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RE: The Monty Hall problem.
January 17, 2013 at 9:34 pm
(January 12, 2013 at 9:07 pm)Rev. Rye Wrote: Well, since this thread's been resurrected and I haven't heard any mention of "The Curious Incident of the Dog in the Night-Time", I'll explain it as Christopher Boone explained it.
Let the doors be called X, Y and Z.
Let Cx be the event that the car is behind door X and so on.
Let Hx be the event that the host opens door X and so on.
Supposing that you choose door X, the possibility that you win a car if you then switch your choice is given by the following formula
P(Hz ^ Cy) + P(Hy ^ Cz)
= P(Cy)·P (Hz ¦ Cy) + P(Cz)·P(Hy ¦ Cz)
= (1/3 · 1) + (1/3 · 1) = 2/3
I would note Monty Hall himself has addressed this problem. He pointed out that the odds are not 2/3 for a very simple reason: he was allowed (and often did) to influence the decision-making (do I change or switch) through manipulation of the contestants.
For his description, see his own Website, Let's Make A Deal: The Monty Hall Problem
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RE: The Monty Hall problem.
January 19, 2013 at 6:59 pm
It can never be genuinely tested because in the actual game no one switches. Computers aren't people. I know the odds come out at 2 chances in three in controlled tests, but those people know what's going on, and switch doors when they wouldn't have in the actual game. I think there is a lot more to this than the question of odds.
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RE: The Monty Hall problem.
January 19, 2013 at 7:15 pm
The "Monty Hall Problem" is loosely based on the gameshow; it is not entirely based on it. The problem is still part of mathematics, and should be as it reveals bad assumptions most people make with probability.
(January 19, 2013 at 6:59 pm)Joy Squeezy Wrote: It can never be genuinely tested because in the actual game no one switches. Computers aren't people. I know the odds come out at 2 chances in three in controlled tests, but those people know what's going on, and switch doors when they wouldn't have in the actual game. I think there is a lot more to this than the question of odds. Of course it can be genuinely tested.
The question wasn't "will anyone ever switch?", it was "is there any advantage to switching?" That question can be answered by using probability, and the answer is most definitely yes.
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RE: The Monty Hall problem.
January 19, 2013 at 7:28 pm
Did you read what I wrote? In the ACTUAL game no one switched. The apparent paradox has been scientifically tested, but the doors were switched and yes, the switchers did win a disproportionate number of times. What I am saying is that this would never happen in a REAL game where the contestants were completely unaware of the Monty Hall problem. Do you not see something else going on here? Can anyone cite a similar situation where the seeming anomaly could be proven in its 'natural' form?
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RE: The Monty Hall problem.
January 19, 2013 at 7:40 pm
What you wrote is that "it can never genuinely be tested." If by genuine, you mean in a televised setting, this is still incorrect. People don't need knowledge of the game to test the probabilities. The probabilities play themselves out whether people know (and/or switch) or not. They just may likely stick with the door, but then they will only win on average 33% of the time, leaving the remaining door as the winning one 66% of the time. Knowledge is not a requisite for probabilities to play out. It's only a requisite for the contestants to come out as a winner the majority of the time.
What more has to be proven? How would that not verify the probabilities by itself?
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RE: The Monty Hall problem.
January 19, 2013 at 7:45 pm
But it doesn't tell you if they would have won 66% of the time if they had switched, because almost no one switched.
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RE: The Monty Hall problem.
January 19, 2013 at 7:56 pm
(January 19, 2013 at 7:45 pm)Joy Squeezy Wrote: But it doesn't tell you if they would have won 66% of the time if they had switched, because almost no one switched.
Because they are retards 
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RE: The Monty Hall problem.
January 19, 2013 at 7:57 pm
(This post was last modified: January 19, 2013 at 8:01 pm by Whateverist.)
Here is a loosely related problem which has fewer distractors.
A bag contains three identical tiles. One is painted with an "X" on both sides. Another is painted with an "O" painted on both sides. The third tile has an "X" painted on one side and an "O" painted on the other.
If you draw a tile and hold it up so that you can only see the side facing you and that side is marked with an "X", what is the probability that the other side is also an "X"?
Correct Solution:
Popular Incorrect Solution:
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RE: The Monty Hall problem.
January 19, 2013 at 7:58 pm
(This post was last modified: January 19, 2013 at 8:02 pm by Golbez.)
(January 19, 2013 at 7:45 pm)Joy Squeezy Wrote: But it doesn't tell you if they would have won 66% of the time if they had switched, because almost no one switched.
So they didn't switch. And they only won 33% of the time. One door remained. What are the odds it was the winning door? I'm pretty sure you can still do the math on that one.
whateverist Wrote:A bag contains three identical tiles. One is painted with an "X" on both sides. Another is painted with an "O" painted on both sides. The third tile has an "X" painted on one side and an "O" painted on the other.
Who taught you what identical means? =P
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RE: The Monty Hall problem.
January 19, 2013 at 8:02 pm
They didn't switch, so it's irrelevant. Sneering at my 'mathS' isn't going to get you anywhere. Many mathematicians were also stumped by the problem.
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