*The game show host presents the contestant with three closed boxes. One contains a check for a million dollars, the other two are empty. The contestant is asked to pick a box, without opening it; if the box he picks contains the check, he will walk away a millionaire. After he has made his choice, the game show host opens one of the two remaining boxes, and finds it empty. Thus, the million-dollar check is in one of the other two boxes.
At that point, the contestant is given the option of sticking with his initial choice, or switching instead to the other box, that is still closed. What should he do, in order to maximize the chance of winning a million dollars ?*

This is a question that teachers of basic probability and statistics are fond of asking to their classes. The expected, “canonical” answer is that the contestant should always go with the other box, for reasons that will become clear below. I remember being in class and hearing that question. I already knew the canonical answer, having heard the quiz before and been explained it. I was ready to make myself look smart before my class mates by repeating it, when from the back of the room a question came “Excuse me, but… the way you phrased the question is ambiguous. You did not specify whether the game show host *knows* that the box that he is opening is empty, or is he himself picking a box at random… and does the contestant know that ? That changes things !”.

I remember thinking ‘OK, who is this guy ? What the heck is this person talking about ? What difference can it possibly make whether the game show host knows or not ?”… Well… it does make a difference, and it may seem really trivial but back then it got me thinking about probability and knowledge [0]. The fact is, the teacher himself had not thought of this problem in full, and as a result did indeed phrase the question ambiguously as written above — in a way that does not allow one to come up with the proper answer.

The canonical answer goes as follows: the probability that the contestant will pick the right box in the first place is 1/3 (number of favourable cases divided by the number of possibilities). Thus, the probability that the check will be in either of the remaining two boxes is 2/3, i.e., the sum of the probabilities that the check will be in one or the other. If the game show host *knows* in which box the check is (and, the contestant *knows that the game show host knows*), the action of opening one of the two is immaterial to the goal of determining probabilities. He is simply making use of his prior knowledge. As far as the contestant is concerned, it remains twice as likely for the check to be in the box that he did not pick, and is still sitting there unopened, than in the one that he did pick. It is therefore to his advantage to go for the other box.

But if the game show host himself is (or, claims to be) unaware of where the check is, randomly picks one of the two remaining boxes and finds it empty, there is *no reason whatsoever for the contestant to change box*, other than superstition or other reasons unrelated to probability. For, the likelihood that the check will be in either the one box that he picked or in the other one still unopened, is the same. By opening a box at random and finding it empty, the game show host has merely increased by identical amounts the probability that the check be in either one of the other two, from the initial 1/3 to 1/2 — but the same for both.

So, this is a situation in which prior knowledge on the part of one of the participants significantly alters one’s course of action, if this is based upon probability. The interplay between prior knowledge and probability is often subtle, and is intrinsically built into Bayesian statistics.

**Notes**

[0] Whether that person was really smart, or had been explained himself the extended version, is something that I never could find out.

Tags: Life, Miscellaneous

March 7, 2010 at 9:45 am |

Am I mistaken, or does this extended version only work for a subjectivist approach, but not for a frequentist?

March 7, 2010 at 10:11 am |

My understanding is that it works in both cases, a frequentist would simply run into the usual ambiguity with the specification of possible outcomes — see, for instance the discussion in the introduction of this book.

March 7, 2010 at 5:30 pm |

Thanks for posting this–I’ve always felt a little uneasy about my understanding of this particular problem, and your post cleared it right up.

March 8, 2010 at 9:52 am |

Check here and here.

Play the game here.

March 8, 2010 at 9:55 am |

Hey, thanks !

March 8, 2010 at 11:25 pm |

Happy to see this discussion of the problem. I’ve been trying to convince various people of exactly this point but found it very hard to convince people that it does matter how the problem is worded as to the operation of the host. The first time I heard the problem I mulled over for long enough to forget what the wording I was told was and do had to solve both the canonical and the extended versions on my own