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Old 04-23-2011, 06:51 PM   #21
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Default Re: Probability question

The only way your choice will not affect the host's choice is if you picked the right door, as then he can arbitrarily pick the other door and no matter what it will be empty. However, if you picked the wrong door, then your choice affects the host's choice because he cannot pick a door at random since then there is a probability that he will eliminate the door with the prize. When you extend this to n doors and let n go to infinity (a general case), then the probability that you picked the right door is ~0 and therefore the probability that the host would eliminate the correct door at random is ~1. This makes the events dependent.
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Old 04-23-2011, 06:55 PM   #22
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Default Re: Probability question

No, it doesn't matter which one you pick. No matter what, there is still and empty door left. And the host will pick it. There is no 2/3 chance. You have 2 choices and 1 prize. You have a 50% chance. You have no idea whether you picked the right door or not first.
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Old 04-23-2011, 07:02 PM   #23
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Default Re: Probability question

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No, it doesn't matter which one you pick. No matter what, there is still and empty door left.
In the 3 door case, there is at least 1 door left. That is the key difference that you're missing.

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And the host will pick it. There is no 2/3 chance. You have 2 choices and 1 prize. You have a 50% chance. You have no idea whether you picked the right door or not first.
If you pick the wrong door, the host can possibly only pick one door. You just affected his decision by limiting him to opening one particular door. This makes the events dependent. Therefore, by what we've established, if the events are dependent the chance is 2/3. What we're arguing now is whether the events are dependent or not. We know that if they are dependent then the probability is 2/3 (if you switch).

If there are 1,000,000 doors and you picked the wrong one. There is only one combination of 999,998 doors that will not eliminate the prize. Once again, your decision has affected the host's decision and the events are dependent.

I can't be any more clearer.

The difference is that you're taking permutations rather than combinations.
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Old 04-23-2011, 07:06 PM   #24
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Default Re: Probability question

No, your first choice doesn't even matter. All that happens NO MATTER WHAT YOU CHOOSE is that the host opens an empty door. He knows which ones are empty, and he will open whichever one you don't choose if you were to choose one of them. All that's left are two choices and one has a prize. I get exactly what the theory is and it's a stupid one.
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Old 04-23-2011, 07:10 PM   #25
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Default Re: Probability question

ok.

I see, it is easier to think of a problem with a billion doors.

when you pick a door you only have a one in a billion chance that you pick the right door, which is like saying it's near certain that you'll pick the wrong door.
(to get an idea of how small a billion to one is, it's 14 million to one at getting all six numbers in the lottery) -and even then you statistically have more likely hood of getting hit by a bus outside your own house (assuming you live on a bus route I guess!).

anyway, with a near certainty that you have the wrong door.

it's a near certainty that the host will have the right door.

then the host eliminates tens of millions of wrong doors,
it's near certain that he's got the right door, so you should pick that.



when saying it like that, it makes more sense...
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Old 04-23-2011, 07:10 PM   #26
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Default Re: Probability question

There is no point in choosing a door before the host. No matter what, the host will open all empty doors so that 2 doors remain.

Look at it this way. You start with 3 doors, the host opens an empty one and asks you to make your choice. What are your chances? It's no different than if you had told the host a door before he picked one. The host knows 1 and 3 are empty, you pick 1, he opens 3. If you pick 3, he picks 1. If you pick 2 he picks 1 or 3.
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Old 04-23-2011, 07:12 PM   #27
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Default Re: Probability question

HOW DOES THE FIRST DOOR NOT MATTER?! If I pick the wrong door he can only pick one of the other 2 doors! His choice is no longer arbitrary and my choice just influenced his! His choice is only arbitrary, and therefore independent of mine, if I pick the right door! This "theory" is not "stupid". It is correct and proved by mathematicians who have accomplished extraordinary things in their life.
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Old 04-23-2011, 07:15 PM   #28
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Default Re: Probability question

Your first choice doesn't matter because it doesnt matter which of the empty doors Monty picks, only that he picks an empty one.

We are under the assumption that he knows which one the prize is behind correct?...
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Old 04-23-2011, 07:20 PM   #29
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Default Re: Probability question

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Your first choice doesn't matter because it doesnt matter which of the empty doors Monty picks, only that he picks an empty one.
There's where you're making an assumption! You're talking about him picking "an empty one". The only way he can choose between empty doors is if you picked the right one. In which case I agree, your choice of picking the correct door does not influence his choice of picking the wrong door. But that only happens if you pick the right door, which is 1/3 of the time. On the other times (2/3), you pick the wrong door. This leaves 2 doors for the host to pick from. 1 has the prize. Because there is only 1 empty door, he has no choice but to open that door. So you're decision to pick a wrong door (unknowingly) just influenced his decision (as he's only able to pick one door).
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Old 04-23-2011, 07:47 PM   #30
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Default Re: Probability question

So the host could pick the right door?
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