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linkdon
Member since Apr2902

Apr2902, 03:29 PM (EST) 

"Monty Hall Problem"

I recently read the book _The Man Who Loved Only Numbers_ by Paul Hoffman (the story of the prolific mathematician Paul Erdős). In his book, Hoffman mentions the Monty Hall problem and the fact that Erdős just couldn't believe the answer that Marilyn vos Savant gave in her weekly column in Parade magazine. Well, neither could I, and even after reading many web sit's on the topic, including this one, I still believe she had it wrong. Summary: We have 3 doors with a car behind one (the prize) and goats behind the other two (nonprizes). You pick one of the doors; Monty Hall then opens one of the remaining doors with a goat behind it (he never opens the door with the prize). The issue is: do you stick with your original choice or switch. Marilyn says you should switch, as do many of the sites I've visited, claiming a 2/3 probability of win if you switch vs. 1/3 if you don't. The case analysis that Marilyn did is shown in Hoffman's book, and I believe it is flawed. She combined the two different action options of my Case A into a single option. My decomposition of cases and their outcomes below shows that the probability of winning is the same (50%) whether you switch or not. The critical case is Case A in which the host (Monty) has two different doors he can open. Here's my analysis: (Note: I apologize for the clumsy table format, but having it in tabular form makes it much clearer. I tried html but couldn't get it to work correctly, and extra spaces are automatically deleted.) I select Door #1, but I consider all cases of prize arrangement, so the same analysis applies for any initially selected door. Option 1: Choose Door & Always Switch: ___________ Door ______ ___________Actions__________  _______ Case __1_____2____3 __ 1stPick  HostOpens  2ndPick _ Outcome A ___ Car _ Goat_ Goat  ___ 1 _______ 2 ______ 3 _____ Lose A ___ Car _ Goat_ Goat  ___ 1 _______ 3 ______ 2 _____ Lose B ___ Goat _ Car_ Goat  ___ 1 _______ 3 ______ 2 _____ Win C ___ Goat  Goat_ Car _ ___ 1 _______ 2 ______ 3 _____ Win Option 2: Choose Door & Always Stick with It: ___________ Door ______ ___________Actions__________  _______ Case __1_____2____3 __ 1stPick  HostOpens  2ndPick _ Outcome A ___ Car _ Goat_ Goat  ___ 1 _______ 2 ______ 1 _____ Win A ___ Car _ Goat_ Goat  ___ 1 _______ 3 ______ 1 _____ Win B ___ Goat _ Car_ Goat  ___ 1 _______ 3 ______ 1 _____ Lose C ___ Goat  Goat_ Car _ ___ 1 _______ 2 ______ 1 _____ Lose
For both options, the probability of win is 1/2. How is my analysis flawed? Thanks, Don Link Columbia, MD 

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alexb
Charter Member
2250 posts 
Apr2902, 03:42 PM (EST) 

1. "RE: Monty Hall Problem"
In response to message #0

LAST EDITED ON Apr2902 AT 03:55 PM (EST) To start with, before the first pick has been made, there were 3 cases:
car   goat   goat 

goat   car   goat 

goat   goat   car 

How did you manage to get 4 cases after the pick? If you decided to distinguish goats at thi stage, why not to distinguish them before the first pick? You would have B and C repeated twice. 

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alexb
Charter Member
2250 posts 
Jul1808, 03:06 PM (EST) 

71. "RE: Monty Hall Problem"
In response to message #70

Monty, hi. Mr. Stikker is quite redundant, no doubt about that. I may only guess that his intention was to preempt a possible question, "And what if you choose the middle door? Or even the right one?" There is no difference of course which door you start with. This may be obvious to you, me and to many others. Some may even insist that those who do not see this simple truth have no business solving puzzles. I reserve a judgement. Would you like me to add your comment to the letter of Mr. Stikker? Thank you, Alex 

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a_tattletale
Member since Apr1602

Apr2902, 09:27 PM (EST) 

2. "RE: Monty Hall Problem"
In response to message #0

I love how this problem seems to fool so many people of all intelligence. An easy way to see this is to first convince yourself, using the strategy of always switching doors, that:  If you selected the winning door to begin with you are guaranteed to lose.  If you selected a losing door to begin with you are guaranteed to win. Then it is easy to see that the you will win with this strategy 2/3 of the time. Your analysis here uses the strategy of exploring all possible outcomes. The flaw in your analysis is that you do not explore all possible outcomes (for probability's sake). Basically there are a total of 6 possible outcomes (you pick 1 of 3, host "picks" 1 of remaining 2). In your case A, the probability of either of the 2 outcomes is actually 1/6 yet for cases B and C the probability is 1/3 (each since the host does not really have a choice). From there you should see how the correct outcome comes to light. 

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linkdon
Member since Apr2902

Apr3002, 06:31 PM (EST) 

3. "RE: Monty Hall Problem"
In response to message #2

Thanks to <a_tattletale> for an illuminating response. I think I see it now. Essentially the flaw with my counting outcomes approach is that with the Always Switch or Always Stay strategies, what the host does in picking the door to open has no impact on the outcomehence my two outcomes under case A should really be collapsed into one. By the way, I don't see how there's 6 possible outcomes. Under the rules as originally stated, the game host will never open the door with the prize. He does have a choice under Case A, where you select the door with the prize, but as I stated above, this choice does not affect the outcome. Thanks for clearing this up for me. Don Link 

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Psychlogic
guest

May0102, 03:54 PM (EST) 

4. "RE: Monty Hall Problem"
In response to message #3

The six initial possibilities are: Car Goat1 Goat2 Car Goat2 Goat1 Goat1 Car Goat2 Goat2 Car Goat1 Goat1 Goat2 Car Goat2 Goat1 Car (Same reasoning as why chances of getting a head and a tail with a toss of 2 coins is 50% and not 33.3%HH HT TH TT) The above 6 starting positons clearly show that staying with the first choice (Column 1)loses 2/6 times and win 4/6 times. 

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jrr7
guest

May0302, 04:18 PM (EST) 

5. "RE: Monty Hall Problem"
In response to message #4

What if the host does *NOT* always give you a chance to switch? What if the host knows the prizes beforehand and decides whether to give you a chance based on his knowledge? What if you suspect he wants you to lose/win? I was reading it where you think of it in terms of game theory. The Contestant wants to win the car. The Host may: want the Contestant to win, want the Contestant to lose, or not care. The Host, though, only gets to choose whether or not to offer the Contestant a switch. If the Host wants the most possible wins, he'll only show a goat and offer a switch whne the player is on a goat. If the Host wants the most possible losses, he'll only show a goat and offer a switch when the player is on the car. What about the other cases? 

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Trent Tatro
guest

May0502, 03:46 PM (EST) 

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

I can't believe what I'm reading, you people could convince yourselves that the earth is flat.The problem  You have 3 doors and choose 1. Then Monty Hall removes one of the wrong answers from the problem. So you have TWO possible outcomes  either stay or switch. One of these is the right answer so you either get it right or you get it wrong. 50% You don't have 6 possible outcomes, or 4, or 3 or whatever you're all saying. You either get it right or wrong (stay or switch) 2 possible outcomes with 1 correct answer, that says 50%. 

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Alan Cooper
guest

May1002, 07:17 AM (EST) 

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

I don't know about the others, but anyone who believes your argument should easily be convinced that although the earth may not definitely be flat it at least has a 50% chance of being flat.Pick a planet at random. Either it is flat or it is not flat. Two possible outcomes. So the probability of being flat must be 50% This illustrates a common fallacy in analysis of probabilities. The fact that an experiment can be described as having two possible outcomes does not mean that both are equally likely. To convince you that the conclusion of equal probabilities is wrong in the Monty Hall problem, consider the situation with a million doors  one car, the rest goats. Monty says Pick one and promises to show you 999998 goats. You have a 1 in a million chance of picking the car to start with. If you didnt choose the car to start with switching will give it to you for sure. If you don't agree that switching is the better strategy I'd like to meet you in Las Vegas.


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anon
guest

May1502, 06:38 PM (EST) 

8. "RE: Monty Hall Problem"
In response to message #7

I don't see how you can make the analogy with the million door example and say Monty will show you 999998 goats. Surely he would show you just 1 and then ask if you would stick or switch. 

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The Pianist
guest

Jul2902, 11:50 AM (EST) 

10. "RE: Monty Hall Problem"
In response to message #9

This problem is new to me, and fascinating. I've read most of the material relating to it on the excellent CTK site, and there seem to be numerous easy traps to fall into. Take for example Peter Stikker's explanation (www.cuttheknot.com/peter.shtml). He draws a decision tree to demonstrate the solution  but omits some of the possibilities! His first column shows the prize behind door A (1/3 probability); he chooses door A (1/3); the host opens door B (1/2)  but then what? Peter assumes that we MUST switch to door C, and does not provide for sticking with door A. If we redraw his tree to show the sticking option in all cases, suddenly the probability of winning becomes 1/2, not 1/3. When the prize is behind door A, the eight possible outcomes are split 4 win and 4 lose. They are not all equally likely, but do fall into pairs of equal probability  so the win/lose outcomes are equally likely. Will that be the end of it? I doubt it! :) (PS: I tried to include the image in this posting  but couldn't get it to work. Sorry.) 

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The Pianist
guest

Jul3002, 01:29 PM (EST) 

12. "RE: Monty Hall Problem"
In response to message #10

I think I've sorted out the graphic now. And despite subsequent postings to the forum, I STILL believe that 50% is the correct answer. As the decision tree shows, there are 24 possible outcomes, 12 with probability of 1/18, and 12 of 1/36. They appear in pairs of win and lose, each of the same probability. So the probability of a win must be 1/2. If someone would show me an error in this tree, I am willing to be convinced that I am wrong. :D 

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Ian Marshall
guest

Aug0703, 00:33 AM (EST) 

29. "RE: Monty Hall Problem"
In response to message #12

I suggest a simple solution, build a simultion of the original Monty Hall problem run it many times, then record the outcome. How often does a switch end in a prize and how often does it end in failure? I side with the the many doors solution as I 'solved' it in my own mind that way. The inital chance of choosing the correct door is tiny, however after the bad doors are revealed your chance of now getting the right answer are far better. Throughout the entire sequence of events it appears that switching is the way to go. After all the entire sequence is what generates the outcome not just the final switch: First choice = 1 in 3 Second choice 1 in 2 (remember the first choice was 1 in 3 this means when you switch you are trading a 'bad' choice of 1 in 3 for a better one of 1 in 2, you were less likely to choose the right door on your first choice, it's a way of looking at it.) Ian 

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Peter Stikker
guest

Jan1005, 08:16 AM (EST) 

58. "RE: Monty Hall Problem"
In response to message #10

Since there is some critics here about my own version, here's a respond: "Peter assumes that we MUST switch to door C, and does not provide for sticking with door A." Sorry for the omitting the variant of sticking, however: "If we redraw his tree to show the sticking option in all cases, suddenly the probability of winning becomes 1/2, not 1/3." Is completely nonsense and not true!!!!! Draw it again and you'll see. The only critics about the general statement might be that often in the story two things are left out: a) The game host knows where the price is b) He will always first open a door that is not the original chosen one and does not contain the price. Kind regards, Peter Stikker p.s. Also a computer simulation will show the same thing. 

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CapriRS302
guest

Jan1005, 01:57 PM (EST) 

59. "RE: Monty Hall Problem"
In response to message #10

You are making a common mistake, and looking at the non winning doors as individual doors, don't do that. The Non Winners, or goats, must be treated as the same, not door B, or C or whatever, or as two separate prizes, but as losing choices, period. That way, when you construct your tree, you don't count the same scenario twice, which would skew your answer. For instance, if door A was the winner, and you picked it, Monty would open either door B or C. If he opened B, and you switched to C, you lose. If he opened C, and you switched to B, you lose. THESE MUST NOT BE COUNTED AS TWO DiFFERENT POSSIBILITIES!!!! It'should be one possibility, you picked, you were shown a loser, and you switched to a loser. That will straighten out your tree. (your way of thinking may be correct if there were three different prizes and you were asked whatr is the probability of getting each prize.)


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Peter
guest

Feb0705, 06:32 PM (EST) 

61. "RE: Monty Hall Problem"
In response to message #59

It does not matter to regard them as 1 door with price and 2 without. I added the two doors in the tree, to show all options, as any good probability tree should. Another approach that leads often to the same conclusion is to think about the following: The probability that my initial chosen door is correct = 1/3 The probability that my initial chosen door is incorrect = 2/3 If my initial chosen door is correct and I switch I will loose, if my initial chosen door is incorrect and I switch I will win. Hence probability of winning with switching = 2/3, loosing 1/3. Not switching will actually 'switch' these probabilities around. As mentioned earlier, the full rules are often left out in order to keep things clear. 'The rules' I mean as in that the host knows behind what door the price is, that he will always first open a door that does neither contain the price, neither the original chosen door etc. If all of this still doesn't convince you, either create a software programe yourself and run it a few times (you could actually do this in Excel even). :) 

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Don Greenwell
guest

Jul3103, 02:07 PM (EST) 

25. "RE: Monty Hall Problem"
In response to message #8

I like this 1,000,000 doors analogy. Suppose there are 1,000,000 doors with a car behind one and goats behind the others. Select one! What is the probability the car is behind the one you selected? What is the probability the car is behind one of the other doors? You know that behind at least 999,998 of the other doors are goats. You know Monty knows where the car is. So now he opens 999,998 doors and you see all goats. Do you know any more than you did before he opened these doors? Has the probability that the car is behind your door changed? Has the probability that the car is behind one of the other doors changed? Do you want to switch? 

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Vladimir
Member since Jun2203

Aug1103, 10:29 AM (EST) 

30. "RE: Monty Hall Problem"
In response to message #26

I pick 3 doors and the probability of success is 3/7. Monty shows 3 of the remaining 4 doors to be empty. either 1. If I stick with my 3 doors and do not act on the new information, the information is irelevant. The probability of my success (that the prize is behind one of my original 3 doors) is still 3/7. The probability of the opposite event (that the prize is behind the one remaining door I did not choose and Monty did not show) is 1  3/7 = 4/7, because the probability of the prize being somewhere is 1  certainty. or 2. I ask myself: What is the probability that the one remaining door I did not pick is empty? The same as the probability that the 4 doors I did not pick were all empty to begin with, that is 3/7. Probability of the opposite event (that the prize is behind the one remaining door) is 1  3/7 = 4/7, again, because the probability of the prize being somewhere is 1  certainty. Always works. I make the 3 for 1 swap to increase my chances from 3/7 to 4/7.


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Michael Klipper
guest

Jul2902, 08:31 AM (EST) 

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

I might have a better analogy than the 1 million goat idea. (By the way, I've tried this on my computer and the answer is definitely 2/3).Most people who claim 50% seem to downplay the effect that Monty has when he shows you the goat. Think of it this way. You make an initial guess with 1/3 probability of being right. Let's say that Monty DOES NOT open any doors, and he asks you if you want to change your guess. If this is the case, then you have a 2/3 chance that the car is behind the door you didn't pick. This 2/3 chance is spread evenly across the two doors you didn't pick, leaving 1/3 chance for each door. This is normal, right? So why don't you switch? It's because you don't know which door to switch to, so your chances don't become any better. Now, when Monty shows a goat, he's actually helping you out a lot. He's saying "You'd like to switch but you don't which door to switch to? Well, DON'T switch here! Now, instead of spreading your 2/3 probability over two doors, it concentrates on one door." If this doesn't please you, here's a reasonable analysis. You can draw a tree for these possibilit's. Without loss of generality, let's say that your initial guess is Door 1. 1) Let's say that Door 1 is in fact the car. This occurs with a 1/3 chance. Monty picks one of the other two doors, each with 1/2 probability (so the chance that of both you being right and him picking Door 2 is 1/3*1/2=1/6; same for Door 3). In each of these cases, you should stay. Thus, you should stay for 1/6+1/6=1/3 of the games. 2) Let's say that Door 1 is NOT the car. This happens 2/3 of the time. Monty's choice of door is forced. You should switch 2/3 of the time. And by the way, don't get so damn annoyed at other people's arguments. I used to think as you do, but then I realized that my two apparent choices were actually not equal. 

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jman_red
Member since Jul202

Jul3102, 09:45 PM (EST) 

13. "RE: Monty Hall Problem"
In response to message #11

For those of you with maybe 10 minutes of spare time, it is fairly easy to convince yourself of the correct answer. Grab a volunteer, three plastic cups and a cotton ball. Have the cotton ball represent the car, and the other empty cups the goats. Have your volunteer turn away while you place the cotton ball under a cup. Have the volunteer first switch all the time, for maybe 2030 trials, and then stay every time for 2030 trials. I did a science fair project on the subject once and my results were as follows: Switching for 1000 trials: 664 wins, 336 losses Staying for 1000 trials: 332 wins, 668 losses jman_red 

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The Pianist
guest

Aug0102, 07:51 AM (EST) 

14. "RE: Monty Hall Problem"
In response to message #4

After 3 days, this is my 3rd posting, and the most humble. We have to distinguish between the probability of winning at all, which is 50%, and the way in which the winning outcomes are split between sticking and switching doors. Anyone who wants to convince themselves of the correct answer should draw a decision tree, like Peter Stikker (www.cuttheknot.com/peter.shtml). But once it's drawn, create a table of the probability for winning and losing against sticking or switching doors. I found it convinced my head  though not my heart. It'still FEELS wrong, but now I know although that it really is twice as likely that you'll win if you switch. Stick & win: (1/36)*6 = 1/6 Stick & lose: (1/18)*6 = 1/3 Switch & win: (1/18)*6 = 1/3 Switch & lose: (1/36)*6 = 1/6 Thank you to all the contributors to this forum, whose postings have finally led me to the truth. :7


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OzExplorer
guest

Aug0202, 07:29 AM (EST) 

15. "RE: Monty Hall Problem"
In response to message #14

If such a remark is allowed here, the Journal of Recreational Mathematics had a nice article on the Monty Hall Problem a few years back. It included possible interesting perspectives. 

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Jack Wert
guest

Aug0202, 04:24 PM (EST) 

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

I have been aware of this problem since Ms. Vos Savant first published it in the Sunday Parade magazine, and have had no problem agreeing with her. Since then, I have also enjoyed following the discussions in various forums espousing views of both the 50/50 and 2/3 camps. What interest me most is the fact that there are so many approaches  many of which are expressed in mathematical lingo that I do not understand. So I just ignore those, and merely mentally register the fact that many "more learned than I" persons disagree with my "novice" viewpoint. This applies to the most recent posting (02 Aug 02), wherein a referenced math author declared the Vos Savant view as wrong. It'seems to me that a simpler review is important easy for anyone to understand. Let's look at the problem from the start. The contestant is standing in front of three doors, behind one of which is a prize  a car. The other two doors each shield a "booby" prize  a goat. The contestant chooses a door, and the host then opens one of the other doors displaying a goat. Ths host then allows tha contestant a second choice: stay with the initial choice or switch to the other door. Which is the best winning strategy? My review follows. The probability of the contestant initially picking the "car" door is 1/3  and the probability or his/her picking a "goat" door is 2/3. I think no one will disagree with that. If he/she picks the "car" door, regardless of which door the host opens, the contestant loses if he/she decides to switch. This means that in 1/3 of the initial choices, a later switch decision results in a loss  and a win if a "stay" decision is made. If he/she initially picks one of the "goat" doors (2/3 probability) the host must choose a a "goat" door, and if the contestant switches, he/she wins  and loses if a decision is made to switch. In other words, not knowing what is behind the door of the initial contestant choice, once that decision is made, and the host performs his chore, the odds of switching  to win the car  are 2/3. No probability "trees" or equations are necessary  just simple, easy to understand, every day logical reasoning. 

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Seesall
guest

Sep1504, 10:07 PM (EST) 

36. "RE: Monty Hall Problem"
In response to message #2

Tattletale is correct and the only when with a coherent anaylsis in this entire thread.Assume door 1 is chosen Another way to look at it is, there are really two games being played. One when the car is beyond is behind door 1 and one when the car is behind door 2 or door 3 The winning percentage for each door for the second game is 100% 1/3 of the time. The winning percentage for each door for the first game is 0% 1/6 of the time (if Monty treats each door the same). The total winning percentage for either door is then 100% *1/3 + 0% 1/6  = 1/3 / 3/6 = 6/9 = 2/3 1/3 + 1/6 1/3 + 1/6


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Whymme
guest

Aug0402, 06:52 PM (EST) 

17. "RE: Monty Hall Problem"
In response to message #0

Back to the original question.. the analysis of Don is flawed in that he gives equal probabilities to all four situations. However, the two situations A both have a chance of one in six, while the situations B and C both have a chance of one in three. In bridge there is a "Principle of Restricted Choice" that says something about where to expect a court card if the adjacent court card has already been played by the opponents. This principle is based on the same logic as the Monty Haul problem. A friend of mine, an avid bridge player, follows that principle almost blind. However, when she was confronted with the Monty Haul problem she stayed adamant that the chances were 50%, even when everybody was explaining to her that she was wrong. Whymme 

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aris
guest

Aug0702, 05:04 PM (EST) 

18. "RE: Monty Hall Problem"
In response to message #17

This has been a very long discussion. I want to put it in simple terms: First, there are two strategy here, choose one door then switch or choose one door then stay. If you want to switch then you want your first choice to be the goat therefore you got 2/3 probability that your first choice will be the goat. On the other hand, if you want to stay then you want your first choice to be the car therefore you got 1/3 probability that your first choice will be the car.And one more thing for those counting the possible outcomes, you can't distinguish the 2 goats from each other even if they have different colors because it's just either you win or you loose. In the 3 doors, 1 door is the prize and the other 2 doors are nothing. 

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jeremy
guest

Aug2102, 07:47 PM (EST) 

19. "here's some logic"
In response to message #0

It'shouldn't make a difference as to how you do it. Statistically it may, but reallife, it won't. There are 3 choices. Good, bad, and bad. (G,B,B) Now, you choose one. You don't know whether it is a G or a B. You decide to choose again. Since you didn't eliminate a choice earlier, then you still don't know what you had, or what you will have. It does not matter how many guesses you take. The last guess is all that matters, and you have a 1/3 chancec of winning. If you chose 1 door, and then decide to take a different one without seeing what was in it, Your first guess did you the same amount of good as me sitting at home watching the TV yelling at you what to choose. Only that one guess matters. I can say "door A, wait no, door B!" and the fact that I said door A before choosing door B will not affect what's behind it. There's my opinion. Jeremy 

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jman_red
Member since Jul202

Aug2202, 06:14 PM (EST) 

20. "RE: here's some logic"
In response to message #19

Your opinion, while right for the scenario you propose, is wrong for the monty Hall dilemma. Look at how the problem is worded. Step 1: Choose a door. Step 2: The host will open a door, revealing it to be a goat, not the prize Step 3: You may either switch to the last remaining door or stay at your own. If you initailly choose wrong, which is 2/3 probability, then the host will eliminate the other door that is incorrect and you will switch to the correct door. That is true both statistically and in real life. Try it yourself. It'seems to me, however, that you simply missed step 2. jman_red 

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mathsonthebrain
guest

Aug2302, 11:35 AM (EST) 

21. "simplification into two games"
In response to message #20

A person has two choices half way through the game: 1) change 2) do not change let us analyse the game for each of two players. Firstly, the easy one. Somebody who does not change. He has already decided not to change and can therefore 'skip' any door being revealed. Therefore the decision is random selection out of three. P(choosing the car with initial door selection)= 1/3 therefore p(win car)=1/3 I am about to demonstrate that the alternative strategy, changing, gives a higher probability of winning. 'Not changing' is now NOT an option. As before, there is a 1/3 probability of choosing the car first. However, for this individual, choosing the car would lead to finally picking a goat after swapping. EVERY OTHER initial choice of door leads to winning as Monty will reveal the other goat, leading the player straight to the car. The only way of not being led to the car is to choose the car the first time!! P(selecting car first time)=1/3 and P(being shown one of the two incorrect doors)=1 thus P(swapping from incorrect door to correct door)=2/3 Hence this is the better strategy as it doubles the expectation of winning. This result becomes clear as soon as you consider new rules as if swapping was required, or the converse. 

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Subjugator
guest

Aug2702, 10:25 PM (EST) 

23. "Two decision trees"
In response to message #21

While I admit I am no supreme expert in statistical analysis, I have to say that I'm seeing a trend of incorrect thinking here. The way this should be defined is as follows: Decision tree 1: C = Car G1 = Goat 1 G2 = Goat 2 Possibilities for Decision Tree 1: C G1 G2 G1 C G2 G1 G2 C The chances of picking the correct door at random are 1 in 3. The contestant picks door number 1. They have a 1 in 3 chance of having picked the correct door. Monty opens door X and shows the contestant a goat. Monty then offers to let the contestant choose another door. This creates a NEW decision tree, where the possibilities are: C G1 G1 C This is an original decision and is not linked in any way other than chronology and prize to be won to the previous decision. Thusly, there is a 50% chance to select the correct door. In this case, staying and changing are equally possible, and are equally beneficial. Note that when considered as a separate decision, the odds of matching are different. Compare this to matching dice. The odds of rolling two six sided dice simultaneously and coming up with the same number are far different than the odds of rolling a single six sided die, recording the first number, and then rolling a second six sided die hoping to match that first number. In this case, the second option is far more likely (1 in 6 rather than 1 in 36). Subbie


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brash barbaloot
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Oct0904, 11:10 PM (EST) 

42. "RE: Two decision trees"
In response to message #23

Actually, for both the dice scenarios you mentioned, the probability is 1/6. The odds of rolling 2 dice simultaneously and coming up with the same number is 1/6 since there are six ways of doing so (1,1; 2,2; 3,3; 4,4; 5,5; 6,6) out of 36 possible outcomes of rolling the dice. 

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Joshua
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Aug2902, 10:25 PM (EST) 

24. "simplification simplification"
In response to message #21

This blew my mind at first, but after thinking it through it is better to switch. It doesn't cross peoples minds that the host KNOWS where the goat is. It's easier to see with more doors also. Someone used an example with one million doors but his logic was flawed because you would be dead before you could open them all. It's easier to see the ansewer if you ten doors with 9 monkeys and a car behind them. It's a little absurd to think that after you have made your 10% chance pick and then the host shows you 8 monkeys that your odds increase to 50%. The answer is more obvious even without using math if you look at two things: the host didn't randomly pick 8 doors and monkeys are more fun than goats. 

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Whymme
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Aug2502, 10:25 PM (EST) 

22. "Is this still being discussed?"
In response to message #0

I talked about this with a few guys from work, and they were all quite convinced that I was completely, totally and utterly wrong. So I proposed a bet. "OK," I said, "let's play the game a number of times. Say twenty times  the more often we play, the more random chance gives way to probability curves. The idea is simple. Each round you pay $25. I have three cards, the ace of hearts and two black kings. You have to find the ace. You pick one card of the three that I put down face down. I will then reveal one of the other cards that is not the ace. If you want to switch after that, you pay another $25  but why would you want to, if you have a 50% chance? I then reveal the card that you finally picked and if it is the ace, I will pay you $60. That must be a neat way for you to earn money; if the breaks are 50/50, as you maintain, then on the average you earn $5 with every game." In spite of their absolute conviction that I was wrong, nobody dared to take the bet. Cowards! :) Whymme


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Vladimir
Member since Jun2203

Aug0203, 04:10 PM (EST) 

27. "RE: Is this still being discussed?"
In response to message #22

LAST EDITED ON Aug0203 AT 09:02 PM (EST) I do not know what reasoning you used on the guys, you can try this:The guy picks a card. The probability of the card being the ace is 1/3. You show him one of the others being a king. 1. If he does nothing, you might have shown the king to pigeons (and not to him) with the same effect, the probability of his success (that his initial choice is the ace) is still 1/3. The probability of the opposite event (that the ace is the card he did not pick and you did not show) is 1  1/3 = 2/3, because the ace surely is somewhere. 2. He asks himself: What is the probability that the other card I did not pick is also a king? The same as the probability that the 2 cards I did not pick were both kings to begin with, that is 1/3. Probability of the opposite event (that the other card is the ace) is 1  1/3 = 2/3, again, because the ace surely is somewhere. If you want just to convince the guys by an experiment, do it with something that does not cost them much, say nuts, pennies, etc.


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MJ
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Aug0503, 00:33 AM (EST) 

28. "RE: Monty Hall Problem"
In response to message #0

I finally "got it," that is, I can now explain why you should switch every time. I think of the three doors as "two groups of doors." One "group" is the ONE door I choose, the other "group of doors"the other two doors. IT IS OBVIOUS THAT: IF the prize is placed randomly, the prize will only be in my "group" one time out of three, or 33.3% of the time. TWOTHIRDS OF THE TIME, THE PRIZE WILL BE LOCATED IN "THE OTHER GROUP." When I finally saw the problem this way, "I got it." I should always jump to the group which has the prize 2/3 of the time. 

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quinn@mit.edu
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Oct0403, 07:57 AM (EST) 

31. "RE: Monty Hall Problem"
In response to message #0

The difficult part of this puzzle is understanding that there is a difference (however subtle) between the door you picked and the door Monty leaves closed. I heard it best described (without mathematics) like so: The key is to realize that Monty has certain information (i.e. the car's location) and that by purposefully avoiding one of the doors, he has given you new information. Say that you choose door A, and he reveals door B to be empty. Before he reveals the door, you know only one thing: 1) The car is behind one of the doors, A, B, or C. When he reveals door B, you have been given two new pieces of information: 2) B is empty 3) Monty *chose* to open B and to avoid opening C. Because of that last fact, you can make a better informed choice, which is to switch to C. Of course, it may be more convincing just to try the game at home or write out the probability tree. 

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Pat
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Oct0603, 03:22 PM (EST) 

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

This problem has been analyzed many times, and the result is this: the probabilities do change. The chance of initially picking the correct door with all doors closed is 1/n where n is the number of doors. The chance of picking the correct door by switching is (n1)/(n*(nk1)) where k is the number of doors opened by the host. (n1)/n is the complimentary chance of 1/n, or in words the chance the player picked incorrectly on the first pick. The two chances of picking correctly and incorrectly sum to unity. (nk1) is the number of doors remaining closed, except for the first one picked. This is also the degrees of freedom in this problem. To satisfy intuition, consider the case with one trillion doors. Pick one. Then Monty opens up all the doors except the one picked, and one other door. Are the odds 50/50? Of course not. It is a quick proof to show that the second fraction is larger than the first whenever k=>1. The solution is all over the web, I am surprised it wasn't posted first and finished this thread quickly. 

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webwizlv
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Sep0404, 07:31 PM (EST) 

34. "RE: Monty Hall Problem"
In response to message #0

I think the part that makes the right answer (i.e., always switching) counterintutitive is the fact that two of the possibilities are being collapsed into one. If you choose a goat initially, there is only one door the host is allowed to show you. (He'll never show you the car.) That creates two cases of pickshowswitch. Two wins. If you choose the car initially, he can choose either of the other doors to show you. People are considering that situation as one scenario, rather than two. We should consider pickshow the first goatswitch and pickshow the other goatswitch. Two losses. 

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Carl Wilson
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Sep1404, 07:17 PM (EST) 

35. "RE: Monty Hall Problem"
In response to message #34

This theory does not take into account that the one of the options is eliminated from the equation. In the simplified Quiz Show example Doors 1 2 3 Contestant chooses a door, host shows what’s behind one of the other doors, which instantly eliminates this door from the probability equation. I am not a mathematician but a wee bit of common sense is all that is required.
This theory cleary shows that even after one of the doors is opened, it is still part of the equation. If the contestant is silly enough to still choose this door then the theory is correct, else, well I think you know the answer


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Dave Dorn
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Sep2304, 06:37 AM (EST) 

37. "RE: Monty Hall Problem"
In response to message #35

Restate the problem thusly, and it all become clear. When you are presented with three doors, you choose one which you do not wish Monty to discard. It doesn't matter what's behind it. You're not trying to pick the prize door. Monty opens a door, which has no prize behind it. You are now presented with a simple choice  and here's what you know: One door has the car, one door doesn't. That statement is 100% absolutely certain. NOW, you make the choice  which is it? Door A or Door B? It's a 50:50 choice... 

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rewboss
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Sep2504, 00:22 AM (EST) 

38. "RE: Monty Hall Problem"
In response to message #37

Yeah, that's what I thought. I was wrong. If you think about it, whether or not it is good idea to switch depends on whether or not the contestant initially chose the door with the prize. If the contestant initially chose the door with the prize, then he must stick with that choice in order to win the prize. That is a nobrainer. If the contestant initially chose a door without the prize, then if he changes his selection, he is certain of winning the prize. Why? Because Monty has opened the only other door that does not conceal the prize. What is the probability that the initial selection was the correct selection? 1/3, yes? And the probability the initial selection was incorrect? 2/3, yes? If you work that through, you will come to the conclusion that the contestant does indeed double his chances of winning by changing his selection. That is backed up by possibly hundreds of experiments. 

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seesall
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Sep2604, 11:13 AM (EST) 

39. "RE: Monty Hall Problem"
In response to message #37

>Restate the problem thusly, and it all become clear. > >When you are presented with three doors, you choose one >which you do not wish Monty to discard. It doesn't matter >what's behind it. You're not trying to pick the prize door. > >Monty opens a door, which has no prize behind it. You are >now presented with a simple choice  and here's what you >know: > >One door has the car, one door doesn't. That statement is >100% absolutely certain. > >NOW, you make the choice  which is it? Door A or Door B? >It's a 50:50 choice... Repeat after me.
Dont' open the door then place the car. Place the car then open the door. Place, then open, don't open then place. If you first place the car in door number one you will see that only half that time is door two opened. When the car is placed in door3 all the time door 2 is then opened. Place then open. Place then open. Don't open and place


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Solomon English
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Oct0404, 11:59 AM (EST) 

40. "RE: Monty Hall Problem"
In response to message #39

1. Probabilities. The probability of something happening is dependent on constraints. Given a ‘free'  unconstrained  choice between two actions, there is a probability of one half of either one being chosen. These probabilities may change if some constraint is put on the choice. 2. In the Monty Hall problem, the contestant chooses on two occasions. In the first, unconstrained, he stands a one third probability of choosing the door hiding the car, two thirds of not doing so. In the second, unconstrained, he has a probability of one half of picking the car, one of the two other options having been eliminated. 3. If the contestant constrains his second choice, this last probability may change. If he decides always to stick to his first choice, the second ‘choice' is now forced. He has essentially eliminated the probability of one half of winning on the second choice; he wins only if his first choice was correct Thus the probability of winning overall remains one third. If he decides always to switch, then he has anulled the one third probability of being right first time. He wins if his first choice was wrong with a probability of two thirds, since Monty has eliminated the other possible wrong first choice. If his first choice was right  with a probability of one third  he has given it up, and loses. (In this case, it might appear that he can lose in two ways, depending on Monty's choice of door, but neither of these possible events alter the contestant's choices, and hence lie within the one in three chance of the his first choice being wrong.) 4. If the contestant decides before his first choice to switch at his second chance, he converts his two chances of choosing wrongly to winning choices, since Monty has to eliminate the other wrong choice. He does this at the expense of losing his one in three chances of having been right first time. 5. So decide to switch every time, and never change your mind at the second choice. If you do, half the time you'll be wrong. 

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seesall
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Oct0404, 03:32 PM (EST) 

41. "RE: Monty Hall Problem"
In response to message #40

>1. Probabilities. The probability of something happening is >dependent on constraints. Given a ‘free'  unconstrained  >choice between two actions, there is a probability of one >half of either one being chosen. These probabilities may >change if some constraint is put on the choice. >Yup, if Monty places the car AFTER the door is opened but not before. >2. In the Monty Hall problem, the contestant chooses on two >occasions. In the first, unconstrained, he stands a one >third probability of choosing the door hiding the car, two >thirds of not doing so. In the second, unconstrained, he has >a probability of one half of picking the car, one of the two >other options having been eliminated. No, only one half if the car is placed AFTER the door is opened. The car is distriuted BEFORE the door is opened. You are opening then placing. You cannot do that! Place the car THEN open the doors in ALL combinations. Place then open. Do not open then place! Do you understand?, you are driving me mad. >
>3. If the contestant constrains his second choice, this last >probability may change. If he decides always to stick to his >first choice, the second ‘choice' is now forced. No sir, the probabilty on the second door is 2/3 even if the choice is made after one door is eleminated. Run a simulation and see, but make sure to label the same doors with the same number each round. He has >essentially eliminated the probability of one half of >winning on the second choice; he wins only if his first >choice was correct Thus the probability of winning overall >remains one third. > If he decides always to switch, then he has anulled the >one third probability of being right first time. He wins if >his first choice was wrong with a probability of two thirds, >since Monty has eliminated the other possible wrong first >choice. If his first choice was right  with a probability >of one third  he has given it up, and loses. (In this case, >it might appear that he can lose in two ways, depending on >Monty's choice of door, but neither of these possible events >alter the contestant's choices, and hence lie within the one >in three chance of the his first choice being wrong.) > >4. If the contestant decides before his first choice to >switch at his second chance, he converts his two chances of >choosing wrongly to winning choices, since Monty has to >eliminate the other wrong choice. He does this at the >expense of losing his one in three chances of having been >right first time. > >5. So decide to switch every time, and never change your >mind at the second choice. If you do, half the time you'll >be wrong. NO, That is wrong, wrong, wrong. If the door would have been 2/3 by "keeping" your switch choice after the door was open, then not switching is 1/2? 1/2+2/3=1? You are crazy


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Beth
guest

Oct2504, 02:27 PM (EST) 

43. "RE: Monty Hall Problem"
In response to message #0

Reading all the responses to this problem I have finally figured out where I was going wrong. It always looked to me like the final decision was a 50/50. I just realized I was assuming that the information given by the host was neutral. (Not in a technical sense my math lingo is back to nonexistent by now.) The constraint on the host to open one of the doors which the contestant did not pick gives the advantage to switching. If the host could open any door the advantage would be lost. For example, doors A & C have goats, door B has a car. I pick door C. The host has to pick one door that has a goat. He picks door C. Now my chance of getting the prize is obviously 50/50. I must choose another door. When the host is constrained to pick a door that I haven't chosen he is giving me extra, though not definitive, information. The original way, I pick a door. My chance of being wrong is 2/3. The host removes a goat that I did not choose. Because the host could not choose my door, it'still has a 2/3 chance of being wrong. I should change doors. If the host could have chosen my door it would have lost that advantage and my chances go to 1/2. I hope this is clear enough to help anyone who still has trouble seeing what is happening. Beth 

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Seesall
guest

Oct2704, 07:19 AM (EST) 

46. "RE: Monty Hall Problem"
In response to message #45

>>Thanks for your imput. Still the best explanation that I >>read was Kevin Devins column, the link to which Alex posted >>earlier. > >The name's Keith Devlin > >>7 chance of winning. See Kevin Devin's column at >>https://www.maa.org/devin/devin_07_03.html > >The correct reference is > >https://www.maa.org/devlin/devlin_07_03.html Actually his explanation is wrong. He writ's: 1. The probability that the prize is behind door B or C (i.e., not behind door A) is 2/3. 2. The prize is not behind door C. Combining these two pieces of information yields the conclusion that the probability that the prize is behind door B is 2/3. Endquote In fact if the contestant chooses door A and Monty never opens C when the car is behind A the chances on B are 100%


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Seesall
guest

Oct2704, 12:48 PM (EST) 

48. "RE: Monty Hall Problem"
In response to message #47

>Seesall writ's: >>In fact if the contestant chooses door A and Monty never opens C when the car is behind A the chances on B are 100% > >That is correct but one third of the time, i.e. when the car >is behind door A, Monty also opens door C (i.e., one of the >wrong doors). Thus Keith Devlin's column is correct. >Have a Good Day >KJ Ramsey Read what I said again. I said that Monty (chooses) never to open Door C when the car is behind A "Monty never opens C when the car is behind A" you then said: "but one third of the time , when the car is behind A Monty also opens door C" Wrong, I JUST said Monty chooses NEVER to open door C when the car is behind A. He opens door B instead! Delvins column is wrong. In fact not enough information was provided to determine the odds on door B which could vary between 50%100%. All that can be said with certainty is that if the contestant always switches to both door B and Door C the overall odds are 2/3. That is a hellavu lot different than saying the odds on door B are 2/3


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Ramsey_KJ
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Oct2704, 06:49 AM (EST) 

49. "RE: Monty Hall Problem"
In response to message #48

Read Devlin's column again. A B and C are used by Devlin as follows. A is the door that the contestant chooses, C is the "wrong" door that Monty opens and B is the other door that "was not opened". You are arguing apples instead of oranges because using Devlin's lingo, door B is never the door opened by Monty. You must use Devlin's lingo if you are to argue an error in his logic. You did not. You are right only if you do not follow the lingo in Devlin's column, but that does'nt mean that Devlin's column was wrong. The probability that the car is behind door B (using Devlin's lingo) is truely 2 in 3 because it can't be behind door C which is a wrong door "The car is not behind door C". 

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Seesall
guest

Oct2804, 04:56 PM (EST) 

50. "RE: Monty Hall Problem"
In response to message #49

>Read Devlin's column again. A B and C are used by Devlin as >follows. A is the door that the contestant chooses, C is the >"wrong" door that Monty opens and B is the other door that >"was not opened". You are arguing apples instead of oranges >because using Devlin's lingo, door B is never the door >opened by Monty. You must use Devlin's lingo if you are to >argue an error in his logic. You did not. You are right only >if you do not follow the lingo in Devlin's column, but that >does'nt mean that Devlin's column was wrong. The probability >that the car is behind door B (using Devlin's lingo) is >truely 2 in 3 because it can't be behind door C which is a >wrong door "The car is not behind door C". Remember in the (2/3) cases when the car is behind door B or C the chance of winning is 100%. Your reasoning is that 2/3 of the time (when the car is behind B or C) the chances of winning is 2/3.
The error in your reasoning is to associate a probablity of losing with "those instances when the car is not behind A". Indeed 2/3 of the time the contestant will win 100% of the time. Half the time being the cases when Monty opens door C. Considering the 2/3 of the time the car is behind B or C, those cases in which door C is opened do nothing to alter the winning percentage. It's still 100%. To figure out the overaall winning percentage for Door C you have to figure out how many times door C lOSES. The losing cases come from when the car is behind door A. Here Monty has a choice, he can open door B or C. If he always opens door B he shifts all the losing cases to door B and door C always wins. half of that (2/3) or 1/2(2/3)=1/3 Monty opens door C and the car is behind B so we know that the contestant wins 1/3 of the time. All we know then is how many times the contestant wins. What fraction of the time Does


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Ramsey_KJ
Member since Sep2304

Oct2904, 10:27 AM (EST) 

51. "RE: Monty Hall Problem"
In response to message #50

>>Read Devlin's column again. A B and C are used by Devlin as >>follows. A is the door that the contestant chooses, C is the >>"wrong" door that Monty opens and B is the other door that >>"was not opened". You are arguing apples instead of oranges >>because using Devlin's lingo, door B is never the door >>opened by Monty. >Remember in the (2/3) cases when the car is behind door B or >C .... > The losing cases come from when the car is behind door A. Here Monty has a choice, he can open door B or C. If he always opens door B he shifts ... Once again you are not using the lettering system used by Devlin where the letters are not applied until after the choices are made. Monty does indeed have a choice between two doors but the labels "B" and "C" are not applied until after the contestant choses door "A" and Monty next opens a door. Regardless of which door Monty opens that door is labeled "C" and the remaining door is labeled "B". It is easy to see from this system of labels that the probability that the car is behind door "A" is indeed 1/3 since the contestant has no idea where the car is and that the probability that the car is behind door "C" is zero since Monty never next opens a door with a car behind it. It follows that the probability that the car is behind door "B" is equal to 1  1/3  0 = 2/3. This agrees with the post by Tattertale where it was explained that the chances of winning are doubled if the contestant switches. However, the contestant would never switch to door "C" because that is the door that Monty opened. Have a Good Day KJ Ramsey 

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Seesall
guest

Oct3004, 06:02 PM (EST) 

53. "RE: Monty Hall Problem"
In response to message #51

>>C >.... >> The losing cases come from when the car is behind door A. Here Monty has a choice, he can open door B or C. If he always opens door B he shifts ... > >Once again you are not using the lettering system used by >Devlin where the letters are not applied until after the >choices are made. That is not true here is what Delvin said: "Suppose the doors are labeled A, B, and C. Let's assume the contestant initially picks door A. The probability that the prize is behind door A is 1/3. That means that the probability it is behind one of the other two doors (B or C) is 2/3. Monty now opens one of the doors B and C to reveal that there is no prize there. Let's suppose he opens door C." Delvin did not say that Monty picks one of the two remaining doors and then labels that door C. >Monty does indeed have a choice between >two doors but the labels "B" and "C" are not applied until >after the contestant choses door "A" and Monty next opens a >door. Regardless of which door Monty opens that door is >labeled "C" and the remaining door is labeled "B".
I agree that such a labeling system changes the problem but that is not the spirit of the orginial question in which three doors each have distinct identities given by A B and C or if you prefer 1,2,3. People can judge for themselves what they think Delvin really meant. 

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Ramsey_KJ
Member since Sep2304

Oct3104, 11:51 AM (EST) 

54. "RE: Monty Hall Problem"
In response to message #53

>>Once again you are not using the lettering system used by >>Devlin where the letters are not applied until after the >>choices are made. > >That is not true here is what Delvin said: > >"Suppose the doors are labeled A, B, and C. Let's assume the >contestant initially picks door A. The probability that the >prize is behind door A is 1/3. That means that the >probability it is behind one of the other two doors (B or C) >is 2/3. Monty now opens one of the doors B and C to reveal >that there is no prize there. Let's suppose he opens door >C." > >Delvin did not say that Monty picks one of the two remaining >doors >and then labels that door C. > > >>Monty does indeed have a choice between >>two doors but the labels "B" and "C" are not applied until >>after the contestant choses door "A" and Monty next opens a >>door. Regardless of which door Monty opens that door is >>labeled "C" and the remaining door is labeled "B". > >I agree that such a labeling system changes the problem >but that is not the spirit of the orginial question in which >three doors each have distinct identities given by A B and C >or if you prefer 1,2,3. People can judge for themselves what >they think Delvin really meant. > > The paradox is that with either labeling system the same fact situation arises and the problem is the same, what is the probability of "B" at that instant. Using "my" labeling system one can easily see that by switching, contestants double their chances of winning. The only way to resolve the paradox is to convince yourself that at that point the probability of "A" is indeed only 1/3 since the contestant had no idea where the car is out of three choices, and the probability of "C" is indeed 0 since in fact there is no car behind it, therefore regardless of your analysis, the probability of "B" remains 1  1/3  0 = 2/3. Have a Good Day KJ Ramsey 

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Seesall
guest

Oct3104, 03:46 PM (EST) 

55. "RE: Monty Hall Problem"
In response to message #54

>>>Once again you are not using the lettering system used by >>>Devlin where the letters are not applied until after the >>>choices are made. >> >>That is not true here is what Delvin said: >> >>"Suppose the doors are labeled A, B, and C. Let's assume the >>contestant initially picks door A. The probability that the >>prize is behind door A is 1/3. That means that the >>probability it is behind one of the other two doors (B or C) >>is 2/3. Monty now opens one of the doors B and C to reveal >>that there is no prize there. Let's suppose he opens door >>C." >> >>Delvin did not say that Monty picks one of the two remaining >>doors >>and then labels that door C. >> >> >>>Monty does indeed have a choice between >>>two doors but the labels "B" and "C" are not applied until >>>after the contestant choses door "A" and Monty next opens a >>>door. Regardless of which door Monty opens that door is >>>labeled "C" and the remaining door is labeled "B". >> >>I agree that such a labeling system changes the problem >>but that is not the spirit of the orginial question in which >>three doors each have distinct identities given by A B and C >>or if you prefer 1,2,3. People can judge for themselves what >>they think Delvin really meant. >> >> >The paradox is that with either labeling system the same >fact situation arises and the problem is the same, what is >the probability of "B" at that instant. Using "my" labeling >system one can easily see that by switching, contestants >double their chances of winning. The two labeling systems are completely different. Your system corresponnds to always switch for both doors which averages out to 2/3. You can't relabel door C door B and say its the same! Suppose the contestant chooses A and Monty never chooses C with a car behind A under "my" system. What are the odds on B and C? Door C can never lose....think about it... 

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Ramsey_KJ
Member since Sep2304

Oct3104, 08:41 AM (EST) 

56. "RE: Monty Hall Problem"
In response to message #55

>>>>Once again you are not using the lettering system used by >>>>Devlin where the letters are not applied until after the >>>>choices are made. >>> >>>That is not true here is what Delvin said: >>> >>>"Suppose the doors are labeled A, B, and C. Let's assume the >>>contestant initially picks door A. The probability that the >>>prize is behind door A is 1/3. That means that the >>>probability it is behind one of the other two doors (B or C) >>>is 2/3. Monty now opens one of the doors B and C to reveal >>>that there is no prize there. Let's suppose he opens door >>>C." >>> >>>Delvin did not say that Monty picks one of the two remaining >>>doors >>>and then labels that door C. >>> >>> >>>>Monty does indeed have a choice between >>>>two doors but the labels "B" and "C" are not applied until >>>>after the contestant choses door "A" and Monty next opens a >>>>door. Regardless of which door Monty opens that door is >>>>labeled "C" and the remaining door is labeled "B". >>> >>>I agree that such a labeling system changes the problem >>>but that is not the spirit of the orginial question in which >>>three doors each have distinct identities given by A B and C >>>or if you prefer 1,2,3. People can judge for themselves what >>>they think Delvin really meant. >>> >>> >>The paradox is that with either labeling system the same >>fact situation arises and the problem is the same, what is >>the probability of "B" at that instant. Using "my" labeling >>system one can easily see that by switching, contestants >>double their chances of winning. > >The two labeling systems are completely different. >Your system corresponnds to always switch for both doors >which averages out to 2/3. You can't relabel door C door >B and say its the same! > The fact situation is the same. The contestant chooses A and Monty opens door C with no car behind it, the probability of the car being behind B is then 1  1/3  0 = 2/3. Note that B can also win if the contestant chooses door C and switches to B but that is not the fact situation. In the one third time that the car is behind A, neither B or C can win regardless of whether the contestant switches to them. >Suppose the contestant chooses A and Monty never chooses >C with a car behind A under "my" system. What are the odds >on B and C? Door C can never lose....think about it... The question is not whether a contestant wins or looses with a particular door but what are the probabilities of the car being behind B given the fact situation. Having Monty open door B is not the fact situation at issue. Have a Good Day KJ Ramsey


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Seesall
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Nov0104, 10:51 PM (EST) 

57. "RE: Monty Hall Problem"
In response to message #56

. In the one third >time that the car is behind A, neither B or C can win >regardless of whether the contestant switches to them. But one can lose more than the other depending on which door Monty picks, suppose he always opens B, switching to B never loses.. >>Suppose the contestant chooses A and Monty never chooses >>C with a car behind A under "my" system. What are the odds >>on B and C? Door C can never lose....think about it... > >The question is not whether a contestant wins or looses with >a particular door but what are the probabilities of the car >being behind B given the fact situation. Having Monty open >door B is not the fact situation at issue. >
Consider the following 6 cases, where Monty never opens Door C with the car behind A. Initially X00 X00 0X0 0X0 00X 00X Now after Monty opens his door given by an M XM0 XM0 0XM 0XM 0MX 0MX Notice that in 4 of the cases door B is opened with a win rate of 50% for switch. XM0 XM0 0MX 0MX In only two of the cases was door C opened with a win rate of 100% 0XM OXM However the average win rate is still 2/3 because (4/6)(50%) + 2/6(100%) = 1/3 + 1/3 = 2/3
So be very careful when talking about the win rate for a particular door without knowing Monty's actions in the car behind door A case. 

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Paul Jennison
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Oct2904, 10:27 AM (EST) 

52. "RE: Monty Hall Problem"
In response to message #50

VERY INTERESTING! From what I see everyone is right  it is both 2/3 and 1/2. It depends from when you are measuring the probability of an event. I seem to remember from my university days that in probability we have the phrase 'given that....' and that there is a difference between measuring the probability of an event from end to end (i.e. from the whole of the group of events) and the final or single event (e.g after all but one of the events have occured). Therefore, we need to decide whether we are measuring the probability from the BEGINNING of the whole chain of events or whether we measure the probability of the final event. So, we have: 'given that someone has chosen a door and the host has shown a goat etc etc' and we are then left with two coices  one is a goat and one is the prize. AT THAT POINT there is a 50/50 chance  whatever has gone before is irrelevent. However, at the begininng of the whole chain of events  i.e. before we have made our first choice  there is a 2/3 choice of winning if we swap as per the example. Both are right....... but it depends whether you measure from the the beginning of the whole game or just the final choice.. Any comments? 

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Abhishek Sharma
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Feb2205, 01:17 PM (EST) 

62. "RE: Monty Hall Problem"
In response to message #52

Correct!!! The answers 2/3 and 1/2 are both correct, depending on when was the event measured from. Further the one thing that is to be kept in mind is there should be no distinction between the goats( both are losses) to win the game if we start the event count after one of the goats is found then the probabilty is 1/2 cause the car is definitly behind any of the two unopened doors.. In case if we start to measure the event form the beginning then : 1.when all doors are closed: probabilty is 1/3 2.when one of the doors are opened the probabilty is 11/3=2/3 where 1 is the event of one door opened and a goat is found behind.


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haamu
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Jan1505, 01:26 AM (EST) 

60. "RE: Monty Hall Problem"
In response to message #0

You've all  all of you  gone a bit astray by ignoring one critical fact. Recent surveys indicate that 2/7 of the Earth's inhabitants would prefer the goat. 

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HELEN TSAI
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Oct0505, 04:25 PM (EST) 

64. "RE: Monty Hall Problem"
In response to message #0

My approach to Monty Hall problem using Baye's Theorem (CONDITIONAL PROBABILITY): P(AB) = P(BA)P(A)/(P(BA)P(A)+P(BA')P(A')) LET A=P(CAR), B=SWITCH (AFTER 1ST GOAT IS SHOWN) GIVEN THAT IT IS CAR/GOAT FROM EACH OF THE DOORS, SAY THE DOOR WITH A GOAT BEHIND IT (G1): P(CARSWITCH) = P(SWITCHCAR)P(CAR)/(P(SWITCHCAR)P(CAR)+P(SWITCHGOAT)P(GOAT) P(SWITCHGOAT) = 1/2 (50% CHANCE THAT ONE OF THE DOOR PICKED HAS A GOAT BEHINDAFTER HOST SHOWS THE 1ST GOAT) P(SWITCHCAR) = 1/2 (OR THERE'S 50% CHANCE THAT ONE OF THE DOOR PICKED HAS A CAR BEHIND IT AFTER THE 1ST GOAT IS SHOWN) P(CAR) = 1/3 (PROB OF GETTING A DOOR WITH A CAR FROM THE BEGINNING) AGAIN, WANT P(CARSWITCH) = (1/2 *1/3)/((1/2*1/3)+(1/2*2/3) = 1/3 (THE PROB FROM DOOR WITH G1 OF GETTING A CAR GIVEN THAT SWITCHING IS DONE) FROM DOOR WITH G2, SIMILARLY, P(CARSWITCH) = 1/3 FROM DOOR WITH C, P(CARSWITCH) = 0. PROB OF GETTING A CAR GIVEN THAT SWITCHING IS DONE IS NEVER, SINCE SWITCHING FROM DOOR WITH CAR ALWAYS YIELDS A GOAT. P(GOAT) = 2/3 (COMPLEMENT OF P(CAR)) P(CARSWITCH) FROM DOOR 1 U P(CARSWITCH) FROM DOOR 2 U P(CARSWITCH) FROM DOOR 3 = 1/3 + 1/3 + 0 = 2/3
THANKS. HELEN TSAI


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Andrew
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Oct1805, 09:15 PM (EST) 

65. "RE: Monty Hall Problem"
In response to message #64

This answer is 1/2, not 2/3, and points out a very obvious flaw in probabilty: that is prespective in interpretting what the event is makes a difference. Consider the following. If you have two doors, one with a car, another with the goat. You're asked to find the car. The probablilty is 50/50 car goat > choose.. p = 50/50
(fine) If you choose one and are given a chance to choose again, the probability is still 1/2
car goat > choose one, think it over, then choose again. p = 1/2
(so far so good?) Now let's say the peron is offered a third incorrect choice which he/she knows is incorrect. Since there is no way he/she will choose it, it will not affect the outcome so the probability is still 1/2
car goat knowngoat > choose one knowing that a door you will never pick is the wrong one, think it over, choose again.. p = 1/2
If you look really carefully, this is the Monty Hall problem.
Once Monty opens the one door with a goat, you're presented with a brand new problem which in no way depends on the previous choice. Like economic sunk costs, the previous decisions were become irrelevant. To visualize this, imagine the person making the "switch, no switch choice" is different. This new person is told the following: one of these doors has a goat, the other one a car. A third door has a goat but really, you don't care about that. One last thing, a guy/girl before you made a guess which you don't know whether it is right or wrong. He/She picked one of the two doors you're given. His/Her guess was harder because I had more doors then. Anyhow he/she guessed door 'x' but I guess you probably don't care since you still don't know if it was right. Now please choose a door." Basically you can see thatn it the Monty Hall probelem isn't one of probabilistic saviness, rather simply one packed with irrelevant information, enough to fool the most ambitious mathematicians. When Monty opens a door revealing one which is wrong. You pat yourself on the back for not having guessed that particular goat but that still does not change the nature of your new question. Now Monty says "there is a goat and a car behind on the two remaining doors. Which one do you think has the car, the one you chose originally(door A) or the other one(door B)?" car goat > choice.. p = 1/2
It doesn't take a lot of intelligence to see through this, it only takes the ability to see beyond your own arrogance.
The root of this problem is that too often, logicians and mathematicians treat randomsness as if it were logical: always following the rules and axioms it is supposed to. In the math world, that's fine. The real world however, where most probability is applied, it's not so simple. Weather predictions are wrong; polls don't tell you who will win the election; having a low city crime rate will not mean you can never get mugged. If people were able to recognize this simple fact, perhaps less time would be wasted fiddling around with numbers and losing money in casinos.


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alexb
Charter Member
2250 posts 
Oct1805, 09:25 PM (EST) 

66. "RE: Monty Hall Problem"
In response to message #65

I just came up with what I believe in my arrogance to be the simplest and hence the best possible explanation to Monty Hall's Dilemma. It's at https://www.cuttheknot.org/Curriculum/Probability/MontyHall.shtml.Who needs Monty?Indeed, assume you (the contestant) are given a chance to point to a door and, after a while, offered to either open this door or the other two, i.e., both of them. Every one knows up front that behind one of these doors there's a goat. Of course, there may be a goat behind the other door as well. You receive the prize provided you open a door behind which it is. What should you do? We may go further, and do away with the need to make a decision: to switch or not to switch. Point to a door, say magic words, and let all three doors open at once. Your only function is to keep scores: would you win if you switched, would you win if you did not. Monty is around there available to do you a favor and open a door with a goat behind, saving you a physical effort. Do you care for his help? 

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Michael Frame
guest

Oct2505, 09:46 AM (EST) 

67. "RE: Monty Hall Problem"
In response to message #66

I may have an even simpler explanation than your "Who needs Monty". It involves no simulation, just a different perspective. I believe a professional gambler would have had an easier time with this. Let's hypothesize the gambler will win something if he guesses correctly, but will loose something if he is wrong. Thus, he is more highly motivated than, for example, I. Consider the gambler's reasoning. "I'm being forced to pick one of three doors and only one is the right door. So, I'm probably picking the wrong door. Ah, now Monty is giving me a chance to change my pick and he's shown me that one of the other doors is not the one I want. Since I was thinking that one of the other two doors was likely the correct one, and Monty has eliminated one of them, I'll choose the other. For my money, that's almost like certainty." I find many of the other ideas for reasoning to the correct result quite enlightening, but I wonder how much effort went into them. I know my first response upon reading about this problem was that vos Savant was wrong. I had to think about the problem quite a bit to admit'she was right. It took even longer to come up with the "gambler presentation". Perhaps Ms vos Savant and some others immediately saw the correct answer. They have my admiration. However, it seems unfair for some to berate latecomers to this problem for not immediately seeing the correct answer, as well as the "proper way" of getting to that answer. When I originally thought about the controversy relating to this problem, I began to wonder how many times someone may have given a proof that seemed obviously correct only to find later that it was totally wrong. My original "proof" was like many others  we know one of the other two doors is a wrong guess. When Monty exposes it, we gain no new information. Therefore, there is no reason to change our guess. Equally, there is no reason not to. The flaw, of course, is the assumption we gain no new information. Once we know the correct answer(!), we can easily state and solve the problem using conditional probabilities. Why not use probabability theory in the first place? Because, it was clear we already knew the "correct" answer. Why go through all that formality? Now we know. I enjoy your web site. Mike


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alexb
Charter Member
2250 posts 
Oct2505, 08:04 PM (EST) 

68. "RE: Monty Hall Problem"
In response to message #67

>I may have an even simpler explanation than your "Who needs >Monty". It involves no simulation, just a different >perspective. I am agreeable to sharing the glory. >I believe a professional gambler would have had an easier >time with this... Looks very plausible. To me, I mean. >However, it'seems >unfair for some to berate latecomers to this problem for not >immediately seeing the correct answer, as well as the >"proper way" of getting to that answer. I agree, except that in this time and day, I would appreciate the latecomer who did his homework. Such an effort would give an indication of an honest interest to resolve the problem. So much information is now available that it is hard to justify missing or overlooking much of it. >I enjoy your web site. Thank you for the kind words. 

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Marcus Bizony
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Dec1407, 07:38 AM (EST) 

69. "RE: Monty Hall Problem"
In response to message #68

or consider this approach: when you nominate a door you split the objects into His and Mine. He has two and you have one, so he probably has the car ('probably' here meaning 'more likely than not'). Now he shows you a goat  you always knew that he had at least one, so you have learned nothing and therefore still think he probably has the car. The only difference now is that you know WHERE the car is if he has it. So you argue: he probably has the car, and he certainly would put it THERE..... so it probably is THERE. Clearly you must change your choice. 

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scottiemac
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Jul2408, 06:07 PM (EST) 

72. "RE: Monty Hall Problem"
In response to message #69

>or consider this approach: when you nominate a door you >split the objects into His and Mine. He has two and you have >one, so he probably has the car ('probably' here meaning >'more likely than not'). Now he shows you a goat  you >always knew that he had at least one, so you have learned >nothing and therefore still think he probably has the car. >The only difference now is that you know WHERE the car is if >he has it. So you argue: he probably has the car, and he >certainly would put it THERE..... so it probably is THERE. >Clearly you must change your choice. That's assuming that you KNOW you chose incorrectly from the beginning. The initial pick, and the choice to stay or switch, are independent events. The first "pick" is a throwaway, unless sometimes he does not offer you the ability to switch. Since that's not the rules of the game, we don't have to worry about that. Think through it like this. Regular "Let's Make a Deal" (LMAD): three doors, one car, two goats. You make a pick, then Monty throws out a bad door showing you it's a loser. You then determine if you'll switch or stay. Now let's test out two more versions of the game LMAD version 2.0: three doors, one car, two goats. Monty throws out a bad door, and then you select what door you want. You can switch or stay, shouldn't matter either way. LMAD version 2.1: only two doors, one car, one goat. Pick a door. Those of you that argue that switching wins 2/3 of the time will still say that regular LMAD has a 66% win rate when switching. You MUST agree that LMAD version 2.0 has a 50% win rate. Same goes for version 2.1 50% win rate. Guess what: all three are the same game. You're just being confused with the extra information in the original game. The game becomes 50/50 when you only have two choices. Don't mix that up with the 1/3 probability of initial success. Let's analyze this one: LMAD version 2.2. One car, two goats. You are ASSIGNED a door (let's say door A) and told that there is a goat behind it. You can now keep the goat, or switch to one of the other doors. Well, anyone with an IQ over 85 will switch, but now the dilemma is door B or door C. Those are 50% odds, my friend, and this is the same game as the original. Here's the probability if it helps: Start of game Given: doors A, B, and C; prize is behind one door P(A) = 1/3 P(B) = 1/3 P(C) = 1/3 P(A) + P(B) + P(C) = 1 Assume one door is always the loser, which it is, we just don't know which one. For sake of example, we'll say it's C. Nobody ever chooses door C in our game, and Monty always opens door C. It's a waste of time to even put three doors out there (much as is a person making an initial pick, if you'll see it). Once we get to phase two of our game, here are the new probabilities, and the only ones that matter. P(A) = 1/2 P(B) = 1/2 P(C) = 0 P(A) + P(B) + P(C) = 1 The fallacy that the 66% crowd believes is that there is not an equal chance of the car being behind doors A and B. They try to draw it up like this... initial choice is A, P(A) = 1/3 *** THAT'S RIGHT, if there's not a second phase to the game *** but they leave P(A) at 1/3, and since P(C) is 0, then P(B) must be 2/3, and that's why it pays to switch. In reality, we are guaranteed that P(A) = 1/2 and P(B) = 1/2, which means this is a 50/50 game. 

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52Card Monty
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Aug0508, 12:48 PM (EST) 

75. "RE: Monty Hall Problem"
In response to message #0

If anyone has still failed to grasp the fact that switching is the better strategy (2/3 chance of winning) then consider this: The host Monty has a standard deck of 52 playing cards which are randomly arranged facedown on a table. You win a fabulous prize if you can select the Ace of Spades from among the cards. If you pick any other card, you get nothing. Now, Monty KNOWS WHERE THE ACE OF SPADES IS before he presents you with your choice. You can choose any of the 52 cards. You make a choice, and the chances you picked the Ace of Spades are 1 in 52  very slim. After you've made your choice Monty will do one of two things: In the 1 of 52 chances you were right, Monty will randomly turn over 50 of the losing cards and ask if you want to switch. MOST OFTEN, however, in 51 OF THE 52 CHANCES WHERE YOU WERE INITIALLY WRONG IN YOUR CHOICE, Monty will not simply turn over 50 random cards, but HE WILL TURN OVER ALL BUT THE WINNING CARD. THUS, in 51 of the 52 possible initial choices you make, an ALWAYS SWITCH STRATEGY will lead to WINNING THE PRIZE, while in only 1 of the 52 possible first choices will an ALWAYS SWITCH STRATEGY lead to a loss. CONVERSELY, an ALWAYS STAY (with your first choice) STRATEGY will lead to defeat 51 out of 52 times and will lead to a win only once out of 52 times. Which strategy do you choose? As you can see, most of the time the winning card will be the one card Monty chose to leave facedown and not the card you initially chose. The same logic applies to the three doors. 

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