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Topic: variable change (Read 3329 times) |
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puzzlecracker
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variable change
« on: Aug 25th, 2008, 6:41pm » |
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You stand in front of 3 doors (A, B, C), behind one there is a car, and behind the other two -nothing. You need to pick one door. Say, you pick door, A, but before, you have a chance to open it, someone opens the door C, which contains nothing. Now, you're offered to change your guess to B or remain with A. What would you do and why?
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teekyman
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Re: variable change
« Reply #1 on: Aug 25th, 2008, 8:43pm » |
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I believe it depends on why door C was opened.
If C was just chosen randomly between B and C, then it shouldn't matter whether you switch or not.
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| « Last Edit: Aug 25th, 2008, 9:01pm by teekyman » |
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towr
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Re: variable change
« Reply #2 on: Aug 26th, 2008, 12:24am » |
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on Aug 25th, 2008, 8:43pm, 1337b4k4 wrote:I believe it depends on why door C was opened.
If C was just chosen randomly between B and C, then it shouldn't matter whether you switch or not. |
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If C was randomly chosen between A, B and C, then it doesn't matter. If the choice was only between B and C then you should switch.
(If A isn't involved, it's probability doesn't change; so B must take on the joint probability of B and C in the latter case.)
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| « Last Edit: Aug 26th, 2008, 12:57am by towr » |
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Hippo
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Re: variable change
« Reply #3 on: Aug 26th, 2008, 12:40am » |
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on Aug 26th, 2008, 12:24am, towr wrote:
If C was randomly chosen between A, B and C, then it doesn't matter. If the choice was only between B and C then you should switch.
(If A isn't involved, it's probability doesn't change; so B must take on the joint probability of B and C in the latter case.)
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No, if it was opened randomly between B and C (so it may happen the car would be behind) it does not matter.
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Grimbal
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Re: variable change
« Reply #4 on: Aug 26th, 2008, 12:40am » |
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The problem is definitely tricky.
If the other person chose randomly, there was a possibility that the door opened on a car. That did not happen, as a result, the problem space after the opening is not the same as it was initially.
The other person had more chances to find the car if you had the wrong door than if you had the correct one. So the fact that he did not find the car is an indication that your door is the correct one.
Simply put, if there are 2 doors, you chose one, someone opens the other door, it is empty. Your reasoning would imply you still have only 1/2 chance to be right.
However, if the other person chose his door specifically to be different from your door and different from the one with the car, then the probability to be right remains at 1/3.
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| « Last Edit: Aug 26th, 2008, 12:45am by Grimbal » |
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towr
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Re: variable change
« Reply #5 on: Aug 26th, 2008, 1:01am » |
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Ok, fine.. 
If C was chosen with a probability independent of the contestants choice, than it doesn't matter. So either randomly choosing from A,B,C or from B,C, or even just always opening C, it doesn't matter.
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rmsgrey
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Re: variable change
« Reply #6 on: Aug 26th, 2008, 2:50am » |
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on Aug 26th, 2008, 1:01am, towr wrote:Ok, fine..
If C was chosen with a probability independent of the contestants choice, than it doesn't matter. So either randomly choosing from A,B,C or from B,C, or even just always opening C, it doesn't matter. |
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The key point is not independence of the contestant's choice, but independence of the location of the car...
(looks like you're having an off day for once)
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towr
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Re: variable change
« Reply #7 on: Aug 26th, 2008, 3:00am » |
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Which is why, if possible, it's best just to write the case your dealing with out..
What if I phrased it as "If the information you get is independent of the information you want, then it's useless" ?
Did I finally get it right 
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puzzlecracker
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Re: variable change
« Reply #8 on: Aug 26th, 2008, 5:48am » |
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The person, who opened the door C, knows where the car is since he hosts this show, so to speak. Hence he didn't open it randomly.
Maybe he is playing a mind game, wanting the contestant to change his mind and pick B instead.
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| « Last Edit: Aug 26th, 2008, 5:52am by puzzlecracker » |
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towr
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Re: variable change
« Reply #9 on: Aug 26th, 2008, 6:10am » |
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on Aug 26th, 2008, 5:48am, puzzlecracker wrote:| The person, who opened the door C, knows where the car is since he hosts this show, so to speak. Hence he didn't open it randomly. |
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That doesn't necessarily follow. He may still open a door randomly even if he knows what's where.
Anyway, it's just Monty Hall problem again.
Quote:| Maybe he is playing a mind game, wanting the contestant to change his mind and pick B instead. |
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For that you'd need knowledge about the host's psychology, and/or statistics about his/her previous dealings. Hard to take into account.
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Grimbal
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Re: variable change
« Reply #10 on: Aug 26th, 2008, 6:25am » |
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on Aug 26th, 2008, 2:50am, rmsgrey wrote:
The key point is not independence of the contestant's choice, but independence of the location of the car...
(looks like you're having an off day for once) |
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And yet...
Suppose the host could have chosen the same door as you, but not the car. He could have chosen door A but didn't. If you have the right door, it certainly happens. If you have the wrong door, it happens only half of the times. The fact that it didn't happen is an indication that you have the right door.
If there are only 2 doors, and the host chooses not your door regardless of the car position, and you see it is empty, you do get information about where the car is!
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Hippo
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Re: variable change
« Reply #11 on: Aug 26th, 2008, 7:13am » |
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And once more
Following variant of Monty hall.
3 persons are invited, they should pick different dors A/B/C.
A person who chooses wrong doors is eliminated "by opening the doors".
The symmetry of the positions of the remaining players guaranties the probabilities are equal.
Of course it differs from the classical variant with only one player and Monty forced to open noncar doors.
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towr
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Re: variable change
« Reply #12 on: Aug 26th, 2008, 7:16am » |
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on Aug 26th, 2008, 6:25am, Grimbal wrote:| If there are only 2 doors, and the host chooses not your door regardless of the car position, and you see it is empty, you do get information about where the car is! |
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But all remaining doors are still equally likely. You don't get distinguishing information.
If there are N doors equally likely to have a car behind them, and the host opens K doors, the remaining N-K doors are still equally likely.
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Grimbal
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Re: variable change
« Reply #13 on: Aug 26th, 2008, 7:54am » |
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Hm... I get confused as to who is claiming what.
Let's summarize:
P is the probability for the contestant 1st choice to be correct after the host opened C.
- If the host chooses any door randomly, P is 1/2.
- If the host chooses any door but not the car, and it just didn't happen that he chooses A, P is 1/2
- If the host chooses any door but A, and it just didn't happen that he found the car, P is 1/2
- If the host chooses any door but not the car and not A, then only P is 1/3.
So taking into accout A's choice, even though it does not tell anything about the car's position, does affect the final choice.
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| « Last Edit: Aug 26th, 2008, 7:57am by Grimbal » |
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