Quote:
Originally posted by punkworld
Simply put, no. This very post of yours in fact makes clear why: there are actually 6 possibilities if you follow your line of thinking. The fact that A being right offers 2 valid ones and B or C being right each only offer 1 valid one, does not change that fact. (this paragraph lacks a bunch of explanations and definitions, and makes some big jumps, so if you don't get it my laziness is probably to blame)
Let me explain the entire issue in the simplest way possible:
A = 1/3 chance
B = 1/3 chance
C = 1/3 chance
Contestant picks A.
If A is right (1/3 chance of that):
then the host reveals B or C
and switching does NOT pay off
If B is right (1/3 chance of that):
then the host reveals C
and switching does pay off
If C is right (1/3 chance of that)
then the host reveals B
and switching does pay off
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The problem with your thinking is the or. That or means there are two possible outcomes if option A is right and the contestant chooses option A. For some reason, you've lumped two outcomes into one to prove your point, when two outcomes are just that, two seperate outcomes.
Explain to me why this isn't the case:
Contestant picks A.
If A is right (1/3 chance of that):
then the host reveals B
and switching does NOT pay off
If A is right (1/3 chance of that):
then the host reveals C
and switching does NOT pay off
If B is right (1/3 chance of that):
then the host reveals C
and switching does pay off
If C is right (1/3 chance of that)
then the host reveals B
and switching does pay off
It makes no sense to me that the first two outcomes i've shown above shouldn't be considered as likely as the rest.