Quote:
Originally posted by chodadog
For one to be more likely than the other, wouldn't it's result have to be more frequent in the pool? I mean, let's take a six sided die. Let's say it has 1, 2, 3, 4, 5 and 5. No six on this dice.
Total possible results= 6.
Chance of a 1 coming up = 1 in 6
....
Chance of a 4 coming up = 1 in 6
Chance of a 5 coming up = 2 in 6 or 1 in 3.
In the situation we're dealing with, there is only one possible combination for each answer, and therefore they are just as likely as eachother, aren't they?
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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