Fanghur Posted March 29, 2012 Share Posted March 29, 2012 I've wondered this for a long time: Why the heck is the famous Monty Hall Problem even to this day so controversial? I mean, I'll admit that when I first heard the problem I was like "Give me a break; two doors left means there's a 50/50 chance, so switching your choice is pointless," but once I grasped it fully I was quite frankly embarrassed that I didn't get it right away, because it's pure common sense. For those of you that don't know the Monty Hall Problem, this is it. A game show host presents a game show contestant with three closed doors and tells her that behind one of the doors is a new car (i.e. one out of three doors is a winner), and behind the other two doors are goats (i.e. two out of three doors are losers). He asks the contestant to pick any door, let's say they pick door number 1. The host, who knows what's behind each of the doors, then opens one of the remaining two doors and reveals a goat. He then asks the contestant whether they would like to switch their original choice to the remaining closed door. Now, at first glance this would seem to be pointless, as most people would think that two closed doors = 50/50 chance of picking the car. But because your original choice has a 2/3 chance of being a goat, and the host always eliminates the second goat, it means that is you stay with your first choice you have a 1/3 chance of picking the car, while if you switch your choice you have a 2/3 chance of winning the car. The average person would find this very counter-intuitive, but I refuse to believe that even people with degrees in mathematics and statistics couldn't see that switching your choice is best. So why the heck is it still such a hotly debated subject when it's just common sense, and high school level mathematics? Link to comment Share on other sites More sharing options...
John Posted March 29, 2012 Share Posted March 29, 2012 (edited) I think at first glance it almost looks like an example of the gambler's fallacy, but most mathematically literate people probably accept it at this point. Edited March 29, 2012 by John Link to comment Share on other sites More sharing options...
Ophiolite Posted March 29, 2012 Share Posted March 29, 2012 I've always been distracted by thoughts of being able to own my very own goat. 5 Link to comment Share on other sites More sharing options...
Xittenn Posted March 29, 2012 Share Posted March 29, 2012 Because it omits a step taken when calculating standard probabilities, which is to combine the probabilities stepwise. The final probability is the same in all cases as a combination, or else you have a 1/3 probability the first time, and a 1/2 the second. If lying to yourself makes you feel better, then philosophically speaking, all the power to you! The chances that the pair group probability is, is also another consideration when putting this idea into perspective but also does not help your case any. Link to comment Share on other sites More sharing options...
Joatmon Posted March 29, 2012 Share Posted March 29, 2012 Simply put, if the contestant picks the goat (to which two of the three doors lead) they will win a car by switching as the other goat can no longer be picked, while if the contestant picks the car (to which one door leads) they will not win the car by switching. So, if you switch, you win the car if you originally picked the goat and you won't if you picked the car, and as you have a 2 in 3 chance of originally picking a goat you have a 2 in 3 chance of winning by switching. I find this explanation quite understandable. It is copied from:- http://en.wikipedia.org/wiki/Monty_Hall_problem 3 Link to comment Share on other sites More sharing options...
Ophiolite Posted March 29, 2012 Share Posted March 29, 2012 That makes sense to me. Is it wrong? If not, why has someone given negative rep? Link to comment Share on other sites More sharing options...
Joatmon Posted March 29, 2012 Share Posted March 29, 2012 (edited) That makes sense to me. Is it wrong? If not, why has someone given negative rep? I would also like to know why. Perhaps whoever gave the -1 (or anyone else for that matter) would explain why the explanation is faulty. Because it omits a step taken when calculating standard probabilities, which is to combine the probabilities stepwise. The final probability is the same in all cases as a combination, or else you have a 1/3 probability the first time, and a 1/2 the second. If lying to yourself makes you feel better, then philosophically speaking, all the power to you! The chances that the pair group probability is, is also another consideration when putting this idea into perspective but also does not help your case any. Are you saying that switching does, or does not, improve your chances of winning the car? Edited March 29, 2012 by Joatmon Link to comment Share on other sites More sharing options...
mathematic Posted March 29, 2012 Share Posted March 29, 2012 I would also like to know why. Perhaps whoever gave the -1 (or anyone else for that matter) would explain why the explanation is faulty. Are you saying that switching does, or does not, improve your chances of winning the car? By switching your probability of getting the car is 2/3. 1 Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted March 30, 2012 Share Posted March 30, 2012 I would also like to know why. Perhaps whoever gave the -1 (or anyone else for that matter) would explain why the explanation is faulty. If it makes you feel any better, a bug in the forum software let someone give you 2 positive rep points for that post. Link to comment Share on other sites More sharing options...
Xittenn Posted March 30, 2012 Share Posted March 30, 2012 I would also like to know why. Perhaps whoever gave the -1 (or anyone else for that matter) would explain why the explanation is faulty. State something mathematical! Although this theorem has an entry on wolfram it doesn't really do anything but state the problem in a rather meaningless way. If tested the idea would fail and under mathematical scrutiny it looks like goo IMO. I negated a bad idea, sorry Jo. I would have talked about it but it's hard to get my points across and I'm very busy with studies. Give me numbers and back it up, show me the money! Link to comment Share on other sites More sharing options...
John Posted March 30, 2012 Share Posted March 30, 2012 When the contestant chooses a door, the probability that the car is behind that door is 1/3. This means there's a probability of 2/3 that the car is behind one of the other two doors. One of the two others is opened to reveal a goat. The probability that the car is behind the contestant's chosen door is still 1/3; therefore, the probability that the car is behind the unopened door the contestant didn't initially choose is 2/3, and the contestant should switch. To put it another way, consider three doors. There are only three possibilities here (considering the goats to be interchangeable): car, goat, goat; goat, car, goat; and goat, goat, car. Let the contestant choose any door, and then let a second door be opened to reveal a goat. In one of the three possible orderings, switching will cause the contestant to go from car to goat, losing. With either of the other two orderings, switching will cause the contestant to go from goat to car, winning. Therefore, the probability that switching will result in a win is 2/3. There are several explanations of the solution in the Wikipedia article. 1 Link to comment Share on other sites More sharing options...
Xittenn Posted March 30, 2012 Share Posted March 30, 2012 This is a trivialization . .. if I am in one of three groups there is a 2/3 chance that I have chosen a group that has both a car and a goat. If the second goat is destroyed either I beat the odds and am in the group that had a 2/3 chance or I am in the group that has a 1/3 chance not the other way round. If I then make a choice between two options I have a 50/50 . . . . and will remain so in my mind until I see a mathematical proof! And not a 2/3 chance of winning the car, but a 2/3 chance of either having picked the car or of being in the group that the car is in . . . where the one third you picked the double goat! Link to comment Share on other sites More sharing options...
John Posted March 30, 2012 Share Posted March 30, 2012 This is a trivialization . .. if I am in one of three groups there is a 2/3 chance that I have chosen a group that has both a car and a goat. If the second goat is destroyed either I beat the odds and am in the group that had a 2/3 chance or I am in the group that has a 1/3 chance not the other way round. If I then make a choice between two options I have a 50/50 . . . . and will remain so in my mind until I see a mathematical proof! And not a 2/3 chance of winning the car, but a 2/3 chance of either having picked the car or of being in the group that the car is in . . . where the one third you picked the double goat! Again, the Wiki: http://en.wikipedia.org/wiki/Monty_Hall_problem#Bayes.27_theorem Link to comment Share on other sites More sharing options...
Xittenn Posted March 30, 2012 Share Posted March 30, 2012 more added notes which I might continue to do now that I've been sucked into the pit of doom . . . . I'm saying that by picking one of a goat, a goat, or a car, without knowing which is which of coarse, x is thereby picking into a pair that is bound--I can do this because you started it and if we are playing that game then I remain on equal grounds whether you like it or not. So now that I have picked an item such that I am bound by two groups I am either bound to two successful or to only one successful in having a car. By eliminating the second goat we actually create this bound state that you are all so adamantly protecting because you all wish it were true. So again you are either in group 'I was bound by the car twice', or group 'I was bound by the car and the sucker prize.' The unbound real chance still remains 1/3 chance of having chosen the right item, and you must now choose whether to stay or to switch. If you switch there is a 50/50 chance that you could now be in the 1/3 group and all items had a 1/3 chance so being that only a goat is removed--which could be a 2/3 or a 1/3 we don't even know unless of course someone tells you it is a specific way in some half assed paradox--then all bets remain on your final decision which is 50/50. Even if you compound this by combinatorics you are still stuck with an equal chance of sucking no matter how bad you wish it wasn't so. No probabilities form positive interferences with itself PLACE BIG DOT HERE Again, the Wiki: http://en.wikipedia.org/wiki/Monty_Hall_problem#Bayes.27_theorem Wiki is a cesspool of stuff that remunerates underneath city streets! Also there is NO MATHEMATICAL FOUNDATION anywhere, wiki or not! Link to comment Share on other sites More sharing options...
Appolinaria Posted March 30, 2012 Share Posted March 30, 2012 Simply put- You're most likely to pick a goat in the beginning. Using that information solely, it is a good idea to switch. Link to comment Share on other sites More sharing options...
Xittenn Posted March 30, 2012 Share Posted March 30, 2012 Is an illusion that doesn't hold true, for the reasons already given. A mirage, is a mirage, is a mirage . . . Link to comment Share on other sites More sharing options...
John Posted March 30, 2012 Share Posted March 30, 2012 (edited) Wiki is a cesspool of stuff that remunerates underneath city streets! Also there is NO MATHEMATICAL FOUNDATION anywhere, wiki or not! Wikipedia is fairly accurate on most topics. Also, the specific part of the article I linked is a mathematical proof that switching is the rational decision. It's also followed, a bit further down, by other mathematical proofs. Below that, there is a list of variants of the problem that lead to different results, including the 50/50 you intuitively expect. Edit: Here is another article detailing programs written to test the idea: http://en.wikipedia.org/wiki/Empirical_solution_of_the_Monty_Hall_problem This is, of course, another Wikipedia article, but in this case you can easily run the code yourself to verify the results. In addition, http://leeps.ucsc.edu/misc/page/monty-hall-puzzle/ also provides references to studies done confirming the result. Edited March 30, 2012 by John Link to comment Share on other sites More sharing options...
Xittenn Posted March 30, 2012 Share Posted March 30, 2012 This is like the infinite improbability drive. And there are no proofs! Link to comment Share on other sites More sharing options...
Schrödinger's hat Posted March 30, 2012 Share Posted March 30, 2012 Xitten, I cannot tell if you are joking, and if you're not, I cannot tell what you are trying to say. One thing that will make the normal conclusion of the Monty Hall problem incorrect is if the host is not forced to use his knowledge and open a goat door. Assuming he knows where the goats are, and he opens a goat door every time, do you then accept the conclusion (ie. it is the host transferring some of his knowledge to you that alters your probability estimate)? An alternative scenario that is often much more clear is the following: There are 1 million doors, with 999,999 goats and one car (assuming you want the car). You pick a door. The host knows where all the goats are. He opens 999,998 doors which all have goats (and he does this every time the game is played). Do you switch (or is switching a good idea if you want the car)? If switching does not improve your odds of getting the car. Please explain why. If switching does improve your odds of getting the car, please explain the difference between this and the 3 door version. Link to comment Share on other sites More sharing options...
Xittenn Posted March 30, 2012 Share Posted March 30, 2012 An alternative scenario that is often much more clear is the following: There are 1 million doors, with 999,999 goats and one car (assuming you want the car). You pick a door. The host knows where all the goats are. He opens 999,998 doors which all have goats (and he does this every time the game is played). Do you switch (or is switching a good idea if you want the car)? If switching does not improve your odds of getting the car. Please explain why. If switching does improve your odds of getting the car, please explain the difference between this and the 3 door version. Precisely! I might be mucking around a bit yes! But this was what I was getting at when I said "No probabilities form positive interferences with itself PLACE BIG DOT HERE" I think I hit on a deeper thought but honestly I couldn't make it any clearer if I tried! Link to comment Share on other sites More sharing options...
the asinine cretin Posted March 30, 2012 Share Posted March 30, 2012 When I first heard of this I supposed the odds would be 1/2 after the goat reveal but now I get it and am slightly disturbed. I'm writing a simulation at this moment just for fun. Link to comment Share on other sites More sharing options...
Xittenn Posted March 30, 2012 Share Posted March 30, 2012 That would be kind of fun! Link to comment Share on other sites More sharing options...
Schrödinger's hat Posted March 30, 2012 Share Posted March 30, 2012 I think I hit on a deeper thought but honestly I couldn't make it any clearer if I tried! This is unfortunate, because I still cannot tell what you are trying to say or even whether you are claiming switching is a good idea. Are you sober and in a state of good mental health? You are normally much easier to understand than this. Maybe break it down step by step and draw some diagrams? 1 Link to comment Share on other sites More sharing options...
Xittenn Posted March 30, 2012 Share Posted March 30, 2012 This is unfortunate, because I still cannot tell what you are trying to say or even whether you are claiming switching is a good idea. Are you sober and in a state of good mental health? You are normally much easier to understand than this. I may not have been clear on how I managed to come to my conclusions but I was very clear about my position, the odds remain the same. Switching does not affect the outcome. I am very sober and I am probably the most sober person you know, I even quite drinking coffee and caffeinated beverages. Maybe break it down step by step and draw some diagrams? I'm quite alright, people see big fancy statements they don't see what is plain in front of them and fancy statements I have none. Link to comment Share on other sites More sharing options...
Joatmon Posted March 30, 2012 Share Posted March 30, 2012 Paradoxically I think I can see the point that almost everyone sees as the weakness in Xittenn's argument and which from the giddy heights of higher mathematical juggling she sees as a strength. She assumes that after the door is opened and you see the goat you have the choice (admittedly stupidly) of choosing that door. In other words the odds against you remain at 1 in 3. The question considers one instant in a game show and on this occasion the door is opened and a goat is seen - whether there is a chance that a car might be seen does not need to be considered as it didn't happen for that contestant on that occasion. The first thing necessary to answer a question is to understand the question. Perhaps the thing most people would take into account is that the human element can be expected to act in a human (sensible) way. In other words you can't enter the human decision as a random factor (IMO) 1 Link to comment Share on other sites More sharing options...
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