has no information about the possible “other” day, T1 and T2 represent different presents, and they are independent contrary to how Brian treats them. The Thirder solution treats them as independent to get 1/3. It address the problem from the viewpoint of either Monday or Tuesday, but not both as Brian does.<br /><br />Both the Alarm-Clock version, and my four-volunteer version, are designed to isolate state where T1 and T2 are independent events. Yes, it is a different problem than the one Brian solves when he treats them as dependent. The question Brian refuses to address is, which is more appropriate to the original question?JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-42550410954070269352015-11-11T05:06:57.462-08:002015-11-11T05:06:57.462-08:00"What part of 'that's the last time I..."What part of 'that's the last time I try to explain it to you' do you not understand?!!" The fact that your explanation does not address my issue. It dodges it. The issue isn't whether my solution is to a different problem than yours, we agree it is. The issues is which corresponds to the SB problem.<br /><br />H2 is identical to SB. H2's answer to the same question is 1/3. <br /><br />I'll avoid ad hominem attacks.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-59017046971523036452015-11-11T04:27:31.962-08:002015-11-11T04:27:31.962-08:00"Beauty must make her decision based on her c..."Beauty must make her decision based on her credence for the jars being filled the Heads way or the Tails way,."<br /><br />No, she really shouldn't. If the coin landed Tails, she will pick two beans, not one. So there is a difference between her confidence that *this* bean is safe, and that the set of beans she will consume on Wednesday will be safe.<br /><br />But Beauty is not asked for what her confidence was on Sunday, or will be on Wednesday (which is what you are solving for). She is asked for her confidence *NOW*. This is exactly the question my variation - the one you ignore because its answer is 1/3 - solves for.<br /><br />AGAIN: In my variation, H2 finds herself awake and in the exact situation SB is in. The other two awake volunteers are in symmetric situations. Exactly one of their labels matches the coin. The confidence any one of them should have, that it is herself, must be 1/3.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-73920826145928284022015-11-11T00:16:24.967-08:002015-11-11T00:16:24.967-08:00For what appears to be an excellent and far-reachi...For what appears to be an excellent and far-reaching survey of the available literature, readers might enjoy perusing http://www.sleepingbeautyproblem.org/tiki-index.php?page=Project%20Portal . Seems like it's a work in progress, so should get even better over time.Creosotehttps://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-31991851326097473892015-11-10T18:01:41.526-08:002015-11-10T18:01:41.526-08:00"So I repeat: Brian, what do you think is dif..."So I repeat: Brian, what do you think is different about the volunteer labeled H2, and the original Sleeping Beauty?"<br /><br />What part of "that's the last time I try to explain it to you" do you not understand?!!<br /><br />I've already explained it to you, again and again. Since you cannot accept my explanation, you counter with:<br /><br />1) "'In your different problem, X1 and X2 are completely independent random quantities.' No, they are not."<br /><br />Oh really? No reason given.<br /><br />2) "When you consider the joint distribution for X1 and X2 with any specific value of X3, the distribution is either identical to the one you describe for SBP (for X3=H2), or symmetric with it."<br /><br />At this point, I seriously doubt you even know what a joint distribution is, but ... (sigh) ... giving you the benefit of the doubt, if the joint distributions were identical then we wouldn't be having this discussion -- would we? We would agree on the result.<br /><br />3) "I can tell you what is wrong with White's analysis as well, but I don't want to expand the issue further."<br /><br />In other words, White's analysis is way over your head.<br /><br />4) "Using 1/3 properly -- and I am too rushed at the moment to do it -- will only tell you the probability that *this* jelly bean will be poison ..."<br /><br />At this point, you have become a parody of yourself. Creosote is correct: you are by definition a Halfer.<br /><br />Please leave us alone now. You are the last person who should be demanding anything from anybody.Brian Mayshttps://www.blogger.com/profile/13962229896535398120noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-12498737163707548652015-11-10T14:56:25.012-08:002015-11-10T14:56:25.012-08:00Beauty must make her decision based on her credenc...Beauty must make her decision based on her credence for the jars being filled the Heads way or the Tails way, i.e. her credence for the coin flip being Heads or Tails. For Thirders, those credences are 1/3 and 2/3 respectively.<br /><br />If you insist that Beauty makes her choice based on credence 1/2 for the Heads way of filling the jars then you are by definition a Halfer.Creosotehttps://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-79160774915399533062015-11-10T12:15:52.135-08:002015-11-10T12:15:52.135-08:00Maybe this will help clarify what I'm trying t...Maybe this will help clarify what I'm trying to say:<br /><br />1) Brain Mays solves the SB problem with a solution with an answer of 1/2. Call this solution B, for Brian.<br />2) I presented a variation of the SB problem that I say, and I think Brian agreed with, has the answer 1/3. Call its solution J.<br />3) Brian insists that solutions B and J address different problems. I agree.<br />4) What the "four volunteers" example is supposed to show, is that the original Sleeping Beauty finds herself in exactly the same set of circumstances as one of the four volunteers, and symmetrically equivalent circumstances to the other three. Brian has not contested this, he has only said it needs a different solution than solution B.<br />5) Those circumstances are addressed by solution J.<br /><br />So I repeat: Brian, what do you think is different about the volunteer labeled H2, and the original Sleeping Beauty? Because she says the probability that the coin matches her label - i.e., landed Heads - is 1/3. A agree that this changes the problem from the one you solved - I'm saying that you solved the wrong problem.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-44605702331801893302015-11-09T09:02:05.152-08:002015-11-09T09:02:05.152-08:00Q Creosote: I must admit I didn’t look at your for...Q Creosote: I must admit I didn’t look at your formula, and I misread your intent. I took your stated proportions to be of poison jelly beans, to all jelly beans. That changed the problem in more ways than one.<br /><br />But the simple answer is that of course thirders should use the prior probability P(Coin=H)=1/2 to determine the makeup of the jar, not the posterior probability (PCoin=H|Awake)=1/3. Using 1/3 properly – and I am too rushed at the moment to do it – will only tell you the probability that *this* jelly bean will be poison, not the probability that you will get one over the course of the experiment. That is the only thing “credence” can apply to.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-70204380881417044772015-11-09T08:22:00.937-08:002015-11-09T08:22:00.937-08:00"In your different problem, X1 and X2 are com..."In your different problem, X1 and X2 are completely independent random quantities." No, they are not.<br /><br />In my problem, there is a third random variable, X3, for the volunteer being considered. And the three are inter-dependent, along the lines you describe. Using that distribution, the answer to my question is clearly 1/3, and I think you have agreed.<br /><br />When you consider the joint distribution for X1 and X2 with any specific value of X3, the distribution is either identical to the one you describe for SBP (for X3=H2), or symmetric with it. But the answer of 1/3 can't change.<br /><br />I'm sorry you seem incapable of even attempting to understand this, but it is a common technique in probability.<br /><br />I can tell you what is wrong with White's analysis as well, but I don't want to expand the issue further.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-46094759236655676422015-11-08T11:00:29.893-08:002015-11-08T11:00:29.893-08:00Sigh ... One more time and that's it. In these...Sigh ... One more time and that's it. In these problems, there are two random outcomes: X1 - the result of the coin toss, and X2 - the day of the week. In the Sleeping Beauty Problem (SBP), the distribution of X2 completely depends on the outcome of X1. This is an essential part of the problem. If X1=Heads, X2 can have only one value (X2=Monday). If X1=Tails, X2 can have two. Thus, X2 cannot be determined until X1 is known.<br /><br />In your different problem, X1 and X2 are completely independent random quantities. Neither can affect the other. That, in itself, makes the two problems different, but you try to pull off a sleight of hand after the fact by introducing "new information" -- i.e., that the combination X2=Tuesday and X1=Heads did not occur, so that there are only three possibilities left corresponding to the possible outcomes in the genuine SBP. Then you adjust the different problem to accommodate this new information and arrive at your answer. This sleight of hand has fooled several people, but in the REAL SBP, that combination of X1 and X2 was NEVER a possibility because that is how the SBP is defined. X1 and X2 were never independent, and there is no "new information" to be discovered that was not already known on Sunday night.<br /><br />This is also why the Thirders get a different answer to White's modified problem. Intuitively, if it is no longer guaranteed that SB will wake each day, then the coin flip takes on extra significance: if the coin is tails, she's less likely to sleep through the entire experiment, because she has two chances, instead of one, to wake up. Thus, the fact that she didn't sleep through the experiment should mean something, and that something should depend on the probability c of waking up each day, but because the Thirders are analyzing a different problem, they end up concluding that it does not. The value of c has no effect on their answer, and it doesn't matter which approach they are using to arrive at this answer.<br /><br />I'm sorry that you are incapable of understanding this. Anyhow, that's the last time I try to explain it to you.Brian Mayshttps://www.blogger.com/profile/13962229896535398120noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-53426899107864580572015-11-08T10:40:35.168-08:002015-11-08T10:40:35.168-08:00In this branch of the comments section, I'll c...In this branch of the comments section, I'll consider only the "beans experiment" as first proposed.<br /><br />I can't quite tell, JeffJo, what calculation you are proposing Beauty uses. By "a prior probability is required", I think you're saying that Beauty proceeds as follows:<br /><br /> (1) survival probability is p+(1-p)/49 for "A" and 36(1-p)/49 for "B"<br /><br /> (2) with prior probability p=1/2 for Heads, those survival probabilities are 25/49 for "A" and 18/49 for "B"<br /><br /> (3) so pick "A"<br /><br />That calculation makes no use of Thirder Beauty's credence, 1/3, for Heads. So, in the Thirder world, "credence" appears to be inconsequential for decision making.Creosotehttps://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-18912551910399681802015-11-08T08:21:49.654-08:002015-11-08T08:21:49.654-08:00@Brian Mays: You keep asserting my version is diff...@Brian Mays: You keep asserting my version is different, yet you refuse to provide a reasoned explanation of what is different. Please tell me which of the following four statement you disagree with, since you must disagree with at least one, providing clear explanations:<br /><br />1) In the version I presented to Creosote, the experiment that volunteer H2 finds herself in identical to the one the original Sleeping Beauty was in. So her answer must be the same as the original's.<br /><br />2) The experiments that H1 and T2 find themselves in differ from H2's only in the value of one random variable. Since the random variable has two equally-likely values, their answers must be the same as H2's. T1's has two such changes (or one, when compared to H1 or T2), and also must be the same.<br /><br />3) Of the three awake volunteers on either day, the answer to "Does the coin flip result correspond to your label?" is "yes" for exactly one of them.<br /><br />4) It does not matter if an awake volunteer sees the other two who are awake, or if they even exist in the experiment.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.com