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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. 13 52.00%
Switch because this increases your probability of winning the big prize. 12 48.00%
Voters: 25. You may not vote on this poll

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Old 20-October-2003, 10:53 PM
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Default Let's Make A Deal

Monty Hall tells you that you can choose one of three closed doors. Behind one of the doors is a fantastic prize, like $64,000 or something. Behind the other two doors is... nothing. So you have a one-out-of-three chance of winning $64,000. Not bad!

So you choose one of the three doors. Before opening the door you chose, Monty (who knows where the prize is) opens one of the remaining doors to reveal that it has nothing behind it. At this point he offers you the opportunity to switch your choice or to keep your original choice. Some sing-song theme music starts playing. The audience starts screaming.... [If you've encountered this problem before, please wait a day or two before commenting....]
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Old 20-October-2003, 11:00 PM
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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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Old 21-October-2003, 01:34 AM
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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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Old 21-October-2003, 01:41 AM
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There's a very interesting diagram that shows exactly how this works, but I'll leave it those who haven't seen it to either come up with it themselves or look it up.
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Old 21-October-2003, 01:47 AM
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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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Old 21-October-2003, 02:52 AM
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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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Old 21-October-2003, 03:12 AM
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According to Marylin vos Savant in Parade magazine.....




Oh, who cares? :wink:



Switch.
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Old 21-October-2003, 11:55 AM
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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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Old 21-October-2003, 11:58 AM
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I already knew that one. Calculus, college year two.

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Old 22-October-2003, 05:43 PM
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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:
Each of the above three options has a 1/3 probability of occurring, because the contestant is equally likely to begin by choosing any one of the three doors. In two of the above options, the contestant wins the car if she switches doors; in only one of the options does she win if she does not switch doors.
As has been pointed out in previous posts, there are lots and lots of different ways to explain this solution - some are better than others. There are diagrams. There's even an applet out there that simulates the situation so you can try it yourself and tally up the results. (That one's listed on the Dr. Math page.)

But everyone agrees - you've got better odds if you switch.

[Final poll tally at this point:
  • 64% (11 votes) say keep first choice - makes no difference.
  • 35% (6 votes) say switch - betters your odds.
What are the odds of that? ]
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Old 22-October-2003, 07:31 PM
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Default Re: Let's Make A Deal

Quote:
Originally Posted by Cougar
So you choose one of the three doors. Before opening the door you chose, Monty (who knows where the prize is) opens one of the remaining doors to reveal that it has nothing behind it.
This bolded statement is the key. Monty's opening of a door is not random. There's always at least one un-chosen door with nothing behind it, and he'll open it.

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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Old 23-October-2003, 12:35 AM
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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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Old 23-October-2003, 01:22 AM
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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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Old 24-October-2003, 08:55 PM
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Robin, Chuck, excellent points. I guess the solution is not so unbelievable and incomprehensible after all....
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Old 24-October-2003, 09:05 PM
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Just one more example of "common sense" giving you the wrong answer! =D>
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Old 24-October-2003, 09:09 PM
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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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Old 24-October-2003, 09:49 PM
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Quote:
Originally Posted by SeanF
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 . . .
Actually, your odds jump up to 75% if you switch.

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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Old 24-October-2003, 09:58 PM
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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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Old 24-October-2003, 10:17 PM
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Quote:
Originally Posted by Eroica
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).
Okay, here's your problem. If A is the winning door, you don't have 12 possibilities. BAB, BAC, CAC, and CAB all need to be removed as possibilities because Monty won't open the winning door.

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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Old 24-October-2003, 10:24 PM