Subject: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 28 Oct 12 - 12:34 PM Some of you will have heard of this. On the game show "Let's Make a Deal," host Monty Hall used to present three giant doors. Behind one was a prize like a car. Behind the other two were booby prizes like a goat. The contestant was asked to pick a door. At that point, of course, the contestant's chance of picking the door that hid the car was 1 in 3. (Or 2 to 1 against, if you prefer.) Whichever door was selected, Monty (who knew where the car was) would open one of the remaining two doors to reveal a goat. He'd ask then the contestant whether they wanted to stick with their original choice or switch to the single unchosen and unopened door, which must contain either a car or a goat. Here's the problem: mathematical geniuses say that switching is a good idea because it raises the odds of winning the car from 1 in 3 to 2 in 3. Others, including myself, cannot follow this reasoning. Wikipedia tries to explain (unsuccessfully in my view) at great length: http://en.wikipedia.org/wiki/Monty_Hall_problem Among the skeptics, by the way, are sophisticated engineers and logicians, who evidently are not sophisticated enough. The mathematicians claim, with equations, that switching changes the odds. Obviously (?) eliminating one door increases the odds that your first choice was correct from 1 in 3 to 1 in 2. But how then does switching your choice boost the odds of winning to 2 in 3? You still have no idea where the car is! I'm prepared to believe the math geniuses because they offer incomprehensible equations and I'm not very smart. They also say the answer is obvious and lament the gullibility of the non-mathematician public. But some of the dissenters also are smarter than I am. Can anybody explain CLEARLY and with a minimum of math how seeing the contents of one of the three doors and then switching your choice (from one unknown quantity to another) increases the odds that you'll win the car? Or do I have to sign up for an advanced course in probability theory? |

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Subject: RE: BS: Monty Hall ProblemFrom:
gnu
Date: 28 Oct 12 - 12:47 PM "Can anybody explain CLEARLY and with a minimum of math how seeing the contents of one of the three doors and then switching your choice (from one unknown quantity to another) increases the odds that you'll win the car?" No. Not even unclearly. It's 50/50. |

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Subject: RE: BS: Monty Hall ProblemFrom:
bobad
Date: 28 Oct 12 - 12:51 PM I too see it as 50/50 but then I'm no math genius. |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Ed
Date: 28 Oct 12 - 01:15 PM Obviously (?) eliminating one door increases the odds that your first choice was correct from 1 in 3 to 1 in 2.I think that this is where you may be getting confused. As you say, in your initial choice of door, the odds are 1 in 3. At this point, you know that at least one of the other doors has a goat behind it. When the host (who knows where the car is) opens one of the other doors to reveal a goat, he is merely confirming what you already know: one of the other doors has a goat behind it. Therefore, and this the counter intuitive part, your odds are still 1 in 3. Nothing else has changed.Hence switching increases your odds of winning to 1 in 2 at this point (or 2 in 3 overall). |

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Subject: RE: BS: Monty Hall ProblemFrom:
DMcG
Date: 28 Oct 12 - 01:22 PM The explanation is not too difficult, but it does depend on understanding how information affects probability. But it is very easy to carry out the experiment: you need, for example one red playing card, two black ones, and a friend to be Monty. Play the game ten times without changing your mind, then another ten where you always change your mind and compare the results. Sometimes doing the experiment is more convincing than any amount of argument! |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 28 Oct 12 - 02:12 PM >Therefore, and this the counter intuitive part, your odds are still 1 in 3. Nothing else has changed. >Hence switching increases your odds of winning to 1 in 2 at this point (or 2 in 3 overall). The "hence" is where you lose me, Ed. Revealing one goat would indeed boost your odds to 1 in 2. But how does *switching* from the original unknown to the remaining unknown raise them to 2 in 3 at the moment you switch? You still have no idea what's behind either remaining door. As you explain it, it almost sounds as though "1 in 3" at the start is somehow mystically equivalent to "2 in 3" later. But "one" never used to equal "two" in my book. I'll do DMcG's experiment later. But if it works, I still won't understand how. Suppose I *wanted* a goat. My chances at the start would be a pretty good 2 in 3. Then Monty opens a door, which I didn't select, with a goat behind it. My chances of winning a goat now have slipped to 1 in 2. Would they increase to 2 in 3 if I switched to the one remaining door? |

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Subject: RE: BS: Monty Hall ProblemFrom:
DMcG
Date: 28 Oct 12 - 02:20 PM I agree doing the experiment won't explain WHY the probabilities are not as you expect, but it should be enough (given enough trials - 10 might be too few) to show the probability is (something like) 2/3 if you change and 1/3 if you do. Once that is clear, you will see that the 'obvious' 50:50 solution can't be right. Then go slowly though your logic to see where you assume it is 50:50 and that's where you are going wrong ... |

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Subject: RE: BS: Monty Hall ProblemFrom:
gnu
Date: 28 Oct 12 - 05:16 PM If you have ONE choice, the odds are 50/50. Previous choices are moot. They are gone, they do not exist. They do not impact subsequent choices. Soooo, you have ONE choice. And the odds are 50/50 that your choice will win or lose. Period. This is not socks in a drawer. This is A OR B and it's defined as such. They are asked to make A choice between TWO doors. One OR... OR... OR... the other door... THAT is 50/50. |

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Subject: RE: BS: Monty Hall ProblemFrom:
DMcG
Date: 28 Oct 12 - 05:47 PM Have you done the experiment? :) Probability is tricky: very subtle changes in definition of the problem can have significant effects. In this case, 50:50 is demonstrably wrong if you spend fifteen minutes or so actually trying it. |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Ed
Date: 28 Oct 12 - 05:48 PM Previous choices are moot. They are gone, they do not exist. They do not impact subsequent choicesBut they do here, gnu. If, for example, your initial choice is Door 1, the host can open either Door 2 or 3 to reveal a goat. Let's assume he opens Door 2. You can then choose between Doors 1 and 3. If your initial choice is Door 2 and the host opens Door 1 you can then choose between Doors 2 and 3. In other words, your initial choice has influenced what your subsequent one can then be. It is no longer a simple 50/50 choice. |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Reverend Bayes
Date: 28 Oct 12 - 07:14 PM To start with, you had a 1/3 chance of picking the car. Opening the door doesn't change that; it's still 1/3 that you have the car. So the probability the other door is the car must be 2/3. If you think this is hard, my friend the Martingale Representation Theorem would like a word. |

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Subject: RE: BS: Monty Hall ProblemFrom:
John on the Sunset Coast
Date: 28 Oct 12 - 09:31 PM http://en.wikipedia.org/wiki/Monty_Hall_problem I know we don't all love and trust Wikipedia, but therein is presented several ways to the solution, from a basic grid to a formal mathematical problem, and they all agree that one should always switch on the second round. This strategy results in winning the car (mathematically) twice as often as standing pat. BTW, you can trust Darrin on the Sunset Coast (if not Wiki) who is a logician/linguist by training. He explained it, sans math, such that even I understood it. |

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Subject: RE: BS: Monty Hall ProblemFrom:
Bee-dubya-ell
Date: 28 Oct 12 - 10:54 PM As I recall it, there were two levels of "real" prizes and one "goat". Let's say the items behind the doors are a toilet brush, a television set, and a new car. There are three possible scenarios: A) If your choice is the door hiding the toilet brush, Monty will reveal the television set you "lost". (He's not going to reveal the car until the last moment, win or lose.) B) If your choice is the door hiding the car, Monty will also reveal the television set. (He's not going to reveal the toilet brush as long as there's the possibility of your changing your mind and getting stuck with it instead of winning the car.) C) If your choice is the door hiding the television set, Monty will reveal the toilet brush. (Again, he's not going to reveal the car until the last moment.) So , there are two scenarios (A & B) in which Monty reveals the television set. Changing your door choice doesn't increase your chances of winning the car in either of them. You're either going to win the car or get conked with the toilet brush. Second prize is out of the running. But, in scenario C, changing doors will DEFINITELY upgrade your prize from the television set to the car. Since you've been shown that you didn't pick the "goat" prize, you know you've picked a winner. But if the winner you'd picked was the car, you'd have been shown the television set, not the toilet brush. So, the probability of your winning the car in scenario A is 50%, whether you change your mind or not. And the probability of your winning the car in scenario B is 50%, whether you change your mind or not. But the probability of your winning the car in scenario C is 100%, if you change your mind. Total those probabilities up and average them out and the overall odds of winning the car if you change your mind are 2/3. |

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Subject: RE: BS: Monty Hall ProblemFrom:
beeliner
Date: 29 Oct 12 - 01:03 AM The best, and clearest, explanation of the problem that I have read is, of all places, in a novel entitled "The Curious Incident of the Dog in the Nighttime" by Mark Haddon. HOWEVER - Haddon's explanation, and all the others that I have read, seem to assume (tho' not state directly) that Monty would offer the choice no matter which door the contestant had initially selected. But what if Monty offered the choice ONLY if the contestant had initially chosen the door with the car? Would that change the percentages? |

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Subject: RE: BS: Monty Hall ProblemFrom:
Rob Naylor
Date: 29 Oct 12 - 04:20 AM Of ther above, Reverend Bayes' explanation is the clearest and most concise. Without any maths, either. As for many of the skeptics mentioned in the original post being "logicians and engineers", if they were *good* logicians and engineers they'd do the experiment suggested by DMcG and see for themselves that they do win more often if they switch....then work on understanding the logic of the statistics.I spend a chunk of my time doing statistical analysis of data to determine its precision and reliability forgeophysical and engineering purposes and it surprises me how often people who are otherwise great engineers have problems understanding statistics. |

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Subject: RE: BS: Monty Hall ProblemFrom:
Megan L
Date: 29 Oct 12 - 06:36 AM Of course the fact is this Monty Hall(whoever that was) did not let them replay the game multiple times so the laws of probability do not affect the outcome it was down to luck. |

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Subject: RE: BS: Monty Hall ProblemFrom:
Nigel Parsons
Date: 29 Oct 12 - 07:58 AM Are we all agreed? On your first pick you have a one in three chance of picking the car. We all agree, good! So on one time in three you would be swapping the car & getting a goat (if you swapped). On your first pick you have a two in three chance of picking a goat. So two out of three times swapping will get rid of the goat, and get you a car. Although swapping may lose the car, this happens only 1/3 of the time. 2/3 of the time swapping will get you the car. I know which way I'd choose! |

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Subject: RE: BS: Monty Hall ProblemFrom:
DMcG
Date: 29 Oct 12 - 08:26 AM Of course the fact is this Monty Hall(whoever that was) did not let them replay the game multiple times so the laws of probability do not affect the outcome it was down to luck. Ummm ... do you want to think that through a little more? Yes, any individual trial is 'luck', but changing your mind is 'luckier' than not. In fact, it is twice as lucky! |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 29 Oct 12 - 08:48 AM > Monty...did not let them replay the game multiple times so the laws of probability do not affect the outcome it was down to luck. Actually even I can understand that no matter how many times you play (i.e., start choosing one door of three) the odds (whatever they are) remain the same each time. Like flipping a coin. It's always 1 in 2 for heads the next time no matter how many times heads has come up in the past. And some of those mathematicians skeptcal of the new 2/3 odds are/were pretty accomplished. According to Wackipedia, which may be reliable in this case, "Decision scientist Andrew Vazsonyi described how Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until Vazsonyi showed him a computer simulation confirming the predicted result." I'm wondering if those who say the odds increase by switching understand that Monty would allow you to switch your choice of doors *whether or not your first choice found the car.* So revealing one goat did not imply anything about the wheareabouts of the car behind the remaining two doors. He didn't care if you won the car or not, at least for the purposes of this discussion. And where is the flaw in the following reasoning? Regardless of what's just happened, you're left with two doors. The car could be behind either one. Somebody coming fresh to the scene would have a 1 in 2 chance of picking the car, whether he switched or not. (Because there would be no spare goat door to open.) Right? No matter how many times he switched back and forth, his odds of picking the car would be 1 in 2. Right? So how is it that *your* odds can be increased to 2 in 3 by switching, while his can't, in precisely the same situation at the same moment???? Didn't your odds increase from 1 in 3 to 1 in 2 the minute the spare goat was revealed? Why would that change? True, you now know that the car wasn't behind the third door. But so what?! That's in the past, like all those coin flips that came up heads! Two doors, one car, one goat, 50/50! |

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Subject: RE: BS: Monty Hall ProblemFrom:
Howard Jones
Date: 29 Oct 12 - 08:54 AM The first choice is a red herring - it isn't really a choice at all - it's just a bit of TV showmanship. The first choice doesn't matter, because it's not acted on. You may have a 1 in 3 chance of selecting the right door, but your chance of winning is nil. Whichever door you choose, the host will open a different one. The only choice which matters is the second choice, after one of the doors has been taken out of contention so it's then a straight 50/50 chance between 2 doors, one of which must contain the car and the other the goat. It's an entirely new choice, which may or may not be the same as the first choice. The first choice doesn't have a bearing on it, it only appears to. Despite the bewildering (to me) maths, the decision tree and diagrams showing the permutations all show an equal number of wins and losses. |

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Subject: RE: BS: Monty Hall ProblemFrom:
TheSnail
Date: 29 Oct 12 - 09:04 AM Nigel Parsons had it right. Taking the original Car/Goat/Goat scenario, remember that Monty knows where the car is. There are three equally likely sequences of events if you switch. 1) You choose the Car. Monty shows a Goat. You switch and get the other Goat. 2) You choose Goat 1. Monty shows a Goat. You switch and get the Car. 3) You choose Goat 2. Monty shows a Goat. You switch and get the Car. So 2 out of 3, you get the car. If you don't switch, there are another three equally likely sequences of events. 1) You choose the Car. Monty shows a Goat. You don't switch and get the Car. 2) You choose Goat 1. Monty shows a Goat. You don't switch and get the Goat. 3) You choose Goat 2. Monty shows a Goat. You don't switch and get the Goat. So 2 out of 3, you get a goat. |

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Subject: RE: BS: Monty Hall ProblemFrom:
Nigel Parsons
Date: 29 Oct 12 - 09:13 AM The first choice is a red herring - it isn't really a choice at all - it's just a bit of TV showmanship. The first choice doesn't matter, because it's not acted on. You may have a 1 in 3 chance of selecting the right door, but your chance of winning is nil. Whichever door you choose, the host will open a different one.The only choice which matters is the second choice, after one of the doors has been taken out of contention so it's then a straight 50/50 chance between 2 doors, one of which must contain the car and the other the goat. It's an entirely new choice, which may or may not be the same as the first choice. The first choice doesn't have a bearing on it, it only appears to. Yes, given a straight choice between two doors, where your only knowledge is that one hides a goat, the other a car, your chance is 50/50. But... This is not a simple choice. You have already selected a door, and are being asked to choose whether to change. This means that the first choice affects the outcome. The door you have already chosen was a choice from 3, where you knew that it was more likely (2/3) that you would have chosen a door with a goat behind. Someone just seeing the two doors and being offered the choice does not have the additional knowledge, so has a 50% chance of finding the car. With your additional knowledge (by swapping) your chance increses to 66%+. No wonder it confuses the experts! Cheers Nigel (recreational mathematician) |

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Subject: RE: BS: Monty Hall Problem (other cases)From:
Mysha
Date: 29 Oct 12 - 09:15 AM Hi, Two questions above about other cases: - Do the odds change if Mr. Hall only opens another door if at that time you've picked the car? = As it happens, they don't, but your strategy changes slightly. If Mr. Hall shows you a goat, then you know for sure that you have the car, so stick with it. If Mr. Hall does not show you a goat, then you know you yourself picked a goat. You'll now have to pick one of the other two doors for a car, but since he didn't open either of them, you'll only have a fifty/fifty chance of getting the right one. As it happens, this gives you the same two out of three odds the original question had. - If you want a goat, and Mr. Hall opens a door showing one, what door should you pick? = The door Mr. Hall just opened. Bye Mysha |

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Subject: RE: BS: Monty Hall ProblemFrom:
Bobert
Date: 29 Oct 12 - 09:25 AM Seems that after eliminating one of the 3 doors (the goat) and given an opportunity to pick between the two remaining doors presents a new probability as if the game was originally a choice of two... So the "new" probability is 1 in 2 or 50/50... It doesn't, however change the 33.33% chance that was originally offered... The wild card here is that the rules change in that, in essence, if the contestant is allowed to make a 2nd choice then the odds of winning the car are 66.66%... B~ |

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Subject: RE: BS: Monty Hall ProblemFrom:
gnu
Date: 29 Oct 12 - 10:23 AM If you have two doors to choose from, what are you odds of picking the right door? |

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Subject: RE: BS: Monty Hall ProblemFrom:
Jim Dixon
Date: 29 Oct 12 - 10:35 AM You don't need be any equations or mathematical expertise if you have enough patience to use the "brute force" method and list out all the possibilities. To begin with, the prize has already been placed behind one of the doors, so there are 3 possibilities, all equally probable: * The prize is behind door 1. * The prize is behind door 2. * The prize is behind door 3. Now you make a choice, which expands the list to 9 possibilities: * The prize is behind door 1, and you choose door 1. * The prize is behind door 1, and you choose door 2. * The prize is behind door 1, and you choose door 3. * The prize is behind door 2, and you choose door 1. * The prize is behind door 2, and you choose door 2. * The prize is behind door 2, and you choose door 3. * The prize is behind door 3, and you choose door 1. * The prize is behind door 3, and you choose door 2. * The prize is behind door 3, and you choose door 3. As the next step, Monte will open one of the doors, but his choice is not random: he will always open a door that does not reveal the prize. You know in advance that this will happen. So there is no point in waiting to see which door he opens; you might as well decide in advance what your strategy will be. You will either switch or not switch. Suppose you decide that your strategy will be to not switch. We now have enough information to predict the outcome: * The prize is behind door 1, you choose door 1, and you do not switch: YOU WIN. * The prize is behind door 1, you choose door 2, and you do not switch: YOU LOSE. * The prize is behind door 1, you choose door 3, and you do not switch: YOU LOSE. * The prize is behind door 2, you choose door 1, and you do not switch: YOU LOSE. * The prize is behind door 2, you choose door 2, and you do not switch: YOU WIN. * The prize is behind door 2, you choose door 3, and you do not switch: YOU LOSE. * The prize is behind door 3, you choose door 1, and you do not switch: YOU LOSE. * The prize is behind door 3, you choose door 2, and you do not switch: YOU LOSE. * The prize is behind door 3, you choose door 3, and you do not switch: YOU WIN. You can now count the wins and losses and see that your chance of winning is 1/3. On the other hand, suppose you decide that your strategy will be to switch. Here are the outcomes: * The prize is behind door 1, you choose door 1, and you switch: YOU LOSE. * The prize is behind door 1, you choose door 2, and you switch: YOU WIN. * The prize is behind door 1, you choose door 3, and you switch: YOU WIN. * The prize is behind door 2, you choose door 1, and you switch: YOU WIN. * The prize is behind door 2, you choose door 2, and you switch: YOU LOSE. * The prize is behind door 2, you choose door 3, and you switch: YOU WIN. * The prize is behind door 3, you choose door 1, and you switch: YOU WIN. * The prize is behind door 3, you choose door 2, and you switch: YOU WIN. * The prize is behind door 3, you choose door 3, and you switch: YOU LOSE. When you count up the wins and losses, you see that your chance of winning with this strategy is 2/3. |

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Subject: RE: BS: Monty Hall ProblemFrom:
Jim Dixon
Date: 29 Oct 12 - 10:53 AM Here's another thought experiment: Instead of 3 doors, suppose there are 1000 doors. Suppose you choose door number 438. Then Monty opens all the doors except numbers 438 and 722, revealing 998 goats. Now, do you stick with 438, or switch to 722? |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 29 Oct 12 - 11:03 AM > The wild card here is that the rules change in that, in essence, if the contestant is allowed to make a 2nd choice then the odds of winning the car are 66.66%... In practical terms it still makes no sense to me. In the end, you still have to choose one door of two. The only probability that seems to count is what exists in fact at the crucial moment of choosing between two doors. If you stick with door 1 it's effectively a new choice. If you switch to door 2 it's another choice. I don't see that the news that door X was a goat is really "new." You knew when you got into this that 2 of the 3 doors had goats. Now you see where one goat was. You still don't know where the car is, or the other goat. You have two doors. What if, given the choice, you switch when given the chance, but then quickly switch back again? Would that increase your odds even further? Obviously not. But how does that hypothetical opportunity for a third choice differ structurally from being granted a second choice? If you grab Monty's mike and switch between the two doors a hundred times, are your chances of winning the car slowly approaching 1 in 1? Please say no. Why is Howard not right when he says that the first choice is a red herring? Whatever door you pick, you'll wind up having to choose between just two doors later. Despite the efforts of well-meaning 'Catters, I don't understand how having made a blind choice from three doors in the past can influence the odds of choosing correctly between two in the present. Are the 2 in 3 odds somehow retroactive? If so, wouldn't this be a vital key to practical time travel? Wait a minute. You know at the start you'll *ultimately* have two choices of three. So in that sense your odds overall are clearly (?) 2 in 3 *at the start* (because you have two shots at 3 doors). But why don't the odds drop to 1 in 2 when the situation resolves into one real car and one real goat? Or is the remaining goat Schroedinger's goat? (To add to the fun, once you made your first or second choice, Monty would offer you a worthwhile amount of cash in lieu of revealing the car or the goat. But let's not get into that. (And, at least in theory, if won a goat, you had to keep it.) |

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Subject: RE: BS: Monty Hall ProblemFrom:
Bobert
Date: 29 Oct 12 - 11:16 AM You are correct, Lighter... If you, in essence, get one bad guess before the count begins then it is 50/50... B~ |

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Subject: RE: BS: Monty Hall ProblemFrom:
Nigel Parsons
Date: 29 Oct 12 - 11:32 AM I don't see that the news that door X was a goat is really "new." You knew when you got into this that 2 of the 3 doors had goats. Now you see where one goat was. You still don't know where the car is, or the other goat. You have two doors. That information is new. You previously new only that the car was behind one of 3 doors. You now know that it is behind one of 2. Before you made your decision you had a one in 3 chance of picking the car. The removal of one door and one goat means that you now only have to choose between two doors. But you know (from the first choice) that your first door has a 1/3 chance of hiding the car, so you accept the swap for a 2/3 chance. Monty Schrödinger's GoatAt the start of this act, behind each of three doors is one third of a car, and two thirds of a goat (or one third of two goats). The compere asks you to choose a door ... |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,TIA
Date: 29 Oct 12 - 11:46 AM If you want to use the "nothing new" argument, try this: At the outset, you have a one in three chance. When Monty opens one of the doors, you are correct that you still have a one in three chance that you have picked the car. But there is only one other door at this point, and your door has an unchanged one in three chance of being the car. Now what are the other possibilities *at this point*? There is only one other possibility - the other door! So the single remaining door must have a two in three chance of holding the car! |

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Subject: RE: BS: Monty Hall ProblemFrom:
Nigel Parsons
Date: 29 Oct 12 - 12:06 PM Sometimes I wonder whether some people just don't get it, or whether they're winding us up to see how many different ways we'll try to explain it. |

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Subject: RE: BS: Monty Hall ProblemFrom:
Bee-dubya-ell
Date: 29 Oct 12 - 12:29 PM Perhaps it would help to imagine more doors. What if there were a million doors? You pick one, Monty eliminates 999,998 of the remaining 999,999. Which door is more likely to have the car behind it? The one chosen randomly, or the one left standing after almost a million others have been eliminated? |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 29 Oct 12 - 01:28 PM > Sometimes I wonder whether some people...'re winding us up to see how many different ways we'll try to explain it. If the solution were as obvious as you seem to think, there'd be no controversy. It would be me against the world. And it isn't. Of a million doors, say I pick No. 1. Monty "picks" No. 17. Obviously the car is behind one of those two doors. At the start of the game, with nothing revealed, the chance that the car is behind one of those two doors would be 1 in 500,000, and that it is behind whichever one I choose is 1 in a million. But Monty doesn't know where the car is either. He's thrown doors open at random and left No. 17 closed at random. *If one of 999,998 opened doors had hidden a car, we'd know it and the game would be over. I'd have lost.* And the odds are tremendous that that's what would have happened then and would happen almost all of the time. (It would be a very boring show, with very few winners.) But if we've got this far without finding the car, it must be behind either 1 or 17. No matter what door Monty ultimately "picked," the chance that it hid the car would have been, at the start, precisely 1 in a million, just like the chance that it's behind door No. 1. And now you're left with two doors: odds of 1 in 2. I'm not insisting that I'm right. Far from it. But it would be nice to know precisely what statements or assumptions I'm making that are so obviously incorrect. What reality alters the 1 in 2 odds when Monty doesn't know where the car is either, it hasn't been revealed, and you're down to two doors? Most counterintuitive things become clear when explained, but this hasn't. I presume I'm not the only one still in the dark. And by admitting continued confusion, if that's what it is, I'm probably losing all credibility in every other BS discussion. So be it. |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Ed
Date: 29 Oct 12 - 01:32 PM But Monty doesn't know where the car is either.But he does! From the Wikipedia article you initially linked to: "the host, who knows what's behind the doors, opens another door" |

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Subject: RE: BS: Monty Hall ProblemFrom:
gnu
Date: 29 Oct 12 - 02:59 PM If you have two doors to choose from, what are you odds of picking the right door? If you have two doors to choose from, what are you odds of picking the right door? Pick one of my questions and answer it. You have a 50/50 chance of picking the right question. |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 29 Oct 12 - 03:36 PM Ed, sorry for the confusion. The W. article is very dense. Stipulating that Monty always knows where the car is changes everything in a way that really is easy to understand. I used to watch the show in the '60s. It appeared that Monty didn't know (as I posted earlier), or at least pretended not to know, and the contestants assumed he didn't know either. Or at least had no way of knowing for sure. I believe he would sometimes open a door and the car would be behind it, without opening a goat door. Probably he knew some of the time - or all of the time, but the contestants didn't know whether he did or not. That was part of the fun. Sure, if he knows, and you know he knows, then obviously if he leaves door No. 17 shut, out of a million, it's pretty suspicious, and that door, selected with insider information, is far more likely to hide the car than your own blindly selected door which would hide the car by pure chance alone. And I hope you're saying that if Monty doesn't know the whereabouts of the car, the 1 in 2 analysis at the crisis point is correct, whether there were a million doors or three. |

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Subject: RE: BS: Monty Hall ProblemFrom:
TIA
Date: 29 Oct 12 - 04:24 PM If Monty doesn't know, *your* odds will not change at all. Sometimes the game will end abruptly when he reveals the car rather than a goat, but *your* odds will remain the same if the opened door reveals a goat. |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 29 Oct 12 - 04:52 PM Exactly. Case closed. Thanks to all. |

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Subject: RE: BS: Monty Hall ProblemFrom:
gnu
Date: 29 Oct 12 - 04:53 PM Oh, BTW... d'ya think it's cheaper for "Monty" to have two goats and one car or to have three goats? Crank out those odds. And, don't tell me Monty showed the car if the wrong choice was made in order to "prove" it was all legit. Ye must be stunned as me arse iffn ya dunno how that ruse was done. Slight of hand is for the slight of mind. |

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Subject: RE: BS: Monty Hall ProblemFrom:
Howard Jones
Date: 29 Oct 12 - 06:34 PM With the first choice, you have no chance at all, because even if you pick the door with the car behind it (1:3) you still don't win it. All the first round does is help the host to choose which of the doors with a goat behind it he opens. But that isn't random - he knows where the prize is. It's just a set-up to raise the tension for the second round, where you have a choice between two doors, one of which must hide the prize. What confuses the issue is that one of the doors must be the door you chose first time. So it feels as if you have to decide whether to switch or stick with the first choice. The host winds up the tension by suggesting that. If you guess right and win you'll feel vindicated, if you lose you'll feel frustrated that your first choice was right all along. But that's all psychology. It's still simply a choice between one door or the other. Why should the choice you made the first time make it more or less likely that you'll guess right the second time? |

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Subject: RE: BS: Monty Hall ProblemFrom:
Bobert
Date: 29 Oct 12 - 06:41 PM What kinda of car is it, anyway??? B;~) |

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Subject: RE: BS: Monty Hall ProblemFrom:
gnu
Date: 29 Oct 12 - 06:43 PM Howard... it doesn't. That's what all the scientists and philosophers and math geeks are hiding behind when they spout their "theories". If this, then that... if that, then this... there ain't none... it's two doors and one choice. Plain and simple. |

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Subject: RE: BS: Monty Hall ProblemFrom:
Howard Jones
Date: 29 Oct 12 - 07:00 PM Ah, I think I've got it. The first round does affect it, because it is twice as likely that your first choice was wrong than right. You're more likely to choose the goat than the car in the first round, and since the other goat is out of contention that makes it more likely that the other door hides the car. |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 29 Oct 12 - 07:02 PM Sometimes it was a luxury car, sometimes a boat, sometimes tickets to Hawaii, etc., etc. It was always valuable. And the "goat" of the puzzle could be any kind of disappointing or worthless item. (Sometimes it was a goat, or a cow, or a dented bowling ball or something like that.) The third door could hide almost anything from junk to something decent to something almost as good as the grand prize. It varied. It's been a long, long time, but as I recall the location of the grand prize was always revealed. Otherwise, of course, the show would just be a variation of three-card monte (get it? "Monty"? "monte"?) with a disappearing "prize." And there was joking and bargaining and costumes and elaborate promotional descriptions of the worthwhile prizes and cash bribes to get people either to switch their choices or not to switch. Silly, but fun at the time. But none of that has anything to do with the odds related to the stylized version of the game we've been discussing. It does seem, though, that many people *don't* think it affects the odds if Monty knows where the grand prize is and doesn't want you to win it. That naivete' *is* a little troubling. |

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Subject: RE: BS: Monty Hall ProblemFrom:
gnu
Date: 29 Oct 12 - 07:14 PM "The first round does affect it, because it is twice as likely that your first choice was wrong than right." There is it! You Got it! Whatwere the chances? |

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Subject: RE: BS: Monty Hall ProblemFrom:
BK Lick
Date: 29 Oct 12 - 08:40 PM I think Jim Dixon's "brute force" method is eminently clear and should be sufficient to convince the most mathematically naive that switching increases the odds of winning from 1/3 to 2/3. Increasing the number of doors in the thought experiment usually makes the result seem more intuitive, but I think it may be more effective (psychologically) to consider increasing the number only to ten; observe that the "not switching" strategy wins the prize one time in ten., while the "switching" strategy wins the prize nine times in ten. Or in the three door problem, consider that switching turns a correct initial guess into a loss but turns an incorrect initial guess into a win. Since the initial guess is twice as likely to be incorrect as correct, switching is twice as likely to win the prize. If you're intrigued by thought experiments that seem to defy intuition, you might lose some sleep over Newcomb's paradox.—BK |

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Subject: RE: BS: Monty Hall ProblemFrom:
John on the Sunset Coast
Date: 29 Oct 12 - 09:56 PM For the game to work properly the two lesser prizes" must be of comparable value", and switching doors wins 67% of the time. If, as has been asserted, there was 1 ea. Megaprize, Goodprize and Clunker, then the game changes. If your first choice is the GP, you will not be be shown the MP...because the game needs to continue; you will be shown the C. Staying Loses If your first choice is the C, you will be shown not your choice or the MP because the game needs to continue. So again you should switch. Staying loses. If you actually pick the MP, you will be shown either GP or C, but you don't know that you actually have chosen correctly. So since you know that in two scenarios out of three switching is correct, you should still switch. Remember you have no way of knowing where the MP is on the first round, but you maximize your chance of getting the MP whenever you are shown a lesser prize. Using the switching strategy you only lose the MP is picked first (1 of 3); you win if you did not pick it first (2 of 3). Where the strategy may break down is if you have picked the MP, but are are shown the C. You might stand because you're sure to get at least |

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Subject: RE: BS: Monty Hall ProblemFrom:
GUEST,Lighter
Date: 29 Oct 12 - 10:08 PM It complicates the game to stipulate that Monty knows where everything is and doesn't want you to win. That's one way of playing. But in the version discussed by me, he doesn't know and he doesn't care if you win or not. And there might be one or two clunkers. I could not understand how, in that situation, switching would increase your odds. And it now looks as if it wouldn't. But if Monty knows and you know he knows, that's a different story. |

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Subject: RE: BS: Monty Hall ProblemFrom:
gnu
Date: 29 Oct 12 - 10:35 PM The contestant picks a door. Monty opens a door with a goat. Then he asks the contestant to pick one of the two doors remaining. Stop changing and twisting the facts to suit your arguements. Two doors, one choice. 50/50. |

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Mudcat time: 19 September 9:42 AM EDT

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