Thread: A simple puzzle (with a lot of controversy)

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.

Hello, Iconoclast!

A box contains four stones: one white, one yellow, and two 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 is the probability that the other stone in my hand is also blue?

The problem is small enough to LIST all the outcomes.

The four stones are:.

There are six possible outcomes:
. .


We are told that one of the two stones is blue.
The sample space is reduced to five outcomes:
. .

The stones are both blue in exactly one case.


Therefore: .

Thank You Soroban! That was the official solution. Some people convinced themselves somehow that we must know whether he chooses one of the stones from his fist at random and then presents it to us (in which case we must include more outcomes in the sample space and the answer is 1/3) or whether he always picks the blue one first. The simplest explanation that I gave was that it is irrelevant once we know one of the stones is blue. I am a layman and I don't know mathematical jargon and I i think I failed to explain it to them. If anyone is interested in modifying the puzzle to accommodate different answers or explaining properly why 1/3 is definitely wrong in this case I would be really grateful.

This is my first time i visit here. I found so many entertaining stuff in this website, especially math-puzzles.

How would you do this problem without listing the possibilities? I can't quite remember the formulas involved.