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pythagoras
guest

Aug2004, 08:08 PM (EST) 

"Monty Hall Problem  logic proof"

Hi everyone. I have found a simple solution/proof for the Monty Hall problem (goats behind two doors, car behind other, etc.) Rather than looking at your changing choice, we shall examine the prize’s different positions. Let us label the doors A, B and C. Let us say you choose door A originally (the probability of selecting the prize originally is the same for each door). Now let us examine each of the cases for the different positions of the prizes. If the prize is behind door A your should stay with your original choice (you of course do not know the prize is behind this door). If the prize is behind door B, door C will be revealed and thus you should change your choice to door B. Similarly, if the prize is behind door C, you should also change your choice to this door. Thus there are two cases in which you should change your choice, and only one in which you should keep it as it remains. Therefore, you should change your choice. I think this is a fairly logical argument. I have developed this myself, and I am very sorry if someone has already thought of this solution. Tell me what you think: david@sydneysystems.com 


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Silas
guest

Aug2504, 06:24 AM (EST) 

1. "Monty Hall: contra Terry Pascal"
In response to message #0

An answer to Terry Pascal's incorrect assessment. Alex, I've been reading the Monty Hall problem and would like to post the answer to demonstrate where most people math is incorrect. Given 2 doors, one with a prize and one without, which should you pick? I think you'd agree that odds are 5050 you'll be right. The fact that I may have had 3 doors initially and opened one, or 1000 doors and opened 998, does not change the current problem. There is still only a 50/50 chance of winning. What is wrong here is that Mr Pascal is attempting to turn the switch into a separate 50/50 choice. But that is not the case, as I think would be evident to anybody that actually played through the test on the Monty Hall problem page. There is only your first choice, which has a 1 in 3 chance of being correct. The supposed second choice of switching or not switching is actually a foregone conclusion dependent entirely upon your first choice. As you are sitting there clicking, you can already tell that you're only getting the prize one time in three. (Or in my case 3 times in 3 for my first 3 clicks, so I would have been very unlucky at this game, since I always switch.) As the switch when you haven't picked the prize is always to the prize, you should always switch. Because the probability of the problem can be expressed in three different ways. P1  1/3 chance prize is door #1, 2/3 chance it is not P2  1/3 chance prize is door #2, 2/3 chance it is not P3  1/3 chance prize is door #3, 2/3 chance it is not Most of the logic I've seen has the following flaw. As soon as the contestant picks door #1, they immediately assume that P2 and P3 are not valid. Well, they are. So they follow only P1. Regardless of which door I pick, all 3 are still valid. It is only when Monty opens a door that things change. By opening door #3 we now have P3  0/0 chance prize is door #3, 3/3 chance it is not. Now here everyone says, "Hey by P1, then the 2/3 chance all goes to door #2' But then why not, "By P2, then the 2/3 chance all goes to door #1"? Both are incorrect. When Monty picks #3, the whole problem is changed to: P1  1/2 chance prize is door #1, 1/2 chance it is not P2  1/2 chance prize is door #2, 1/2 chance it is not P3  0 chance prize is door #3, 1 chance it is not But it is not a flaw to assume that P2 and P3 are not valid. They are not valid by dint of your not having picked #2 or #3. They become intrinsically meaningless since they do not reflect the reality of the situation. P2 and P3 are still valid probabilities, but having picked #1, you cannot apply "by P2 then the 2/3 chance all goes to door #1". You didn't pick #2, so P2 is meaningless. Look at it another way. What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? What Mr. Pascal forgets here is that Monty Hall knows which one contains the money. If he picks #1 and Monty shows #3, 2/3 of the time #2 will win. But the reason #3 is shown is that it doesn't have the money in it. He then says "If I had picked #2 to start with and then monty opens #3 then the odds go 2/3 to #1." This is looking at it back to front. If Monty in this case is always opening #3, then the prize was either in #1 or #2. But this case is specifically excluding all the cases where the prize was in #3 in which case Monty would not open it. That is an incorrect way of looking at probability. What if I don't have to tell monty? What if I could write it on a secret ballot? Terry Pascal
But in that case, Monty would every now and then open up your choice  which is not what happens in the Monty Hall problem, and therefore is irrelevant. The basic flaw in Mr. Pascal's analysis is as follows: Two times out of three, Monty Hall has no choice in which box to open. He only has a choice when you picked right first time in which case he randomly picks the box to open up. So in the analysis in which the pick was #1 and Monty Hall opened #3, in the 1/3 chance you picked the correct box first time, Monty Hall could equally have picked #2, but you are excluding all those events. On the other hand, in the two thirds chance that the wrong one was picked, Monty Hall only opens #3 when the prize is in #2. The occasions that the prize is in #3 are excluded. Again, this is not the right way to look at probability. 


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Silas
Member since Aug2504

Aug2504, 07:44 AM (EST) 

2. "Monty Hall Problem  logic proof"
In response to message #0

>Hi everyone. I have found a simple solution/proof for the >Monty Hall problem (goats behind two doors, car behind >other, etc.) > >Rather than looking at your changing choice, we shall >examine the prize’s different positions. Let us label the >doors A, B and C. Let us say you choose door A originally >(the probability of selecting the prize originally is the >same for each door). Now let us examine each of the cases >for the different positions of the prizes. If the prize is >behind door A your should stay with your original choice >(you of course do not know the prize is behind this door). >If the prize is behind door B, door C will be revealed and >thus you should change your choice to door B. Similarly, if >the prize is behind door C, you should also change your >choice to this door. Thus there are two cases in which you >should change your choice, and only one in which you should >keep it as it remains. Therefore, you should change your >choice. > >I think this is a fairly logical argument. I have developed >this myself, and I am very sorry if someone has already >thought of this solution. >Tell me what you think: >david@sydneysystems.com No problems in the logic. It'simply raises the whole problem of the Monty Hall Problem, which is the simplest and most logical way to explain the right answer to people who continue to "intuit" the other answer. My post above is emphatically not an attempt to do this, but to point out flaws in logic in one of the common responses. I think your way of explaining it is the simplest and clearest I've ever seen. I don't like seeing Truth Tables, since some people (including me) switch off when they see a diagram, and I don't like the argument about "What if it was 1000, you pick one and Monty Hall opens 998 other doors  isn't it obvious you switch then?" because although the 10001 option is obvious, it'still isn't obvious when you take those extra boxes away whether it is reducing to 1 in 3 or 1 in 2 at the final choice. 


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rewboss
guest

Aug2604, 12:12 PM (EST) 

3. "Monty Hall Problem  logic proof"
In response to message #2

I received a very clear and logical answer in the thread I started on this very subject. Here's my version: Suppose the host is particularly bad at this game, and whenever he is faced with any kind of choice, he dithers (much to his wife's exasperation, we can imagine). Everyone notices this except for the host himself. Now imagine a game in progress. The contestant has just selected door A. Now, there is a 1 in 3 chance that he has selected the winning door, and the host will then have to choose whether to open door B or C, and will dither. But there is a 2 in 3 chance that the contestant has selected the wrong door; in that case, the host has no choice (he must open the one door not chosen and not concealing the prize), and so he doesn't dither. If the host dithers, the contestant knows that he has already selected the correct door and doesn't switch  he wins the prize. If the host doesn't dither, the contestant knows he must switch  and so wins the prize. That way, he wins every single time. So, what are the chances he will switch? He switches when the host does not dither, but unhesitatingly opens a door. That happens in 2 out of every 3 cases, on average. And sooner or later, the host is fired. 


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Graham C
Member since Feb503

Sep1104, 09:25 AM (EST) 

5. "RE: Monty Hall Prob.  logic proof"
In response to message #3

Your answer only confirms my own long and deeplyheld conviction that in its original form it was wrong to treat the Monty Hall situation as a probability problem in the first place. It's a game theory problem, in which the state of knowledge of the contestant has to be taken into account, including his assessment of the motivations and likely actions of the host. The question put is not 'what should you do, knowing all this background', but 'what should the contestant do, unexpectedly presented with the second choice?' The first question is a nobrainer. The second is interesting. Like should you draw to an inside straight, really  or, mor accurately, how should you respond to a confident raiser who has just drawn one to four cards. 


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neuroglider
guest

Mar0105, 10:14 AM (EST) 

6. "RE: Monty Hall Problem"
In response to message #5

My assessment of the Monty Hall problem is this: By the very nature of the Description, there were always only two doors to choose from. The way the story is told, the AUTHOR (not Monty Hall) takes away the third choice. The scenario in which the contestant chose correctly the first time is not an option. Therefore, the answer is 50:50 from the outset. For the contestant to switch his/her choice makes no difference. To pretend otherwise is like sneaking a live cobra in the box with Schroedinger's cat. This argument is similar to those in which the motives or integrity of the gameshow host are questioned. Except those arguments are made by addressing the RULES by which this game could be generally played in a REPEATED fashion. What I'm saying is that the problem as originally posed is a single instance, a specific subset of the possibilities, regardless of motives or rules.



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Latecomer
guest

Aug0105, 05:38 PM (EST) 

8. "RE: Monty Hall Problem  logic proof"
In response to message #0

I think I understand the generally accepted analysis of the Monty Hall problem, but for more than a decade, I've been bugged by a variation which seems to me to call that answer into question. I'm assuming that the prize is placed before any choices are made, that Monty Hall knows where the prize and the zonks are, that he will open a door with a zonk, and that he always offers a contestant a chance to switch. My variation: suppose the contestant, instead of being in an open studio, is in an isolation booth, where he can see the doors and hear Monty's offers when allowed, but doesn't know anything else that's going on. And suppose there is a second contestant playing the game simultaneously. Neither contestant knows of the other, and neither is limited by the other's choices  if they each pick the winning door, they'll each get the prize. Contestant 1 chooses door A. The monitor in his isolation booth is switched off. The monitor and headphones in contestant 2's booth are turned on, and he chooses door B. Monty then opens door C for each in turn, revealing a goat. He offers each contestant in turn the chance to switch doors. If we only knew about contestant 1 (which, I believe, is the original situation), the conventional wisdom says he improves his chance of winning from 1/3 to 2/3 by switching. The same would be true of contestant 2, if we only knew of his existence, wouldn't it? But contestant 1 can't be doubling his chances by switching from A to B while contestant 2 is also doubling his chances by switching from B to A, right? I'm assuming that somebody has posed this variation and had whatever flaws are there pointed out many moons ago; I just haven't been able to find it. If anybody can direct me to a place where this twist has been discussed, I'd be grateful! 


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mr_homm
Member since May2205

Aug0105, 08:10 PM (EST) 

9. "RE: Monty Hall Problem  logic proof"
In response to message #8

> >But contestant 1 can't be doubling his chances by switching >from A to B while contestant 2 is also doubling his chances >by switching from B to A, right? > >I'm assuming that somebody has posed this variation and had >whatever flaws are there pointed out many moons ago; I just >haven't been able to find it. If anybody can direct me to a >place where this twist has been discussed, I'd be grateful! I don't know of any such discussion, but I think I can clarify it here. In order for the problem to be a proper probability problem, you have to have a well defined set of possiblities, from which the various outcomes are selected by chance. By assuming that Monty Hall can open a door neither contestant chose, you are assuming that that door did not contain the big prize. Therefore, your analysis applies only to the set of cases in which the unselected door definitely did not contain the prize. Among these cases, either contestant is equally likely to have picked the prize, so switching will not help either one. You can't take the conclusion of the usual analysis over to this new case, because the set of possible outcomes has been reduced, and the probabilities change when you do this. For example, your chances of getting a full house in poker change if you play with a deck with no spades. Because of this, we don't know in advance that there is an advantage in switching doors, and your analysis does not lead to a contradiction, but rather to a perfectly valid solution to a different problem, in which the answer is that there is no advantage in switching. Stuart Anderson 


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mr_homm
Member since May2205

Aug0205, 11:29 AM (EST) 

12. "RE: Monty Hall Problem  logic proof"
In response to message #11

>>I don't know of any such discussion, but I think I can >>clarify it here. > >1. This message is part of a thread >2. https://www.cuttheknot.org/hall.shtml Well, yes, of course. I have read these, but I cannot find in them a discussion of this particular variation of the problem, which is what Latecomer was looking for. By the way, it has always seemed to me that the clearest way tp explain the original problem is to note that when you make your choice, 1/3 of the time you have the prize and a 2/3 of the time it is behind one of the other doors. If Monty Hall always opens one of them to show a goat, you now have new information, which lets you pick the right door in every one of those 2/3 of the cases. This is more or less the standard explanation, I'm just mentioning that it is the one that appeals to me the most. As a sidelight, for people whose mathematical intuition works better with dollar signs in front of the numbers, this whole Monty Hall problem provides a simple example of why insider trading on the stock market is illegal. Suppose there are two investors X and Y considering buying stock in three companies A, B, and C. A, B, and C are in direct competition, and only one will be profitable. In an effort to avoid prosecution, X makes a deal with Monty Hall (who has insider information about all 3 companies). The deal is that after X has invested in one company, Monty Hall will give him a tip about which of the other companies is not profitable. After X invests in A, Monty Hall tells X that company B will not be profitable. X then sells his stock in A and buys C. Investor Y does not receive such a tip. When they are brought before the Securities Exchange Commission, Monty Hall argues that he hasn't given X any information about the company X is currently invested in, nor has he told X which company is profitable. Now the question is, should this be illegal? Does the tip give X an unfair advantage? Yes it does. From an investment point of view, the additional information allows X to avoid some risk, namely the risk of investing in B. If X acts on the tip, X can get the same potential profit at lower risk, so X's chances of profit go up. However, if X does not act on the tip, he has the same chance of making a profit as Y does. Therefore X should act on the tip. However, the only possible action is to sell A and buy C, since buying B would be stupid. Therefore, switching from A to C must increase the chance of profit. Stuart Anderson 


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Latecomer
guest

Aug0205, 02:50 PM (EST) 

13. "RE: Monty Hall Problem  logic proof"
In response to message #9

Thanks for the feedback. I think I'm following . . . but let me bounce something else off of you: It'seems to me that my suggested variation is a subset of the possibilities allowed by the original definition of the problem, rather than a completely different problem. From the point of view of contestant 1, who never knows about contestant 2, I think the setup assumptions can stay the same. And from his point of view, all he knows is that he's picked door A, and that there's a goat behind door C. And yet in this admittedly limited subcategory of the original definition, without him learning anything more, it'seems that another person's discovery of the same information trips up the conclusion that he will always double his chances of winning by switching. As long as the original problem's definition can include the variation, and the variation leads to the possible inapplicability of the general analysis, I'm struggling . . . I'm going to pose another variation a little farther down this chain. 


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alexb
Charter Member
1664 posts 
Aug0105, 08:13 PM (EST) 

10. "RE: Monty Hall Problem  logic proof"
In response to message #8

The problem is in these three patagraphs: "Contestant 1 chooses door A. The monitor in his isolation booth is switched off. "The monitor and headphones in contestant 2's booth are turned on, and he chooses door B. "Monty then opens door C for each in turn, revealing a goat. He offers each contestant in turn the chance to switch doors." The choices of the two contestants can't be independent if Monty is still able to open door C with a goat behind. 


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KL
guest

Aug0205, 02:50 PM (EST) 

14. "RE: Monty Hall Problem  logic proof"
In response to message #10

Call the doors A, B and C. Let's assume the car is behind door A. If the players choose independently then there are 9 possible combinations (AA, AB, AC, BA,…). Two of these (BC and CB) do not allow Monty to open a door that is both losing and unchosen. We need to define what our host is going to do. Consider two scenarios: 1. If he can open a losing door, then he always does, and then offers the contestants the opportunity to change their choice. We will ignore the 2 out of 9 cases when he cannot open a losing door. 2. He opens a losing door iff the contestants have chosen different doors, and he can open an unchosen door. We will ignore the 2 out of 9 cases when he cannot open a losing door and the 3 out of 9 cases where the contestants chose the same door. In the first scenario, opting to switch wins on 4/7 occasions, and remains the winning strategy. The second scenario, which is the one presented above, gives a 1/2 probability of winning whether you switch or not.



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Latecomer
guest

Aug0205, 02:50 PM (EST) 

15. "RE: Monty Hall Problem  logic proof"
In response to message #10

You're right; my variation can't always work with the "Monty always shows a goat" and "Monty always offers a switch" assumptions; it can only work if one of the contestants has chosen the prize. But thinking of that, I was thinking of the way Let's Make A Deal actually used to work, without any of the isolation booths and such that I mixed into my variation. So let me try this one on you: Suppose contestants 1 and 2 are together in the audience. Monty offers 1 first choice, he picks door A. Monty offers 2 the next choice, he picks door B. Monty then opens door C, revealing a goat. He then offers 1 and 2 the chance to switch doors. I don't think it matters whether it's a zerosum, only one winner, both must agree to switch situation, or whether it's an either can switch, either can stay, bothcanwin situation. The standard analysis is that "my original pick had a 1/3 chance, and the other two doors had a 2/3 chance, which I can now have in a single door," right? Contestant 1 thinks the 2/3 chance was contained in doors B and C. Contestant 2 thinks the 2/3 chance was contained in doors A and C. Both have learned that the 2/3 chance excludes door C, so both should switch, right? But that can't be right, right? I realize that I've changed the game, but I don't think the changes I've made in either case have changed any of the elements that seem to be underlying the standard answer to the original Monty Hall problem as posed  the original problem says that your initial choice had a 1/3 chance, that the remaining two doors had a 2/3 chance, and that eliminating one of the other doors gets you that same 2/3 chance all of the time. Don't those principles apply to my suggested variations? I don't think I've added any additional revelations, just different views of the same problem. What do you think? I realize that this was a hot topic more than a decade ago, but I'm slow . . . hence my screen name! 


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alexb
Charter Member
1664 posts 
Aug0205, 03:22 PM (EST) 

16. "RE: Monty Hall Problem  logic proof"
In response to message #15

>You're right; my variation can't always work with the "Monty >always shows a goat" and "Monty always offers a switch" >assumptions; it can only work if one of the contestants has >chosen the prize. It is not what I said. I said that, in so far as the experiments are not independent you cant' apply to each of them in separation the theory of an independent experiment. At least, this is what I meant. If you require a stipulation that "one of the contestants has chosen the prize," the experiments again are not independent. The only way you may meaningfully consider two contenstants unaware of each other and each other's choices is to allow as much arbitrariness in one's choice as in the other's. 


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Latecomer
guest

Aug0205, 06:08 PM (EST) 

17. "RE: Monty Hall Problem  logic proof"
In response to message #16

I agree. I was wondering, though, why adding a second contestant needed that contestant to be independent or unaware of the first contestant in order to illustrate my point about my problem with the initial Monty Hall solution. And on reflection, I don't think it matters whether contestant 2 knows about contestant 1, or can or cannot choose the same door, to demonstrate where I'm trying to go. Which is what I was getting at in restating my variation with a known and knowing contestant 2. I think I'm correct in believing that one premise of the original solution is that the initial guess has a 1/3 chance of success. It'seems to me that this is a baseline premise of the standard solution for contestant 1 whether or not there is a second contestant, and whether or not the second contestant is known. Similarly, contestant 2, whatever door he picks, would have the same 1/3 chance in being correct, according to the analysis that's at the foundation of the accepted answer to the original problem. We can set the problem so that he has the same number of choices, or we can set it'so he knows of contestant 1's choice and is locked out of that one. That doesn't matter, does it? Isn't he, according to the accepted analysis, faced with the same 1/3 probability for whatever choice he makes? Now Monty opens door C, revealing a goat. According to the accepted analysis, "the doors I didn't pick" held a 2/3 probability of being the winning door, correct? And that 2/3 probability is now contained in "the doors I didn't pick, excluding C," right? What I'm not grasping is why a second contestant can't also say, "the doors I didn't pick" had a 2/3 chance of success, and why, if the second contestant didn't pick C and Monty opens it up to reveal a goat, he's not also covered by the standard analysis: my pick was 1/3, the doors I didn't pick were 2/3, that 2/3 is now down to one door, so I'm twice as well off by switching to that door. The way I've seen the statement of the initial principle stated doesn't suggest that it depends on anything other than the idea that your initial guess was a 1/3 shot, and that the remainder was a 2/3 shot, whose payoff potential is now narrowed to one door. On principle, the way I've been reading it, nothing prohibits its application to a second contestant. And yet in the instance where there is a second contestant, and he does pick a different door, and there is a zonk door for Monty to open, the application doesn't work  I don't think. Since I think I'm understanding the premise of the original problem, and understanding the principle of the solution, applying the principle to a subset of the original without changing the facts of the original (haven't varied what contestant 1 knows, when he knows it, or where the prize is), when I get to that subset and the principle doesn't work, I'm thinking there's a problem with the principle. Contestant 2's appearance, or his guess, whether he tells it to contestant 1 or not, shouldn't affect the right answer to a probability question, should it? 


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eddieboy
guest

Oct1205, 04:45 PM (EST) 

18. "RE: Monty Hall Problem  logic proof"
In response to message #0

Hi. I'm glad to have found this site/thread because I have a problem regarding the Monty Hall paradox and hope that some of you chaps can help me out. It is this: In another forum I frequent (not maths related) a poster posted the Hall puzzle. After a while he offered this by way of an explanation.  We have three boxes, which I'll call L (left hand), M (middle) and R (right hand). There are three equallylikely arrangements: 1: L has prize : M has a goat : R has a goat 2: L has a goat : M has prize : R has a goat 3: L has a goat : M has a goat : R has prize Let's suppose you pick out the L box (but cannot open it yet). If I know that we have arrangement 1, then I can open the R box and show you the goat. This leaves the M box unopened. Similarly if I know that we have arrangement 2. If I know instead that we have arrangement 3, then I can open the M box and show you the goat. So, after I have opened a box containing a goat, the three possible arrangements are as follows: 1: L has prize : M has a goat : R is open 2: L has a goat : M has prize : R is open 3: L has a goat : M is open : R has prize So, remembering that these three possible arrangements are equally likely from your perspective, what should you do? (i.e.switch or stick)  I read it and thought the explanation was flawed. The poster (a well known UK academic, by the way, so he should know) maintains that it is correct. My reasoning is this: Once the contestant picks a box and the host opens one (containing a goat) surely that then becomes an individual game completely separate from the other two arrangements. If that is correct and I think it must because they can have no bearing on the game being played, how can the three possible arrangements 1, 2 and three still be equally likely from the contestants perspective? I hope I have explained the situation well enough for it to be understood. Am I correct? Or is "the man who should know" correct? I would appreciate any and all opinions. Thankyou 


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ramsey2879
guest

Oct1405, 00:10 AM (EST) 

19. "RE: Monty Hall Problem  logic proof"
In response to message #18

>In another forum I frequent (not maths related) a poster >posted the Hall puzzle. After a while he offered this by way >of an explanation. > > > >We have three boxes, which I'll call L (left hand), M >(middle) and R (right hand). There are three equallylikely >arrangements: >1: L has prize : M has a goat : R has a goat >2: L has a goat : M has prize : R has a goat >3: L has a goat : M has a goat : R has prize >Let's suppose you pick out the L box (but cannot open it >yet). > >If I know that we have arrangement 1, then I can open the R >box and show you the goat. This leaves the M box unopened. >Similarly if I know that we have arrangement 2. If I know >instead that we have arrangement 3, then I can open the M >box and show you the goat. > >So, after I have opened a box containing a goat, the three >possible arrangements are as follows: >1: L has prize : M has a goat : R is open >2: L has a goat : M has prize : R is open >3: L has a goat : M is open : R has prize >So, remembering that these three possible arrangements are >equally likely from your perspective, what should you do? >(i.e.switch or stick) > > > >I read it and thought the explanation was flawed. The poster >(a well known UK academic, by the way, so he should know) >maintains that it is correct. > >My reasoning is this: Once the contestant picks a box and >the host opens one (containing a goat) surely that then >becomes an individual game completely separate from the >other two arrangements. If that is correct and I think it >must because they can have no bearing on the game being >played, how can the three possible arrangements 1, 2 and >three still be equally likely from the contestants >perspective? > Yes the game is changed once the contestant choses a box and the host host opens a box. Remember that the box which has the car has not been opened yet and it could still be in one of the three boxes. Proof: there is a one third chance that the car is in the left box (case 1 with one sixth chance that the middle door is open and one sixth chance that the right door is open), one third chance that the car is in the middle box (case 2) and one third chance that the car is in the right box (case 3). As you can see each of these cases is equally possible regardless of the fact that Monte opens a box with the goat. Thus there is a two thirds chance that the car is not in the left box (case 1) but in the middle (case 3) or right box (case 2) which Monte did not open. > >Am I correct? Or is "the man who should know" correct? You are incorrect since it appears that you don't attach a equal probability of the car being in one of the three boxes. >I would appreciate any and all opinions. Thankyou Your welcome. Note that similar logic applies if the contestant opens the middle or right door instead of the left door. Hope this clarifies things. 


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eddieboy
guest

Oct1405, 04:44 AM (EST) 

20. "RE: Monty Hall Problem  logic proof"
In response to message #19

"Yes the game is changed once the contestant choses a box and the host host opens a box. Remember that the box which has the car has not been opened yet and it could still be in one of the three boxes." That is part of my problem. How can it be so? If one box has been opened and shown to contain a goat, how can it contain both a goat and a car? "You are incorrect since it appears that you don't attach a equal probability of the car being in one of the three boxes." I don't, not once a box is open. Again I say it is impossible for a box to contain both a goat and a car. Isn't it? Remember that one box is open and shown to contain a goat. Therefore the car must be in one of the other two, must it not? I don't get this. Thanks for your reply. 


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ramsey2879
guest

Oct1605, 11:27 AM (EST) 

33. "RE: Monty Hall Problem  logic proof"
In response to message #20

>"Yes the game is changed once the contestant choses a box >and the host host opens a box. Remember that the box which >has the car has not been opened yet and it could still be in >one of the three boxes." > >That is part of my problem. How can it be so? If one box has >been opened and shown to contain a goat, how can it contain >both a goat and a car? > This is because Monte is not a magician, The car has equal probability of being in one of three boxes as there are three boxes originally. Because Monte has no power to change this fact by merely opening a door, you have to take into account the fact that there are two doors other than the door that the contestent has chosen. The posibility that the car is in one of these two boxes is and will always be 2/3 regardless of which box Monte opens. Merely opening a door doesn't move the car from one box to the other two boxes with equal probility so that it is in the unopen boxes with equal probability, rather it means that the car hasn't moved but the possibility that the car in in the other of the two boxes has changed from 1/3 to 2/3 as the probility that the car is in the opened box is 0. Look at the original posibilities, assuming that contestant is removed to a back room and Monte opens the two doors that the contestant did not chose. He has a 2/3 possibility of finding the car since the car could be in either of three boxes regardless of which box the contestant chose. Now assuming that Monte found the car he closes the door having the car, if not he merely closes either door. When the contestant is brought back to the room, he tells the contestant that he opened two doors and may have found the car but to keep the game to appear fair he closed one of the two doors. Of course Monte wouldn't leave open the door that had the car, but Monte's chances of having found the car remain 2/3. Closing a door to hide the car or change the appearances dosent change this. If the contestant switches he is assuming Monte's possibilities of finding a car which is 2/3 rather than the possibility of the car being in the left door which is 1/3. This may seem strange, but the fact is that Monte had two doors which he could choose from where as up until the time the contestant return to the room he could only chose 1 of three. Thus if the contestant switches he is giving up his 1 in 3 possibility for Monte's 2 in 3 possibility. >"You are incorrect since it appears that you don't attach a >equal probability of the car being in one of the three >boxes." > >I don't, not once a box is open. Again I say it is >impossible for a box to contain both a goat and a car. Isn't >it? Remember that one box is open and shown to contain a >goat. Therefore the car must be in one of the other two, >must it not? I don't get this. > Yes but since the probability that the car is in the left door is 1/3 then the probability that the car is in the unopened door of the other two doors is 2/3!! >Thanks for your reply.



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alexb
Charter Member
1664 posts 
Oct1405, 04:59 AM (EST) 

21. "RE: Monty Hall Problem  logic proof"
In response to message #18

>If I know that we have arrangement 1, then I can open the R or box M >box and show you the goat. This leaves the M box unopened. >Similarly if I know that we have arrangement 2. If I know >instead that we have arrangement 3, then I can open the M >box and show you the goat. > >So, after I have opened a box containing a goat, the three >possible arrangements are as follows: >1: L has prize : M has a goat : R is open >2: L has a goat : M has prize : R is open >3: L has a goat : M is open : R has prize I would say there is another possibility 1': L has prize : R has a goat : M is open >So, remembering that these three possible arrangements are >equally likely from your perspective, what should you do? >(i.e.switch or stick) (1 1') are equiprobable with 2 and 3. >I read it and thought the explanation was flawed. The poster >(a well known UK academic, by the way, so he should know) Marilyn would disagree. A good many academicians scolded her after she published her solution. >maintains that it is correct. He is almost, with the above caveat. >My reasoning is this: Once the contestant picks a box and >the host opens one (containing a goat) surely that then >becomes an individual game completely separate from the >other two arrangements. It does not. This has been chewed over and regurgitate so many times that I wonder whether your question is in good faith. Please read this thread, or https://www.cuttheknot.org/hall.shtml or just make a search on the web. Your quandary has been resolved many times over.



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eddieboy
guest

Oct1405, 08:41 AM (EST) 

22. "RE: Monty Hall Problem  logic proof"
In response to message #21

With all due respect it WAS asked in good faith. I have no problem with the solution to the Monty Hall paradox as you seem to imply by pointing me to this thread.https://www.cuttheknot.org/hall.shtml Perhaps I could try to explain my point by using a three horse race analogy? Three horses "A", "B" and "C". The race starts. "A" falls down dead and is therefore out of the race (the open box containing a goat,it obviously cannot contain the car) As "A" to all intents and purposes is last (dead) that then surely must give us just TWO possible outcomes to the race. (1)"B" wins, "C" is second, "A" is last. Or (2) "C" wins, "B" is second, "A" is last. Is there a third? I hope you can now see that my problem is not with the Monty Hall paradox as such but wih the wording of the explanation I was given which says: "after" I have opened a box containing a goat. That then (in that game) is surely unalterable, like the dead horse. I am not a mathematician so please bear with me. 


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alexb
Charter Member
1664 posts 
Oct1405, 09:05 AM (EST) 

23. "RE: Monty Hall Problem  logic proof"
In response to message #22

>Perhaps I could try to explain my point by using a three >horse race analogy? > >Three horses "A", "B" and "C". The race starts. "A" falls >down dead and is therefore out of the race (the open box >containing a goat,it obviously cannot contain the car) > >As "A" to all intents and purposes is last (dead) that then >surely must give us just TWO possible outcomes to the race. >(1)"B" wins, "C" is second, "A" is last. Or (2) "C" wins, >"B" is second, "A" is last. Is there a third? No, there are just two of them. The problem is the analogy is not accurate. Instead consider this: assume under the saddle of one of the three hourses a chance concealed a $1,000 bill. You jump on one of them and are about to start looking under the saddle while another one loses its saddle revealing nothing beneath it. It drops dead if you wish. The other two hold steady. Or you can think of a situation where it is known up front that two of the horses are going to drop dead and not finish the race. You jump on one of them. On the first loop one of the other hourses drops dead. You are given a chance to change horses. Should you? >I hope you can now see that my problem is not with the Monty >Hall paradox as such but wih the wording of the explanation >I was given which says: "after" I have opened a box >containing a goat. That then (in that game) is surely >unalterable, like the dead horse. I did not use the word "after" in the two scenarios. However, each of them is clearly sequential. Horses start droping dead after you chose one of them. Your scenario does not model the Monty Hall problem. What is the horse metaphor? A door? A prize? If you say "door", I reply "Is it a door that you are after or the prize?" If you say "prize", I reply "How do you know which horse has died? Could not it be the prize?" 


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eddieboy
guest

Oct1405, 06:59 PM (EST) 

24. "RE: Monty Hall Problem  logic proof"
In response to message #23

"What is the horse metaphor? A door? A prize?" Neither really. A more accurate comparison would be the box itself and it's contents. The car/prize is represented by winning of the race. So horse "A" (the dead one) would be = The open box containing the goat. In (1)"B" would be = the box containing the car/prize. "C" would be = the box containing the second goat. "A" would be = the open box containing the other goat. In (2)"C" would be = the box containing the car/prize. "B" would be = the box containing the second goat. "A" would be = the open box containing the other goat. "The problem is the analogy is not accurate." Any better?



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eddieboy
guest

Oct1505, 07:58 AM (EST) 

25. "RE: Monty Hall Problem  logic proof"
In response to message #24

It'seems my horse race may have confued the issue a little. OK let's go back to Monty Suppose I am the contestant. Three boxes Left, Middle and Right. Let's say I pick L. Monty opens M to show a goat. That means the car/prize is in either L or R. It cannot be in M. That also means there are two possibilities. M contains a goat. L contains a goat. R contains the car. or M contains a goat. R contains a goat. L contains the car. Now my man says "So, after I have opened a box containing a goat" and "So, remembering that these three possible arrangements are equally likely from your perspective, what should you do? (i.e.switch or stick)" How can that be possible? For it to be so, the third arrangement would have to be: M contains the car/prize. R contains a goat. L contains a goat. As I already said that is not possible because we know that M contains a goat. Monty has showed us it. Unless you know different? 


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alexb
Charter Member
1664 posts 
Oct1505, 08:25 AM (EST) 

26. "RE: Monty Hall Problem  logic proof"
In response to message #25

>Suppose I am the contestant. Three boxes Left, Middle and >Right. Let's say I pick L. Monty opens M to show a goat. >That means the car/prize is in either L or R. It cannot be >in M. That also means there are two possibilities. > >M contains a goat. >L contains a goat. >R contains the car. >or >M contains a goat. >R contains a goat. >L contains the car. >Now my man says "So, after I have opened a box containing a >goat" and "So, remembering that these three possible >arrangements are equally likely from your perspective, what >should you do? (i.e.switch or stick)" I do not see three possibilities, just two. But they are not equally likely. They occur in ratio 2:1. This is because, before you made your selection of L, the third possibility was present. If you went with it, Monty would not open M, but R instead. This possibility has been implicitly folded into your #1. Try to switch a view point. Assume, after Monty opens a door, you are not required to do anything but just compute the probabilities. Surely the probability of having the prize in L is 1/3, whatever Monty does afterwards. So the probability of not having a prize in L is 2/3, whatever Monty does or does not do.



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eddieboy
guest

Oct1505, 09:08 PM (EST) 

27. "RE: Monty Hall Problem  logic proof"
In response to message #26

"I do not see three possibilities, just two." Thankyou. That is what I have been saying all along. Both here and at the other forum. "But they are not equally likely." I know they are not. I have never said they were. Either here or at the other forum. "Surely the probability of having the prize in L is 1/3, whatever Monty does afterwards. So the probability of not having a prize in L is 2/3, whatever Monty does or does not do." I agree. I have never argued against it. Either here or at the other forum. So, would you now agree with me that my U.K. academic's explanation, where he says that after a box is opened and shown to contain a goat, there are three possible arrangements, is incorrect?



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alexb
Charter Member
1664 posts 
Oct1505, 09:27 PM (EST) 

28. "RE: Monty Hall Problem  logic proof"
In response to message #27

>"I do not see three possibilities, just two." > >Thankyou. That is what I have been saying all along. Both >here and at the other forum. Do not know about other forums, but this is not what you were saying on this one. >So, would you now agree with me that my U.K. academic's >explanation, where he says that after a box is opened and >shown to contain a goat, there are three possible >arrangements, is incorrect? No, I just checked your original post at this forum. There is not too subtle a difference between what you posted originally and what you said lately. You also changed roles. At the beginning "I" meant Monty; later you converted it into the contestant. The wording of the fellow you can't agree with may be unfortunate, but the reasoning is correct. After a door is opened by Monty there are indeed three equally likely possibilities: 1: L has prize : M has a goat : R is open 2: L has a goat : M has prize : R is open 3: L has a goat : M is open : R has prize This was the original statement. Later, you formulated it thus Suppose I am the contestant. Three boxes Left, Middle and Right. Let's say I pick L. Monty opens M to show a goat. That means the car/prize is in either L or R. It cannot be in M. That also means there are two possibilities. M contains a goat. L contains a goat. R contains the car. or M contains a goat. R contains a goat. L contains the car. You see "Monty opens a door" and "Monty opens door M" are two different things. (Note that, for the sake of juxtaposition, it would have been less confusing had you said "Monty opens R to show a goat," not "M".) 


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eddieboy
guest

Oct1605, 06:11 AM (EST) 

29. "RE: Monty Hall Problem  logic proof"
In response to message #28

"After a door is opened by Monty there are indeed three equally likely possibilities:" That is not so. The act of opening the door tells me, you, Monty, whoever that that one contains a goat. Since we know that Monty always opens a door with a goat then we also know that that leaves only two possibilities after the door has been opened. It matters not which one it is. "You see "Monty opens a door" and "Monty opens door M" are two different things." I know they are but as I said above it doesn't matter. The end result is the same. "(Note that, for the sake of juxtaposition, it would have been less confusing had you said "Monty opens R to show a goat," not "M".)" OK we/I can do that but it doesn't alter the fact that there will still only be two possibilities. I am begining to think you and my man are one and the same. :) 


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alexb
Charter Member
1664 posts 
Oct1605, 07:07 AM (EST) 

30. "RE: Monty Hall Problem  logic proof"
In response to message #29

>"After a door is opened by Monty there are indeed three >equally likely possibilities:" > >That is not so. You should read this as "With what me know after Monty opened a door there may be three equally likely possibilities." The possibilities do not reflect the physical state of affairs which is "one door open, two doors closed," but are those that might have led to that physical state. As I said before and which you prefer to ignore, the wording  the usage of the word after  may be unfortunate, but this does not change the author's intention, viz., Before Monty opened a door, we had such and such possibilities. The fact of Monty's opening a door made some of them unrealized in principle, so that after he opened a door only such and such possibilities remain. This is what the fellow wanted to say.



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eddieboy
guest

Oct1605, 10:02 AM (EST) 

31. "RE: Monty Hall Problem  logic proof"
In response to message #30

"Before Monty opened a door, we had such and such possibilities. The fact of Monty's opening a door made some of them unrealized in principle, so that after he opened a door only such and such possibilities remain. This is what the fellow wanted to say." Finally, we seem to have arrived at what seemed to me to be the unavoidable and obvious conclusion. Albeit by a rather tortuous route. At times I felt I was flogging my own dead horse. Thankyou. Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true. Bertrand Russell (1872  1970), Mysticism and Logic (1917) ch. 4 Just a joke, bud. 


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alexb
Charter Member
1664 posts 
Oct1605, 10:03 AM (EST) 

32. "RE: Monty Hall Problem  logic proof"
In response to message #31

>Finally, we seem to have arrived at what seemed to me to be >the unavoidable and obvious conclusion. Albeit by a rather >tortuous route. At times I felt I was flogging my own dead >horse. > >Thankyou. You are welcome. >Mathematics may be defined as the subject in which we never >know what we are talking about, nor whether what we are >saying is true. But always strive to understand each other, right? 


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