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October 11th, 2015, 12:01 PM   #1
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Can someone explain the Monty Hall problem to me?

After one door was opened and thus eliminated, why exactly isn't there a 50% chance for the car to be behind either door?

I know for a fact it's not behind the opened door so only 2 are left and I have no more information about which of these 2 doors has the car.

How can that be?
And why do the solutions always continue taking 3 doors and chances out of 3 into account when there are only 2 doors left?
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October 11th, 2015, 02:52 PM   #2
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Which would you rather have - what's behind one door or what's behind two doors?
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October 11th, 2015, 02:59 PM   #3
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Try the problem with 100 doors. You pick one and the host opens 98 to show nothing. What do you do?
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October 12th, 2015, 07:05 AM   #4
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Suppose you chose one door and were then given the choice of switching to both of the two other doors. You'd take that, right? If so, why are things different when one of those doors is opened?
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October 12th, 2015, 12:37 PM   #5
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Quote:
Originally Posted by Ryuhou View Post
After one door was opened and thus eliminated, why exactly isn't there a 50% chance for the car to be behind either door?

I know for a fact it's not behind the opened door so only 2 are left and I have no more information about which of these 2 doors has the car.
Sure you do. To make things easier, let me label the doors. But not with the numbers we assume are on the doors, but with labels that I'll put on them after you choose a door, but before Monty Hall opens one.

Call the door you originally chose C (for Contestant). Call the door to its right R, and the door to its left L. (Assume they are in a circle, so the door to the "right" of Door #3 is Door #1.)

If the car is behind R, Monty must open L. If the car is behind L, Monty must open R. If the car is behind C, he can open either R or L. But he can never open C.

Let's assume he opens L (you can reverse R and L in the following argument in he opens R).

What confuses you is thinking that all the information you have is that L does not have the prize. But you know more than that: you know that Monty choose to open L. He must open R if the car was behind L, so you eliminate all those games. But he can also open R if the car is behind C, so you also should eliminate half of the games where the car is behind C BUT MONTY HALL WOULD CHOOSE TO OPEN R.

What's left is the 1/3 where the car is behind R, and the 1/6 where the car is behind L BUT MONTY HALL WOULD CHOOSE TO OPEN L. Since there are twice as many cases left where the car is behind R, than cases where it is behind C, you should switch.

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why do the solutions always continue taking 3 doors and chances out of 3 into account when there are only 2 doors left?
It's called conditional probability, and people find it easier to try to avoid using it than to explain it. The problem is, these explanations have holes in them, as you noticed, and are often wrong even if they get the right answer.

Technical:
Say something random happens. You can make a list of all the possible outcomes, and the probability of each. Those probabilities will add up to 100%, right? An example is rolling two six-sided dice, and adding them. You can get a sum of {2,3,4,5,6,7,8,9,10,11,12}, with probabilities {1/36,2/36,3/36/,4/36,5/36,6/36,5/36,4/36,3/36,2/36,1/36}.

Say that somebody tells you something about what actually happened. You need to "eliminate" all of the things that you now know didn't happen. And the remaining probabilities you have don't add up to 100%. Since you need to make them add up to 100%, you divide each by the sum of what remains. If I tell you that the sum in my example is 5 or less, you are left with {2,3,4,5} and {1/36,2/36,3/36/,4/36}. This sums to 10/36, so the new probabilities are
{1/10,2/10,3/10/,4/10} which you find by multiplying by 36/10.
The point is that you can only do this correctly if you start with all of the possibilities that existed before you learned anything. That is, all 3 doors, or all 36 combinations of 2 dice.

The mistake people make is eliminating only those cases that couldn't produce the information you have, when they need to also eliminate those that wouldn't. In Monty Hall, eliminating "couldn't" only takes care of the cases where the car is behind L so he had to open R. Eliminating "wouldn't" also takes care those cases where the car is behind C and Monty Hall chose to open R.
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October 13th, 2015, 07:58 AM   #6
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Quote:
Originally Posted by CRGreathouse View Post
Suppose you chose one door and were then given the choice of switching to both of the two other doors. You'd take that, right? If so, why are things different when one of those doors is opened?
Because it matters if Monty Hall doesn't choose randomly between them, when he opens one. This is an analogy only, and it fails to convince people because they sense there is something wrong, even if they can't identify it.

EXAMPLE: Monty always asks if you want to switch to both doors, and then opens the closest door to him that he can. If he walks past a door to open the furthest door, you know the prize is behind the door he passed. If he opens the one that actually is closest, your chances are 50% with either of the other two.

Now, I know this doesn't happen in the problem as it is stated. But the fact that it could requires more of an explanation than you gave.

+++++

But I really am replying to correct a typo in my reply:

What confuses you is thinking that all the information you have is that L does not have the prize. But you know more than that: you know that Monty choose to open L. He must open R if the car was behind L, so you eliminate all those games. But he can also open R if the car is behind C, so you also should eliminate half of the games where the car is behind C BUT MONTY HALL WOULD CHOOSE TO OPEN R.

What's left is the 1/3 where the car is behind R, and the 1/6 where the car is behind C BUT MONTY HALL WOULD CHOOSE TO OPEN L. Since there are twice as many cases left where the car is behind R, than cases where it is behind C, you should switch.
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October 13th, 2015, 09:55 AM   #7
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By "Monty chooses at random", do you mean he may open the door you want to choose?

One can list all the cases to obtain a correct answer, but some people don't find that sufficiently clear either.
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October 13th, 2015, 11:26 AM   #8
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Go do some investigating....

https://www.google.ca/?gws_rd=ssl#q=monty+hall+problem
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October 13th, 2015, 01:36 PM   #9
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Quote:
Originally Posted by skipjack View Post
By "Monty chooses at random", do you mean he may open the door you want to choose?
We assume he is restricted to not opening your door, or the door that has the car.

That leaves one, or two, choices. A correct solution has to use the probability that he would open the door he did when you picked correctly (and he had a choice of two). If that probability is Q, the probability you win by switching is 1/(1+Q).

Of course, if you aren't told how he decides, the best assumption is Q=1/2 and so a 2/3 probability you win if you switch. But some people refuse to acknowledge that this assumption is being made here, or in similar problems.

Quote:
One can list all the cases to obtain a correct answer, but some people don't find that sufficiently clear either.
But that requires knowing Q. In other words, "counting" those cases you listed is incorrect. You need to sum the probabilities that they would occur.

Quote:
Originally Posted by Denis View Post
And if you do, you will see the solution I just gave at the top of the list.
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October 13th, 2015, 03:52 PM   #10
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Quote:
Originally Posted by JeffJo View Post
A correct solution has to use the probability that he would open the door he did when you picked correctly. (and he had a choice of two). If that probability is Q, the probability you win by switching is 1/(1+Q).
That's incorrect. If you picked correctly, you lose by switching, and if you picked incorrectly you win by switching, regardless of the value of Q. Hence the probability that you win by switching is independent of the value of Q.
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