I recently read "The Curious Incident of the Dog in the Night-time" by Mark Haddon. I rather enjoyed this interesting novel (a sort of detective story written from the POV of someone with Asperger's Syndrome, which is a form of mild autism).
In the book the "hero" propounds the Monty Hall Problem, most adequately explained here http://www.faisal.com/docs/monty.html (and elsewhere all over Usenet and the Internet).
The problem is that I don't believe the logical argument. I have been to dozens of sites, learned and stupid, some even pornographical, but I simply don't buy the statistical theory.
The problem I have with it is this. Once one of the doors has been eliminated, the choice is functionally identical to the choice one would have faced were there only two doors and one prize to start with. I simply don't accept that removing one of the dummy doors is anything other than a red herring.
The logical proof reminds me Zeno's Paradox wherein Achilles couldn't overtake a tortoise.
You may very well ask why I am posting this here. Well, the mathematicians can't convince me, but their primary language is Boolean, whereas mine is English. I am hoping that someone here more polyglottal than I can translate successfully.
Edward

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I recently read "The Curious Incident of the Dog in the Night-time" by Mark Haddon. I rather enjoyed this interesting ... language is Boolean, whereas mine is English. I am hoping that someone here more polyglottal than I can translate successfully.

You do see that when you first picked the one door out of three, you were probably wrong, don't you? That's important. 1 of 3 is less than half. One of the other two doors probably will win, even if we don't know which, at the beginning.
The host kindly shows you which of the other doors is definitely* a loser, because he knows which of the two it is. Do you see that that makes the remaining one *probably the winner? That's the trickiest part. Why is that so? Because as I already said at the start, your own choice probably isn't (only 1 of 3), and, as you said, his showing the losing gate doesn't really change anything including the fact that you started off with only a one-in-three chance. The winner was likely to be one of the other two, and now we know which of the other two. Now that is useful info.
The other door is not a guaranteed winner, of course, but it's more likely than the one-out-of-three odds you started with. That's what the chart at the bottom of the page you gave says - out of nine possible outcomes, six of them have the "other" door being the winner.

It will never* be the one Monty Hall shows you. That's what skews it. It would be quite a different game if he had to *randomly show any of the three gates! Or even one of the two that you didn't name.

(It took me a long time, quite some time ago, to understand this problem, so don't feel bad. If my explanation doesn't work for you, maybe someone else's will.)

Best Donna Richoux
I recently read "The Curious Incident of the Dog in the Night-time" by Mark Haddon. I rather enjoyed this interesting ... language is Boolean, whereas mine is English. I am hoping that someone here more polyglottal than I can translate successfully.

Variant:
There are 1000 doors. You pick one you know that your chance of being right is 1 in 1000.
Monty opens 998 doors.
If you stick with your original door, your chance of being right is still 1 in 1000. Therefore switching to the other door must improve your chance to 999/1000.
Back to the original problem: the chances are 1:3 and 2:3 instead.

Matti
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I recently read "The Curious Incident of the Dog in the Night-time" by Mark Haddon. I rather enjoyed this interesting ... start with. I simply don't accept that removing one of the dummy doors is anything other than a red herring.

Argh! Flee in terror! (Sorry. I've seen this come up regularly since about 1983.)
I put a deck of cards face down. You pick one, leaving it face down. I then pick up the remainder and, looking at them, chose one to place face down in front of me, turning over the others to demonstrate that none of them are, in fact, the ace of spades. I then offer you a thousand dollars if you can correctly decide whether the ace of spades is the one in front of you or the one in front of me. What do you do?

Is there any material difference between this and my presenting two cards, one of which is the ace of spades, and allowing you to pick one?

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( Monty Hall problem )
The problem is that I don't believe the logical argument. ... the dummy doors is anything other than a red herring.

How about, try the Semantic theory.
First, you really do need to understand the rules. Some folks don't follow the logic because they imagine that the rules are something different.
I found it hard to believe that the rules were what-they-are, way back when I first heard the problem, because it seemed so obvious to me. It was rather simpler than the problems that Martin Gardner was offering every month in Scientific American.
Ignoring the words for a minute, the rules allow you to (a) select one door, or
(b) select two doors.
Now, which is going to have the better outcome?
Suppose the doors are R, S, and T, and you want to select one door: say, R. You pronounce the magic formula, in two parts, "I take R; I stay with R." That leaves you, always, with R.

Now suppose you want to be greedy, or opportunistic, or smart; so you want to select two doors, say, R and T. For this, you pronounce the magic formula, in two parts, "I take S; I switch." That leaves, you, always, with the better choice of R and T.

Rich Ulrich, (Email Removed)
http://www.pitt.edu/~wpilib/index.html
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In the book the "hero" propounds the Monty Hall Problem, most adequately explained here http://www.faisal.com/docs/monty.html (and elsewhere all over Usenet ... have been to dozens of sites, learned and stupid, some even pornographical, but I simply don't buy the statistical theory.

Is your point that you don't believe it is correct, or that you accept the correctness but are dissatisfied with the explanation?

As I am an engineer, when I first encountered (Marilyn Vos Savant's column) I tested it with a computer program. I was convinced of its correctness through that empirical method. At that point, I was more receptive to the logical explanations and came to understand them.
Brian Rodenborn
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I recently read "The Curious Incident of the Dog in ... that someone here more polyglottal than I can translate successfully.

You do see that when you first picked the one door out of three, you were probably wrong, don't you? ... than half. One of the other two doors probably will win, even if we don't know which, at the beginning.

I do realise that, according to the laws of probability, my first choice had only a 1 in 3 chance of being right, and is therefore probably wrong.
The host kindly shows you which of the other doors is definitely* a loser, because he knows which of the two it is. Do you see that that makes the remaining one *probably the winner?

No, because it seems to me that, notwithstanding my earlier admission, I might have chosen the correct door. At this point, only the host knows, and I am assuming that he (for it is he, being Monty Hall) is being even-handed, and is not attempting to weight the odds one way or t'other. So, if I have the correct door, then ... hang on, I think I might just be getting it. I'll just chug another glass of that rather nice Cahors that I opened earlier and see.
That's the trickiest part. Why is that so? Because as I already said at the start, your own choice probably ... be one of the other two, and now we know which of the other two. Now that is useful info.

Well, that's only true if I didn't choose the winner. Which, as you say, is only a 1 in 3 chance. So I am beginning to see that it is actually useful information. Do you know, by all that's Holy, I think you've managed to get through my thick skull. The problem is that I haven't totally like grokked it yet.
The other door is not a guaranteed winner, of course, but it's more likely than the one-out-of-three odds you started ... page you gave says - out of nine possible outcomes, six of them have the "other" door being the winner.

Yeah, well, charts schmartz I always say.
It will never be the one Monty Hall shows you. That's what skews it. It would be quite a different ... ago, to understand this problem, so don't feel bad. If my explanation doesn't work for you, maybe someone else's will.)

No, your explanation was better and more lucid than a Usenet full of maths monkeys. Many thanks. I have suffered two nights of disrupted sleep. Statistically, I may tonight have a 1 in 3 chance of dreamless slumber.
Edward

The reading group's reading group:
http://www.bookgroup.org.uk
I put a deck of cards face down. You pick one, leaving it face down. I then pick up the ... this and my presenting two cards, one of which is the ace of spades, and allowing you to pick one?


I just KNEW that if I posted this here instead of rec.puzzles I would get an answer that satisfies. Yourself, Matti and Donna have done the business. I can get on with my life, or I would if I had one.

Edward

The reading group's reading group:
http://www.bookgroup.org.uk
I recently read "The Curious Incident of the Dog in ... the dummy doors is anything other than a red herring.

Argh! Flee in terror! (Sorry. I've seen this come up regularly since about 1983.) I put a deck of cards ... this and my presenting two cards, one of which is the ace of spades, and allowing you to pick one?

That's is a good explanation, Evan. I know you're exasperated, but you have to admit that this problem is usually fascinating to people who encounter it for the first time, not least for the fact that normally sensible, logical people, including maths teachers, find it hard to be convinced.

Adrian
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