Many of us grew up watching Monty Hall on Let’s Make a Deal. It turns out that the show presents something of a mathematical conundrum: does switching your choice at the last minute between the two remaining rooms increase your chances of success? The video below presents the answer and it may surprise you.
The so-called Monty Hall paradox is a veridical paradox that is similar to the Three Prisoners problem and a less known Bertrand’s box paradox.
The paradox was first presents by Steve Selvin in a publication in the American Statistician in 1975. He argued that a player who stays with the initial choice wins in only one out of three times while a player who switches wins in two out of three times. Here is another explanation from a math site:
Imagine that Monty opens a door and shows that there’s only a goat behind it. Consider that the car is more likely to be behind a door other than the one you choose. Monty has just shown that one of those two doors – which together have the greater probability of concealing the car – actually conceals a goat. This means that you should definitely switch doors, because the remaining door now has a 2/3 chance of concealing the car. Why? Well, your first choice still has a 1/3 probability of being the correct door, so the additional 2/3 probability must be somewhere else. Since you know that one of the two doors that previously shared the 2/3 probability does not hide the car, you should switch to the other door, which still has a 2/3 chance of concealing the car.
The idea is so counterintuitive that hundreds of experts wrote into the magazine contesting the theory. However, this video from 2009 shows the theory in execution.
I will be trying this tonight with the kids as part of the Turley Blog Research Center.