Three boxes with a check inside I recently saw a math video in the online edition of a Spanish divulgation magazine. In the video, they had three boxes, one yellow, one red and one blue. A 1,000,000€ check was put inside one of the boxes and then a girl was asked to choose one of the boxes. She chose the red box, and then the host opened the yellow one that was empty. He then asked the girl if she wanted to change her mind and choose the blue box. He, then, said that changing the box was the correct answer, because at the start, the red box had 1/3 of being the correct box, and the yellow or blue had 2/3, and after opening the yellow one this was still 2/3, so it was more probable that the blue box was correct.
This strikes me as a statistical fallacy, but I'm not sure. Am I right or is there something here I can't understand?
 A: This is the classic form of the Monty Hall problem. It is true that changing your choice gives you a 2/3 result, provided that the host opens an empty box whether or not you chose the right one first.
The key is that, assuming the host always opens a box, the host has given you additional information: the host will never open a box which has the check in it, so your action in changing is equivalent to opening both of the boxes you didn't choose originally (because if one of them has the check, you will always be choosing that one).
A: The guest had a 1/3 chance of getting it right with the first guess. Note that the host can always choose an empty box (the colours are a distraction). Then sticking gives the guess 1/3 as it always was, changing therefore gives the remaining 2/3. It seems weird, but write down all the possibilities carefully, and you'll see it's true. Look up "Monty Hall Problem" but be careful, there are many examples of false and confusing analysis out there on the internet.
