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I place four balls in a hat: a blue one, a white one, and two red ones.
Now I draw two balls, look at them, and announce that at least one of
them is red. What is the chance that the other is red?
Well, it's not 1/3!
What was the puzzle?
It's not 1/3. It would have been 1/3 if I had taken first ball out, announced it as red and then taken second ball out. But I have taken pair of ball out. So, there are 6 possible combinations.
red 1 - red 2
red 1 - white
red 2 - white
red 1 - blue
red 2 - blue
white - blue
Out of those 6, last is invalid as I already announced the first ball is red. That leaves only 5 valid combinations.
And out of 5 possible combinations only first has desired outcome i.e. both are red balls.
Hence, there is 1/5 the chance that the other is red
