

Title: Re: Sleeping Beauty Post by Icarus on Apr 8^{th}, 2003, 4:58pm on 04/08/03 at 16:30:07, THUDandBLUNDER wrote:
I am a "If she has complete memory loss, how is she going to understand his question, much less make any sense of it and figure out an answer?"er. 

Title: Re: Sleeping Beauty Post by Icarus on Apr 8^{th}, 2003, 5:09pm on 04/08/03 at 16:30:07, THUDandBLUNDER wrote:
Did you perhaps mean he would wake her on both wednesday and thursday for tails? 

Title: Re: Sleeping Beauty Post by aero_guy on Apr 8^{th}, 2003, 6:46pm The way you currently have the question worded he will explain the situation to her, meaning she knows what day it is. If the day she awakens is Tuesday then she has to make a 50/50 guess. If the day is Wednesday she knows with 100% accuracy. Since she knows what day it is you can't really combine these results into an overall probability. If it is Tuesday she has a 50/50 shot. If it is Wednesday she knows for certain. If you interpret the wording that she has no idea what day it is, than I am a halfer. Again, you cannot combine the probabilities in a normal way. Look at the case where she is woken up 99 times with tails. The problem comes in determining the total possible outcome. Even with 99 tail wake ups there are still only two scenarios. 

Title: Re: Sleeping Beauty Post by aero_guy on Apr 8^{th}, 2003, 7:12pm Hmm, on second thought, I think there may be some difficulty in interpreting the question, "how much credence she gives to the proposition that the coin came up heads." If this were a game we were to play several times and you bet a dollar each time you awoke on tails, them 50% of the time you would lose a dollar and 50% of the time you would win two, for a net increase of $1, which seems to state there is a 100% chance of it being tales, odd. As I was getting at before, this differs from a standard probability riddle in that you are really dealing with a base of more than 100%. You have a 50% chance there will be a heads wake up, a 50% chance there will be a first tales wake up, and a 50% chance there will be a second tales wake up, for a total of 150%. So, IMO, the way the riddle is stated it is 50%, but if we play this game a bit and I am asked what is my bet to maximize profits, I go with tails. 

Title: Re: Sleeping Beauty Post by towr on Apr 8^{th}, 2003, 11:43pm the question isn't wether it was heads or tails, but wether it was heads or tails given that she was awoken. She is always awoken, but 1/3 of the time she is awoken when it was heads, and 2/3 she was awoken when it was tails. So Bayes makes me a thirder.. Also it is clear to see that on average she's awoken 3 times per two coin tosses. Once for head, twice for tails. 

Title: Re: Sleeping Beauty Post by Chronos on Apr 9^{th}, 2003, 7:57pm Quote:
I say that the answer (from Slumbering Lovely's perspective) is 1/3. 

Title: Re: Sleeping Beauty Post by rmsgrey on Jul 24^{th}, 2003, 3:23am In puzzles like this, I often find it helps to consider modified cases, so without further ado, what I call the Arabian version: As the original, except that Sheherazade takes part in the experiment. Rather than sleeping for 1001 nights (and being woken up each day if the coin came up heads) instead she simply has her memory of the previous day wiped each day, and on the first day is required to tell a story. On subsequent days she is only required to tell a story if the coin came up tails. I guess I should probably hide the rest of this in case anyone is still trying to decie which side they're on... ::[hide] It seems to me that the real problem lies in how you incorporate the fact she's required to tell a story (or woken up in the original)  which provides additional information. If she isn't required to tell a story (or wasn't woken up) she knows (or would if awake) the coin came up heads. The fact that hasn't happened means that there is additional information. I think the correct solution is to consider the situation as the product of two independent uniformly distributed random variables: the coin toss and the day. In the original, there are four possible outcomes (2002 in the Arabian) all of which are equally probable  Tu/H, Tu/T, We/H and We/T of which We/H can be ruled out by circumstance. A common mistake in probability questions is to overlook the cases that don't arise. An alternative question to try and help the unconvinced: if you knew about the experiment, and knew she was being woken up for an hour at noon on those days she was being awoken, but forgot whether it was being run so heads had her woken up on the Tuesday or the Wednesday, what credence would you give to the proposition that you would find her awake if you dropped in at half past midday?[/hide] :: 

Title: Re: Sleeping Beauty Post by turtler7 on Aug 20^{th}, 2003, 7:32am Reading through this I dont know that anyone said whay feelings I have on it. As in all riddles there is generally extra info to get you off track. Which when modified only to contain needed info it would say this "Sleeping Beauty agrees to participate in an experiment. On Monday she is put to sleep and the scientist conducting the experiment then tosses a (fair) coin. He then awakens her, explains the situation to her, and asks her how much credence she gives to the proposition that the coin came up heads." Which as always with probability to a coin flip it is 1/2. The extra times being awakened is only there to get you off track. The question is simple matter of heads or tails. Not its tuesday and it was heads, tuesday and tails, wendsday and tails. It was not asked that way. Only the probability of a coin flip is asked. 

Title: Re: Sleeping Beauty Post by towr on Aug 20^{th}, 2003, 7:58am No, not the probability of the coinflip being heads is asked, but the probability of the coin flip being heads given that sleeping beauty is woken up. There's a subtly yet important difference there. Suppose they put her under for a years, and flip a coin, if it's tails they wake her up januari first, if it's heads they wake her up every day of the year. So now sleeping beauty is awoken and asked what's the chance the coin flip was heads? on average she'll be woken up every day of the year one year, and just januari first the next. So chances are when sh'es woken up the coin was heads. 

Title: Re: Sleeping Beauty Post by turtler7 on Aug 20^{th}, 2003, 8:03am THUDandBLUNDER can likly clear it up... I see your point and thats why he listed both but I still go with just regular probability as the answer. He has seceded and making u think harder than is nessasary. I could be off i dont say im right nessasarilly but this is my veiw and y it is most likly in my mind set. 

Title: Re: Sleeping Beauty Post by rmsgrey on Aug 20^{th}, 2003, 9:38am A question for the 'halfers': if SB doesn't wake up, how much credence should she give to the proposition that the coin came up heads? OK, so that doesn't quite work, but suppose instead of letting her sleep on Wednesday if the coin came up Heads, the researcher awakens her and tells her that the coin came up Heads. After all the key point of the puzzle is that SB knows that she isn't in the position where it's Wednesday and the coin came up heads... Another variation: After reading this paragraph to two people, I put them in separate (isolated) rooms and then toss a coin. If the coin comes up Heads, I toss it a second time to randomly pick one of the two people to ask which way up they think the coin landed. If it comes up Tails, I ask both of them. What should they guess, and what is their chance of being right? 

Title: Re: Sleeping Beauty Post by turtler7 on Aug 20^{th}, 2003, 12:42pm Their best guess is tails. As if they are asked then its more likly to be tails. Although in this above senario there are not two people. Just diferent days. She doesnt know the day or if she has been asked before. She has to be asked 100% So the guess only has tails and heads are options making it 50/50. The senario doesnt fit the same as yours. If it was the same I would agree with the 33% heads 66% tails. But it is not that way. I see how it is confusing, but I dont see how it would not accually BE 50/50. 

Title: Re: Sleeping Beauty Post by towr on Aug 20^{th}, 2003, 1:10pm do it a million times.. on average she's woken up 1.5 times per week, twice on average per two weeks when it's tails, once per two weeks on average when it's heads. 

Title: Re: Sleeping Beauty Post by James Fingas on Aug 20^{th}, 2003, 1:20pm What we are asking Sleeping Beauty to calculate is the probability that a head was flipped. If she (or he) is smart, the she (or he) will calculate the probability given that she (or he) was awakened. Bayes will give us the straightforward answer here: p(heads given awakened) = p(heads and awakened)/p(awakened) Obviously there are two possible days on which Sleeping Beauty can be awakened, and two possible values for the coin. We'll assume that tails happen as often as heads, and that Tuesdays happen as often as Wednesdays: p(heads and Tuesday) = 1/4 p(heads and Wednesday) = 1/4 p(tails and Tuesday) = 1/4 p(tails and Wednesday) = 1/4 Therefore: p(heads and awakened) = p(heads and Tuesday) = 1/4 p(awakened) = p(heads and Tuesday) + p(tails and Tuesday) + p(tails and Wednesday) = 3/4 p(heads given awakened) = (1/4)/(3/4) p(heads given awakened) = 1/3 No doubt about it ... Bayes is a thirder! And with that recommendation, I'm a thirder too ;) 

Title: Re: Sleeping Beauty Post by James Fingas on Aug 20^{th}, 2003, 1:29pm Here's another formulation that doesn't require amnesia: Bill, an evil customers of your company's, will flip a coin. If he gets heads, he'll call you at home for product support on Saturday. If he gets tails, he'll call you on Sunday. Ned, another evil customer, will do likewise. Early Saturday morning, one of these two morons calls you. What are the odds they flipped the same thing? The probability is 1/3 here as well. 

Title: Re: Sleeping Beauty Post by aero_guy on Sep 23^{rd}, 2003, 6:21am James, Sorry to jump back at this so long after your last post, but I have a problem with the last thing you said. I don't think your evil customers are like the OP as you are given the information that one of them called you, and you know what day it is. This tells you one of them flipped heads. They are no longer part of the problem. The only question now is did the other. Though I agree with the 1/3 for the original problem, I do not see how this relates. 

Title: Re: Sleeping Beauty Post by James Fingas on Sep 23^{rd}, 2003, 8:19am I had to change the puzzle slightly so it didn't require amnesia. In the reformulation, as soon as they call you on Saturday, you know that they couldn't both have flipped tails. There are three equally likely possibilities remaining, and only one of those three has them flipping the same thing. 

Title: Re: Sleeping Beauty Post by aero_guy on Sep 23^{rd}, 2003, 8:35am Ah, they key to that being the same is that you are asking the question when you hear the phone ring, before you pick it up (and therefore know who it was). Does this mean that the act of hearing who it was who called you reduces the chance of them both having flipped heads to 50/50? That would be quite odd. 

Title: Re: Sleeping Beauty Post by James Fingas on Sep 23^{rd}, 2003, 9:16am on 09/23/03 at 08:35:45, aero_guy wrote:
I don't think so. Even after the phone call, all you know is that at least one person flipped heads. The key to these types of questions (e.g. the one I asked, the one head/two head question, Monty Hall, age of death of rock stars, the original question in this thread, etc.) is the statistical problem of "self selection" (maybe that's not exactly what it's called, but I'm too lazy to Google). "Self selection" is where your measurement system implicitly prefers a certain result, either by overreporting (the key to Sleeping Beaty), or by not allowing certain results (the key to the one head/two head question, Monty Hall, and most others). This dramatically affects the probabilities. Basically, you are fooled into thinking that what was a uniform probability distribution is still uniform, when in fact the measurement system has already changed the results. The way to analyze a problem like this is to look at the probability as a probability given the stated measurement system. Using Bayes' theorem, you just add up the probabilities of the various outcomes, keeping in mind how much each outcome would tend to get measured. [edit] I just realized that I never said how selfselection is relevant to this problem. Examining whether a specific one of the two calls you on Saturday is only useful if you pick that person before they call. As soon as he call you, he has picked himself. [/edit] 

Title: Re: Sleeping Beauty Post by rmsgrey on Sep 23^{rd}, 2003, 9:17am on 09/23/03 at 08:35:45, aero_guy wrote:
No odder than in the "not both girls" problem, where seemingly trivial differences in phrasing the puzzle change the probability of both children being boys from 1/3 to 1/2... In fact, the two problems are very closely related  though Sleeping Beauty's phrasing seems rather more robust. 

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