Quote:
Originally posted by punkworld
Wrong? There's nothing wrong in the version I posted. But look at the probabilities of the several options:
A is right and B is revealed => 1/6
A is right and C is revealed => 1/6
B is right and C is revealed => 1/3
C is right and B is revealed => 1/3
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I know where you're getting that from. You think that the 1 in 3 "contains" both possibilities and therefore each one is only a 1 in 6 chance. I totally disagree, and also, this is way out of line with the explanation given on all these websites these other idiots keep linking to. They seem to think it's a 50% chance for one, and a 33% chane for the other.
I think if i'm wrong, then you're right, but those other 'tards are way out of it.
But anyways. On to my explanation. Sure, there is a 1 in 3 chance that the person is going to choose the right one initially, but there are actually four possible outcomes, and thus, you're dealing with a 1 in 4 chance of each outcome.
A is right and B is revealed => 1/4
A is right and C is revealed => 1/4
B is right and C is revealed => 1/4
C is right and B is revealed => 1/4
It doesn't matter how many boxes there are. It matters how many outcomes there are. I'm starting to confuse myself. =\