Daily Archives: Sat, 31 December 2016

Monty Hall problem

As it’s the last day of 2016 and my mind isn’t quite into physics mode today I thought I’d post something frivolous. Here are a few thoughts on the Monty Hall problem, which has reputedly puzzled even some Nobel prize winners. (The problem also featured in episode 4×08, Skyfire Cycle, of the popular TV comedy Brooklyn Nine-Nine.) The problem is named after Monty Hall, who was the host of an American game show called Let’s Make a Deal. In the climax of each show, the winning contestant was faced with three doors. Behind two of the doors was a worthless prize (symbolized in the usual statement of the problem by a goat) while behind the third door was a major prize such as a car. The contestant initially chooses one of the three doors. At this point, Monty Hall opens one of the other two doors to reveal one of the worthless prizes (one of the goats). The contestant is then offered the chance to switch his/her choice to the other unopened door. The question is: should the contestant stay with their first choice or switch to the other door?

At first glance, the answer appears obvious. After Monty opens one of the two remaining doors to reveal the location of one of the two goats, the contestant has a choice between two doors, so the odds appear to be 50-50 as to which door hides the car. Thus it would appear that there is no advantage to be gained by switching, so it doesn’t matter what the contestant does.

However, this isn’t the correct way of looking at the problem. The key is that Monty knows where the car is, since he always opens a door hiding a goat. Thus if the contestant’s initial choice is wrong, they are guaranteed to win if they switch doors. The problem then reduces to: what is the probability that the contestant’s initial choice was wrong? The answer here is obvious: since the initial choice is a blind one-in-three choice, the chance that their initial choice is wrong is {\frac{2}{3}}. When offered the chance to switch doors, they should therefore do it, as they are twice as likely to win the car than if they stayed with their original choice.

It’s surprising that this relatively simple problem has given rise to such controversy (see the Wikipedia article for details).

Anyway, hopefully readers have found my posts from 2016 to be interesting and/or helpful, so here’s looking forward to a few more posts in 2017.