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| View Poll Results: What should you do? Does it matter at all whether you switch your choice or keep your original pick? | |||
| Keep your original pick because there's no advantage in switching. |
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13 | 52.00% |
| Switch because this increases your probability of winning the big prize. |
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12 | 48.00% |
| Voters: 25. You may not vote on this poll | |||
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When he offers you the chance to change, there is essentially a 50-50 chance that your original choice was right. Therefore there is no point in changing. If you do and your first choice had the prize, you'll kick yourself even more.
However, I don't say that with excessive confidence. I may be wrong. This guy sounds as bad as Chris Tarrant.
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Freedom For Fission A breath of fresh Iodine-131 |
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Oops, I didn't read all of the first post before posting myself. How do you get rid of an unwanted post?
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Life is like a box of chocolates. All of your choices are bad for you. |
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You don't really need a diagram to show how it works, you can give a basic description with some quick math and an explanation.
This of course shows that the answer is [deleted by Rantz]
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People who think they know everything are a great annoyance to those of us who do. |
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I said keep orginal choice because i believe in just going with ur "gut". besides if i changed to the wrong one i would feel really really stupid!
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"Astronomy is not merely a hobby or a study, but a way of truly seeing and understanding the vast and magificent universe in which, by some extraordinary event, we exist." |
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Let's Make a Deal is a rigged game. The producers decide whether you win or not. No matter which door you choose originally and no matter whether you change your choice later, whether you win or lose has already been decided. The area behind the doors is open and the prizes are on wheels. If you choose the right door and they don't want you to win, it is simple to move the prizes around while Monty is talking. Conversely, if they want you to win, they can move the prizes around to make sure that you do.
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Any day you wake up on "the right side of the dirt" is a good day. T. Anderson |
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Well, OK, time's up.
It's rather remarkable, but Google says there are roughly 4,500 websites that mention the Monty Hall Problem. That's likely because most people, including some mathematicians, cannot imagine that it would make any difference if one keeps the original pick or if one switches and takes the other remaining door. But in fact it does; it makes a big difference. You should always switch; this massively increases your odds of winning. (It does not guarantee your chances of winning, but hey, life can be cruel.) So Ask Dr. Math is one of the 4,500 sites talking about this problem. As is wise with many a problem in probability, the math doctor considers the sample space, that is, the set of all possible outcomes. This boils down to three options. After a quick tally, the good doctor points out.... Quote:
But everyone agrees - you've got better odds if you switch. [Final poll tally at this point:
]
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Everyone is entitled to his own opinion, but not his own facts. |
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Quote:
If Monty's opening of the door were random, then there'd be no point in switching - there's a 33% chance the chosen door is the winner, a 33% chance the unchosen and unopened door is the winner, and a 33% chance the door Monty opens is the winner. With a 0% chance of the last possibility, that leaves the others at 33% and 67%. When I originally saw this question, the writer did not make it clear that Monty was intentionally opening a losing door, and I had an e-mail argume..., er, discussion with him. I don't think he really understood why it was important, which meant (to me) that he didn't really understand the math, either. ![]()
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SeanF "Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher The contents of this post are ©2008 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF |
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Here's an analogy I like:
Suppose that, instead of 3 doors, there are 100. You pick, say, door number 27. Monty (who knows where the prize is) opens 98 doors, carefully avoiding door 27 and - for some reason - door 54. All the doors he opens are empty. Do you really think there's just a 50% chance the prize is behind door 54? Yes, this really is the same situation, just with the numbers scaled from 3 to 100. In the case I described, switching increases your chances from 1% to 99%. In the original problem, it increases you chances from 33.3% to 66.7%. |
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If you decide in advance that you're always going to switch then the only way you can lose is if the prize is behind the first door you pick since you're going to switch away from it later and you're not going to switch to the door that Monty opens. There's a 1/3 chance that the prize is behind the first door you pick since you're choosing randomly, so you have a 1/3 chance of losing. If you have a 1/3 chance of losing then you must have a 2/3 chance of winning.
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Life is like a box of chocolates. All of your choices are bad for you. |
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On the other hand . . . consider "Who Wants to Be A Millionaire". You've gotten past the $64000 question, so the $125000 is one you won't lose any money on if you miss.
The question is one you have absolutely no idea on. Knowing you've got nothing to lose, you say, "I'm just going to randomly guess 'A'." What are your odds of being right? 25%. Before accepting that as your "final answer," Regis reminds you that you still have your 50:50 lifeline, and you decide to use it. Two wrong answers are removed - B and C, leaving your original choice of A and D. Now, what are the odds that A is the right answer? Is it still 25%, or is it 50%? Are you better off changing to D for your final answer? ![]() The fact that the computer is randomly removing wrong answers (it could remove the one you've identified as your guess) while Monty is not randomly opening a door (he will not open the door with the prize behind it) makes all the difference in the world . . .
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SeanF "Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher The contents of this post are ©2008 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF |
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Quote:
Here's a simple way to look at it. You have three choices A, B and C, and you have chosen A The host is going to remove either B or C now. What is really happening is that by switching your choice, you're actually picking more than one choice at a time. Example. B is removed. If you stick to A, you have a 33% chance of winning. If you switch to C, in effect you have chosen B or C as your answer and thus you have a 66% of winning. With the 50/50 Lifeline, you have A, B, C and D. You choose D, then B and C are removed. Sticking with D means that you still have a 25% chance of being correct. However, switching to A means that in effect you are choosing A, B and C at the same time and thus a 75% chance of winning. So the equation works out to (X - 1) / X chance of winning if you switch choices.
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People who think they know everything are a great annoyance to those of us who do. |
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This problem is starting to do my head in. Okay, I know, the boffins have all done their sums, remembered to carry the one, etc, and it all adds up to the same result: suddenly you have a 67% chance of winning if you switch and only a 33% chance of winning if you stick. But I'm not buying it.
Look at it this way: at the start you have a 1/3 chance of picking the right door. Then Monty throws one door away, and asks you to pick again. Surely this is the same situation as if there never was a second empty door. By removing one and giving you a second chance, Monty has reduced your odds from 1/3 to 1/2. Switching doesn't alter those odds one iota. Still not convinced that Dr Math has got it wrong? Well, let's look at the sample space. Let's call the doors A, B and C. In each scenario there are three events: the contestant picks a door, Monty picks a door, and the contestant once again picks a door. Here are the possible outcomes: A B A A B C A C A A C B B A B B A C B C B B C A C A C C A B C B C C B A In the first, the contestant picks door A, Monty opens door B, and the contestant sticks with door A. And the rest should be obvious. Now do the math. If the winning door is A, the contestant has 4/12 chances of winning. In 2 of those he switches (BCA and CBA), in 2 he sticks (ABA and ACA). Ditto if B is the winning door. Ditto if C is the winning door. Ergo: it makes no difference whether or not you switch. I should point out, however, that I have a history of making stupid mistakes in cases like these. If that is the case here, I would be grateful if someone would point it out, please.
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However, the remaining 8 possibilities are not equally likely. If you choose A, he could open B or C. However, if you choose B, he can only open C. Therefore BCA and BCB combined are as likely to occur as ABA, ABC, ACA, and ACB combined. Refigure your possibilities like that.
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SeanF "Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher The contents of this post are ©2008 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF |