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 MONTY HALL LOGIC - Not adequate for real life situations. Goto page 1, 2, 3, 4  Next
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Ghost Post
Icarian Member

 Posted: Fri Mar 10, 2000 2:56 pm    Post subject: 1 well the monty hall logic is based on Probability im mathematics and the logic is that the probability factor is good for large samples . Infact the "good prize" can be behind any of the two remaining doors. BUT I Think it doesnot improve your CHANCES of winning the prize.
Rollercoaster
Guest

 Posted: Fri Mar 10, 2000 8:52 pm    Post subject: 2 The fact is, your odds are DOUBLED by switching. I will take those odds in mathematics and real life.
HyToFry
Drama queen

 Posted: Fri Mar 10, 2000 9:14 pm    Post subject: 3 look at it from a view of more choices cracker, if i told you to pick a card out of a standard deck of 52 cards (no jokers), and I would bet you that wouldn't be able to get the ace of spades, in fact i would bet you 100 dollars. And after you chose the card, i said now don't look at your card yet, at which point i looked through all of the remaining cards, and kept one for myself, and sat the rest of the deck off to the side, and told you that while going through the cards i was looking for the ace of spades and if i found it, i would keep it out (the card in my hand) and if i didn't i would simply pick a card at random and hold it out, then offered to trade you cards, Would you take the trade? If you would, then you have proven that the monty hall puzzle is truth (switching will increase you chances) although in the case of the cards it takes your chances from 1/52 to 51/52 and in the case of monty hall it only takes the chances from 1/3 to 2/3 or 33% to 66% Comments?
friskythepig
Guest

 Posted: Sun Mar 12, 2000 2:43 pm    Post subject: 4 Hiya, I found a variation on the Monty Hall problem at: http://sunsite.org.uk/usenet/news-faqs/news.answers/puzzles/archive/decision and I've included it below. It is essentially the same problem but this time there are three prisoners A,B and C on death row. One of them is to be executed the next day and the other two set free. Prisoner A goes to see the warden and is informed that B is to be set free. As in the Monty Hall problem, A now has a 1/3 chance of being executed and C has a 2/3 chance of being executed. What I was wondering was, if C also goes to see the warden and is told that B is to be set free, C will have a 1/3 chance of being executed and A has a 2/3 chance. So what are the probabilities of A and C being executed? ==> decision/prisoners.p <== Three prisoners on death row are told that one of them has been chosen at random for execution the next day, but the other two are to be freed. One privately begs the warden to at least tell him the name of one other prisoner who will be freed. The warden relents: 'Susie will go free.' Horrified, the first prisoner says that because he is now one of only two remaining prisoners at risk, his chances of execution have risen from one-third to one-half! Should the warden have kept his mouth shut? ==> decision/prisoners.s <== Each prisoner had an equal chance of being the one chosen to be executed. So we have three cases: Prisoner executed: A B C Probability of this case: 1/3 1/3 1/3 Now, if A is to be executed, the warden will randomly choose either B or C, and tell A that name. When B or C is the one to be executed, there is only one prisoner other than A who will not be executed, and the warden will always give that name. So now we have: Prisoner executed: A A B C Name given to A: B C C B Probability: 1/6 1/6 1/3 1/3 We can calculate all this without knowing the warden's answer. When he tells us B will not be executed, we eliminate the middle two choices above. Now, among the two remaining cases, C is twice as likely as A to be the one executed. Thus, the probability that A will be executed is still 1/3, and C's chances are 2/3.
Ghost Post
Icarian Member

 Posted: Sun Mar 12, 2000 8:51 pm    Post subject: 5 From A's perspective: A requests "Of B and C, I know at least one will go free. Tell me the name of one of the two who will go free (whether there is only one, or both will go free)." A is told B will go free. A then should conclude, from the information available to A, that he has a 1/3 chance of execution, and C has a 2/3 chance of execution. From C's perspective: C requests "Of A and B, I know at least one will go free. Tell me the name of one of the two who will go free (whether there is only one, or both will go free)." C is told B will go free. C then should conclude, from the information available to C, that he has a 1/3 chance of execution, and A has a 2/3 chance of execution. Furthermore, it is possible that: B requests "Of A and C, I know at least one will go free. Tell me the name of one of the two who will go free (whether there is only one, or both will go free)." B is told A will go free. B then should conclude, from the information available to B, that he has a 1/3 chance of execution, and C has a 2/3 chance of execution. Of course, having overheard all three conversations, we know with certainty who will be executed: C. If we had only overheard the first two conversations, we would conclude A and C each have an equal 1/2 chance of being executed.
friskythepig
Guest

 Posted: Mon Mar 13, 2000 11:39 am    Post subject: 6 Thanks for the explanation. That's what I thought the answer ought to be, but intuitively it didn't seem right.
Ghost Post
Icarian Member

 Posted: Thu Mar 23, 2000 7:11 am    Post subject: 7 You guys are out to lunch. I understand the Monty Hall solution but your prisoner example doesn't apply. Look at this table: CASE A B C 1 Die Live Live 2 Live Die Live 3 Live Live Die So, like you said, A finds out B is going to live. So, we can eliminate CASE 2 from our chart. So, please explain to me how A now has a better chance of living than C. You said up above that A has a 1/3 chance of dying and C has a 2/3 chance of dying. Looks to me like A and C have an equal chance of dying.
Ghost Post
Icarian Member

 Posted: Thu Mar 23, 2000 1:51 pm    Post subject: 8 Three equally likely cases, initially: 1) 1/3 chance A will die (B and C live) 2) 1/3 chance B will die (A and C live) 3) 1/3 chance C will die (A and B live) A asks which of B and C will live, knowing at least one will live. There are now 4 unequally likely cases (before the warden answers the question): 1a) A will die, and warden says "B" (1/6) 1b) A will die, and warden says "C" (1/6) 2) B will die, and warden says "C" (1/3) 3) C will die, and warden says "B" (1/3) If you doubt that this much is correct, consider: First, the above cases are the only possible cases, and there are four. Second, merely asking the question does not change anyones probability of being executed. When the warden answers "B", we can eliminate the first and last cases, leaving cases 1b and 2, with B twice as likely to die as A.
Ghost Post
Icarian Member

 Posted: Thu Mar 23, 2000 8:05 pm    Post subject: 9 It is important to note the major difference between this problem and the Monty Hall problem. Let's rephrase this in the context of Monty Hall. Prisoners A, B, and C are in front of you. One has been chosen to be executed and the others will go free. You guess that A will be executed. Monty reveals to you that B will go free. If you now change your guess to C, you are twice as likely TO HAVE GUESSED CORRECTLY. But, the probabilty that a specific prisoner was chosen for execution is still 1/3. A and C are equally likely to be executed. You can just improve your chance of correctly guessing which will die.
Ghost Post
Icarian Member

 Posted: Thu Mar 23, 2000 8:50 pm    Post subject: 10 If you know B will go free, how can A and C each have a 1/3 chance of being executed? The probabilities (of all possible outcomes - in this case two: A dies or C dies) have to add up to 1.0 The important point is that correct probability estimates are based on available information, and after being told that, of B and C, B will go free, we have different information than when we started. There is certainly no longer a 1/3 chance that B was chosen for execution. If there is a 2/3 chance that my guess that C was chosen to die is correct, then there is a 2/3 chance that C was chosen to die. There can't be a difference.
Ghost Post
Icarian Member

 Posted: Thu Mar 23, 2000 9:08 pm    Post subject: 11 The chance that C was chosen is determined at the time the choice is made. Once you know that B was not chosen, you have a 50/50 chance of being right if you guess that C was chosen. It is the difference between "the chance that C was chosen" and "the chance that you can correctly guess who was chosen." The extra information about B affects your ability to guess correctly, but doesn't make a difference in the chance that C had been chosen in the first place. If I flip a coin and you see that it came up heads, does that mean the coin had a 100% chance of coming up heads? No, but you can be 100% sure of you guess about the outcome of the past event. [This message has been edited by Drew (edited 03-23-2000).]
worm
Guest

 Posted: Thu Mar 23, 2000 9:15 pm    Post subject: 12 extro, you said quote: 1a) A will die, and warden says "B" (1/6) 1b) A will die, and warden says "C" (1/6) 2) B will die, and warden says "C" (1/3) 3) C will die, and warden says "B" (1/3) If you doubt that this much is correct, consider: First, the above cases are the only possible cases, and there are four. Second, merely asking the question does not change anyones probability of being executed. When the warden answers "B", we can eliminate the first and last cases, leaving cases 1b and 2, with B twice as likely to die as A. I agree w/ all of this but the last statement. If the warden answers "B" then B has zero chance of dying. Rather we can eliminate the middle 2 cases, which means C is twice as likely to die as A.
Ghost Post
Icarian Member

 Posted: Sat Mar 25, 2000 6:01 am    Post subject: 13 Thank you Drew. Like you have stated, this Monty Hall Logic problem has gone from a problem of logic to a problem of semantics. I said in my first post that people are "out to lunch" because they are applying the MH problem on everything under the Sun and the results of this are quite strange. I didn't know exactly WHY they were wrong, but like I said, the results, especially with the prisoner/death scenario, were silly. Thanks for correctly explaining the flaw to everyone.
Ghost Post
Icarian Member

Ghost Post
Icarian Member

 Posted: Mon Mar 27, 2000 8:17 pm    Post subject: 15 Sorry, I meant to say "a fair coin" and not simply "a coin." You say: "Before the toss, I don't know how it will land, and I estimate a 50% chance of heads. After it lands, it is either heads or tails, but if I don't know which, I estimate a 50% chance it was heads." This relates to my basic point. The probability that a particular outcome will occur and the probability that I can correctly predict the outcome are not the same thing. If I randomly pick a card out of an ordinary deck I have a 1/52 chance of picking the ace of spades. If I show you the card (but do not look at it myself) you can "guess" which card I actually picked with 100% accuracy. I would still only have a 1/52 chance of being correct. You have extra information which greatly increases your chance of being right. But the origianl chance of me picking the ace was still 1/52. If we repeated this a large number of times and recorded the results, we would see that a) I actually pick the ace of spades about 1 time out of 52. b) I correctly guess the card I picked 1 time out of 52. c) You correctly guess the card I picked 100% of the time. Of course I am assuming that you are trying to guess correctly. (You wouldn't guess 2 of clubs after seeing the 5 of diamonds.) Back to the prisoners: A doesn't get any extra info untill after the decision has been made. He still had a 1/3 chance of being picked. Now that he knows one of the people who was not chosen, he can guess who will die with 50% accuracy. If you repeat this many times, you'll see that: a) A actually gets executed 1/3 of the time b) A correctly guesses who gets executed 1/2 of the time. The important thing to see is that A actually dies 1/3 of the time. If his chances of being executed were 50/50 he would only die 1/2 of the time.
Quailman
His Postmajesty

 Posted: Mon Mar 27, 2000 8:36 pm    Post subject: 16 Drew: Are you saying that A has a 1/3 chance of dying, but if he guesses that he is the one who will die (now that he knows B won't), he will be correct 1/2 the time? This doesn't make sense. If he guesses before asking that either B or C will die, he will be correct 2/3 times. If he guesses that at least one of them won't die, he will be right 100%. Now that one of them is revealed as a survivor, he can still guess that one of them (B or C) will die and be right 2/3 of the time. There is NO 50-50.
Ghost Post
Icarian Member

 Posted: Mon Mar 27, 2000 8:59 pm    Post subject: 17 Drew: In an earlier post you wrote: "Once you know that B was not chosen, you have a 50/50 chance of being right if you guess that C was chosen." So let's repeat the experiment until we have 100 times when we know B was not chosen, and let's always guess that C was chosen. You are saying we will be right about 50 times out of 100 ("50/50 chance of being right"). So of those 100 times, about 50 times C dies, and about 50 times A dies. The same should then be true (with B and C reversed) for those times when we know C was not chosen. But in your later post you wrote: "The important thing to see is that A actually dies 1/3 of the time."
Ghost Post
Icarian Member

 Posted: Mon Mar 27, 2000 9:30 pm    Post subject: 18 Quailman - When I said that A would be correct 1/2 of the time I meant that if he knew which of the other two would go free he could guess with 50% accuracy. If he guesses before asking he would have a 1/3 chance of being right. By guessing I mean that he guesses a specific prisoner to die, not that he guesses "either B or C will die." His choice fof his guess are: 1) A will die. 2) B will die. 3) C will die. Once he knows that B will live, he must guess between 1 and 3, which are equally likely. Extro - Yes, A will die half of the times that B does not die. However, if you take into account all of the possibilities, including those where B dies, A will only die 1/3 of the time. In both cases, A guesses who will die with 50% accuracy. It is a case of conditional probability. P(X) = probabilty that event X happens P(X|Y) = probability that event X happens, given that event Y has happened. For example, if I roll a fair 6-sided die, P(1) = probability that I rolled a 1 = 1/6. P(1|odd) = probability that I rolled a 1 given that the number I rolled was odd = 1/3. Let A = A is chosen to die. B = B is chosen to die. C = C is chosen to die. P(A) = P(B) = P(C) = 1/3 P(A|not B) = P(A|not C) = 1/2 P(A|(not B)or(not C)) = 1/3 because P((not B) or (not C)) = 1 The chance that A dies = P(A) = 1/3. The chance that A dies if we know that B does not die = P(A|not B) = 1/2.
Ghost Post
Icarian Member

Wonko the Sane
Daedalian Member

Ghost Post
Icarian Member

 Posted: Tue Mar 28, 2000 1:40 am    Post subject: 21 I agree almost completely with the last post by Extro. Once A has eliminated one of the possible outcomes he has a 50% chance of being correct when he assumes that he will die. But, in the big picture A will still die 1/3 of the time. I think what we are really disputing is the following: If 50% of the time A can correctly guess if he will die does that means that he has a 50% chance of being executed? You say yes. I say no. Our math is the same. The reason he can guess with higher probabaility than his actual chance for execution is because the warden removes a choice for him. Without this help, A could only guess with a 1/3 chance of being correct. If you really believe that A has a 50% chance of being executed then you believe that half of the time the following sequence is carried out, A will be killed. 1. randomly select a prisoner to die. 2. tell prisoner A which of B or C will live (choose randomly if both will live) 3. Kill the prisoner selected in 1. If we skip step 2, it seems obvious that A dies 1/3 of the time, yes? But does step 2 have any bearing on who dies in step 3? No. Step 2 just changes what A knows about who will die so he can make a better guess. [This message has been edited by Drew (edited 03-27-2000).] And, Wonko, thanks for the support but I didn't understand most of what you said. [This message has been edited by Drew (edited 03-27-2000).]
Ghost Post
Icarian Member

Ghost Post
Icarian Member

 Posted: Tue Mar 28, 2000 3:09 am    Post subject: 23 Wonko wrote: "Guess what, you guys are making a very dumb mistake." No. The chance of any given prisoner living is 2/3, and for three prisoners it does not add up to 100%. There is no reason why it should, as the events are not mutually exclusive. The rule is (and I know, because I'm making it up), the probabilities for some number of possible outcomes must add up to 1 (100%) if it is known that one of those outcomes must occur, and only one of those outcomes will occur. For the probabilities of the prisoners being executed, we know one and only one will be executed. The combined (by addition) probabilities of each prisoner being executed must add up to 100%. You also wrote: "Feel free to flame, I've used some intentionally flawed logic in places." Why? The question is, what percentage is intentionally flawed, what percentage is unintentionally flawed, and do the percentages add up to 100?
Ghost Post
Icarian Member

 Posted: Tue Mar 28, 2000 3:50 am    Post subject: 24 Oh my. You are quite right. Way back in this discussion I tried to make the point that the chance of A being executed was different than the chance of A being able to guess if he will be executed. At one point I mentioned that A was just as likely to die as C was, meaning that they both had a 1/3 chance of be selected for execution. This post actually made sense and makes no mention of 50/50 probability. My next post was written quickly without a lot of thought and I turned "equal chance" into 50% probability, but got the context messed up. I was after all mainly arguing that the chance of each prisoner being executed was always 1/3 and that the extra info just gave you better odds of guessing which one would die. I never bothered to go back and check my previous "facts." I mean, I was right and everyone else had too be wrong. Why should I double check myself It turns out to be really easy to convince yourself of something once you are possitive it is true. Stupid stubborn me.
Ghost Post
Icarian Member

 Posted: Tue Mar 28, 2000 4:39 am    Post subject: 25 This is hilarious! Extro, you have created so many new formulations of the argument that you deserve a prize. These problems convey so much about human psychology!
Ghost Post
Icarian Member

 Posted: Tue Mar 28, 2000 4:43 am    Post subject: 26 Aw shucks, I thought this might go on a while longer. Seriously though, I still differ with you about your basic notion of probability and chances. I don't think events, future or past, have "real" probabilities associated with them. Probabilities are just our way of estimating likelihoods of outcomes. If I toss a coin or dice in the air, once it is in the air, the outcome is fixed. It has some trajectory, velocity, spin, etc, and the immutable laws of physics will take it from there. Or at least it might be that way. But the outcome, fixed as it may be, is unknown to us. We just know from experience that all outcomes (for fair coins and dice) seem to occur in equal numbers in the long run, and without a pattern (i.e. randomly). Now, before the coin or dice is in the air, is another story. There's the whole thing of human free will, which is beyond the scope of this discussion. But I don't think it matters anyway. Whether or not these kinds of outcomes are fixed or not (and they would generally be considered fixed for past events, though you'll even get arguments there - the "Schroedinger's Cat" thing, for instance), the point, I think, is what information we have about the outcomes (past, present or future). A probability is an estimate based on available information. If we gather more information, we may update our estimate of a probability. I don't think there is a "real" probability. I think the probability of X is just exactly the probability that we would be right if we guessed X. Have you seen the old "Sleeping Beauty" problem? THAT was a good one. And one which was never quite settled, in my opinion.
Ghost Post
Icarian Member

 Posted: Tue Mar 28, 2000 6:12 pm    Post subject: 27 I have seen that one. The solution says, there is a problem in the wording because it doesn't really define what is meant by the probability in this case. Are you trying to get the most questions right or trying to be right most of the time (tails comes up half the time, but yields two questions)? If you are trying to get the most right answers out of the total questions asked, it seems to me that you should just always say tails. When the coin is heads (50% of the time) you are wrong once. When it is tails (also 50%) you are right twice. Thus you give the right answer to 2/3 of the questions, but only on 1/2 of the flips. Ok, I waited 5 minutes and then reread that. It still makes sense to me, so I'll post it...
HyToFry
Drama queen

 Posted: Tue Mar 28, 2000 8:30 pm    Post subject: 28 Okay, Did ALOT of thinking on this one, and this is the solution i've come up with as far as these prisoners go. The problem with the prisoners, is that there is no Initiall choice, so there fates are set from the get go. Each has a 33% chance of dying, and a 66% chance of living. If one of them asks the wardon which of the other two will go free, then the one that wardon dosen't say suddenly has a 66% chance of dying.. THIS IS TRUE TRUST ME, if you don't believe me, write a script, and you'll see, the one who asks will die 33% of the time, and live 66% of the time, HOWEVER this dosen't change the original odds of being the one that lives or dies, its still 33/66. So givin this let me give you an example... Prisoner A asks, which of the other two will die? There are three 4 possible outcomes here. A will die, so A is Told B 1/6 A will die, so A is Told C 1/6 B will die, so A is Told C 1/3 C will die, so A is told B 1/3 So there is a 1/2 chance A will be told B, and if A is told B, 2/3 times C dies and 1/3 times A dies. (still 33%) There is a 1/2 chance A will be told C, and 2/3 times B will die, and 1/3 times A will die. (yup 33%) so there is a 3/6 chance that B will die 66% of the time, and C will die 66% of the time the other 3 out of 6 times (when A is the person asking), this gives C and B 33% chance of loseing, and A the Same So if A is told that B will go free, C has a 66% chance of being the unlucky prisoner (this is because if it wasn't C, there would have been a 2/3 chance that A would have been told "C" not "B".. ie. there is only a 1/3 chance of the loser being "A" when "A" is told "B" and a 2/3 chance of the loser being "C") However this dosen't change the odds of C liveing from the begining, because there was a 50/50 chance of A being told B or C, C still had a 33% chance of living, all the riddle is saying is that 66% of the time C will die when A is told that its not B. Look at the chart below if you don't follow PC = Person Chosen to DIE ?! = this is the prisoner that is told who will be free ?@ = this is the prisoner that the ?! prisoner is told will be free PC|A!|B!|C!| A |B@|C@|B@| A |C@|C@|B@| B |C@|A@|A@| B |C@|C@|A@| C |B@|A@|A@| C |B@|A@|B@| Sorry, the system font is not fixed width, so the table isn't lined up Now as you can see if A is told that B will go free, there are only Three possible scenerios, two of which C dies in, however if C is told that A will go free, then there are only 3 scenerios, 2 of wich A dies in, and if A and C both get told that B goes free, there are 2 possible scenerios (assuming that A and C both know that the other knows that B will be free), so if A and C know that B will go free, THEN, AND ONLY THEN are there chances 50/50, but that still doesn't change their original chances of 33/66, it's only to say that there are only two possible cases that A and C would both be told B, and in one A dies, and in the other C dies. If just A is told that B will go free however there are three possibles that can happen, It is A and C will be told B also, it is C and C will be told B, or it is C and C will be told A, (2 out of three times C DIES, when A is told B) I hope i've got this right To put it into simpler terms, saying that the probablility of A dieing is 1/3 and C 2/3 is INCORRECT.... You would HAVE to say "The Probablility of A dieing is 1/3 and C 2/3 WHEN AND ONLY WHEN, A is told that B will be set free, and NO OTHER INFORMATION IS TOLD TO ANYBODY... And yes QuailMan I'm SURE hehe J/K [This message has been edited by HyToFry (edited 03-28-2000).]
Quailman
His Postmajesty

 Posted: Tue Mar 28, 2000 9:10 pm    Post subject: 29 But are you sure?
worm
Guest

Wonko the Sane
Daedalian Member

Ghost Post
Icarian Member

 Posted: Wed Mar 29, 2000 2:04 am    Post subject: 32 Wonko, you wrote: Your probability breakdown: 50% Guard says B and C dies 25% Guard says B and A dies 25% 50% Guard says C and B dies 25% Guard says C and A dies 25% Where the heck do you get that from??? This is how it is: 50% Guard says B and C dies 33.33% Guard says B and A dies 16.66%% 50% Guard says C and B dies 33.33% Guard says C and A dies 16.66% Is that what you have? That's what I have. You also wrote "I'm sorry guys. You're wrong." Who is wrong? The question is, IF the warden says B will live (when asked the particular question that has been so thoroughly discussed), THEN what is the chance that C will die? 2/3? If you agree with that, then who are you arguing with??? Nowhere did anyone assume the warden always says B. It was discussed, to a painful extent, that there were two possible answers to the question. You wrote "Also...here's a modified version of the script that only counts deaths when the guard says B. The second part will give that C dies about twice as much." That's what most of us have been saying most of the time, and some of us all of the time. When the answer is B, C dies twice as often as A (C is twice as likely to die as A).
Wonko the Sane
Daedalian Member

 Posted: Wed Mar 29, 2000 2:44 am    Post subject: 33 Actually, I think I misread the actual question. I was interpreting it, "Does A's chance of survival increase by the guard saying B?". The answer to that is no. However, What are A's chances? has a definate answer. I agree with you on that point. Though A isn't actually increases his chances of survival by asking who of the other two will survive. In the grand scheme, he still only has a 1/3 chance of survival. I think the real point I'm making is the same as Drew's. You can't really seperate the big picture from the detail. Although asking the question later gives A a good idea of who will survive. Everyone still has a 1/3 chance of dying. The question isn't really about who has higher chance of being executed, but who has a higher chance of guessing correctly who will die. At this point, 1 person has a 100% chance of dying, and the other two have a 0% chance of dying in reality because the person who will die has already been chosen. That, I think, is the point i've been trying to make with all of this.
Ghost Post
Icarian Member

 Posted: Wed Mar 29, 2000 4:01 am    Post subject: 34 What is going on here? Seriously, what is going on here???? My buddy works in a jail. I told him to line up three prisoners. After lining them up side by side I said I'm going to randomly kill 1 of them. Needless to say, they were upset. So then I told the guy in the middle he can go free. He ran off. Using the LOGIC of you guys, I told the third guy that he has a greater chance of dying than the first guy. He asked me why, it doesn't make sense if I choose at random. EXACTLY! It makes no sense. Why don't we look at a few posts above where someone wrote a friggin computer program to come up with this solution. All prisoners have a 1/3 chance of dying. All prisoners have a 2/3 chance of living. Gee, an unexpected solution to a problem where I'm randomly killing 1 prisoner out of 3. To all these other people who are refuting this and coming up with all these percentages of the third guy dying like 50%, 66%, pi, and every other stupid number they can think of, I would just like to say that this whole discussion of the problem is the saddest thing I have ever heard. My nephew is in Grade 2 and he can solve this problem.
Ghost Post
Icarian Member

 Posted: Wed Mar 29, 2000 4:18 am    Post subject: 35 Here is the solution to the problem in a form that will HOPEFULLY make sense to everyone. At the beginning, all 3 (A,B,C) have an EQUAL chance of dying. Of course, this is 1/3, since 1/3 x 3 = 100% When B is removed (set free), A and C NOW have a 50% chance of dying. But, the removal of B has not improved the situation at all for A and C because they still both have an EQUAL chance of dying 1/2 x 2 = 100% For some strange reason, some of you people seem to think that A's knowledge of B being set free has given A an advantage over C. How is this an advantage for A? It makes NO DIFFERENCE. As I stated above THE REMOVAL OF 1 PERSON FROM OUR PROBLEM DOES NOT IMPROVE THE SITUATION FOR ANY REMAINING MEMBERS. The only thing I can think of is that you guys are comparing this to something like Russian Roulette where at the start I have a 1/6 chance of dying, and after 1 round, I have a 1/5 chance of dying....and of course this continues. But this is hardly the same situation as the prisoner problem. And as someone stated above in an earlier post, A might as well have kept his mouth shut because his knowledge does not increase his odds of living unless he is the one chosen. Why is this so hard to comprehend???
Ghost Post
Icarian Member

 Posted: Wed Mar 29, 2000 6:38 am    Post subject: 36 Ok. I've had it with this stupid problem. I went to a mathematics wizard with more letters behind his name than the alphabet. I posted a new thread with the solution. I hope this is the end of this problem.
Ghost Post
Icarian Member

 Posted: Wed Mar 29, 2000 11:48 am    Post subject: 37 Banana: I'll explain it for you. A initially has a 1/3 chance of being executed. (That's the state of what we know, initially) FACT: A will either learn that B will be set free, or he will learn that C will be set free. (that's because he asks "tell me one of the others who will go free") CLAIM: This will happen no matter what, and does not increase A's chance of execution. So, if learning B will be set free does not change A's chance of execution (1/3), then it must change C's chance of execution (the two chances must add up to 1). And similarly, with B and C reversed. Now, if you don't agree with the "claim" above, consider: If learning B will be set free increases A's chance of execution, then learning C will be set free would increase it also. But how can both of the only possible outcomes increase A's chance of execution? That would obviously be absurd, no? A asks a question, and no matter what answer he gets, his chance of execution increases? Don't you see a problem with that?
Wonko the Sane
Daedalian Member

Ghost Post
Icarian Member

Wonko the Sane
Daedalian Member

 Posted: Thu Mar 30, 2000 3:20 am    Post subject: 40 You've misinterpreted a lot of things extro. And I'll try to go through them all. But everything you criticized you didn't understand. I'm saying that in order for this puzzle to work right in real life. If the prisoner asks the warden which of the two will live, he says B. Well, he only says B 50% of the time. Namely, any time that B will die and half of the time that A dies. If B dies he doesn't say B. Rerun the trial. If A is schedualed to die and the guard "randomly" chooses who to say. Half the time he'll say C. Well, he can't say C and have the puzzle still work. So the trial has to get rerun. I'm talking about the grand scheme. You see, every time means that if you ran a large number of trials, you have to throw out the trials where the guard says C. Well you can't just throw out a trial in real life. You'd have to reset the problem and recreate all of the random varibles. Otherwise it's not actually random. Well, this isn't actually entirely random either unless you set up new trial data for trials where the random choice of the guard is C. I'm talking about results for a large number of trials. And in these, the guard has to say B...every single time. Hence why 1/3 of the time, you need new data. And 1/3 of that time you need new data. And 1/3 of that time you need new data. ( I was wrong there, A's probability is .50, but the reset prob is 1/3, not 1/2). So to get how many times that A dies that the trial data is wrong (and ends up as C, which is similar to being thrown out, but is more realistic). The model for this equation is .5 - the the sum of (1/3)^n for integral values of n between 2 and infinite. That basically gives you .5 - 1/6 or 2/3. That's A's new chances of survival. Now, extro, on to you. Just because you don't understand something gives you NO RIGHT to go on and flame it to hell. I'm arguing your case. Some people don't understand what you're saying because it's intuitively wrong, so I'm giving the hard math on why in a real life situation A will survive 2/3 of the time if you remove 2 of the cases (B dying and A dying and A being told C will live). If you dont' understand it then you can just keep your mouth shut because all you did was look like a dick. Oh yeah, one other thing. I said explicitely that it doesn't matter if B is the second prisoner or the third. A and B are variables. I think that's obvious. [This message has been edited by Wonko the Sane (edited 03-29-2000).]
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