This puzzle has been posted recently on Richard Wiseman's blog:
I have a box and inside the box are four stones. One stone is white, another is yellow, the third is blue and the fourth stone is also blue. I put my hand into the box and pick up two stones. I bring my hand out in a fist, look inside my fist and remove a blue stone. What are the chances of the other stone in my hand also being blue?
It has caused a lot of controversy so I decided to post it on a proper mathematical forum.
So people are saying that it depends on the "protocol" used to remove one of the stones in the fist.
And I think that once we know that one of them is blue and the puzzle only asks for the probability of the other one also being blue the "protocol" is irrelevant.