They say things come in threes so here’s a third blog, extracted from my 2014 book Fingers to the Keyboard, on another aspect of mathematics. Stay with me; this one will drive you crazy!

This puzzle is a classic and cropped up one evening at a dinner party I attended. The party nearly ended in fisticuffs!

Steve Selvin, an American professor in biostatistics, first posed the question in a letter published in the American Statistician journal in 1975. Here is his question:

Suppose you’re on a game show, and you’re given the choice of three (closed) doors: behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, ‘Do you want to pick door No. 2?’ Is it to your advantage to switch your choice?

At the time, an American game show called Let’s Make a Deal and hosted by Monty Hall was popular and the question acquired the sobriquet Monty Hall Paradox.

First, some extra detail:

• When offered the chance to stick or swap the contestant can do either but whichever door he or she finally selects is then opened to reveal the prize. There’s no going back!
• Before the game starts, the game-show host knows where the goats and car are concealed.

Intuitively, it would seem that there is no point in changing your mind when offered the chance to do so. For your initial choice, you had a one-in-three chance of picking the door concealing the car. After the game-show host has opened another door to reveal a goat, you now have a one-in-two chance of winning the car. This means there is an equal probability that the car is either behind the door you’ve already picked or behind the third door. That being so, why swap? You will not improve your chance of winning the car. Right?

Wrong. If you change to the other door, you will improve your chance of winning the car. Here’s why.

The following proof is based on an enumeration of all possible outcomes. Alternative solutions exist, based on an understanding of probabilities and how these probabilities change once you know the location of one of the goats.

Here’s the exhaustive enumeration solution. I’ll assume door 1 conceals the first goat, door 2 conceals the second goat, and door 3 therefore conceals the car. The following argument holds for any other goat and car arrangement.

Case 1. You pick door 1 concealing the first goat.  The game-show host opens door 2 revealing the second goat. Door 3 conceals the car.  If you stick with your original choice, door 1, you win a goat. If you swap to door 3, you win the car.

Case 2. You pick door 2 concealing the second goat.  The game-show host opens door 1 revealing the first goat. Door 3 conceals the car.  If you stick with your original choice, door 2, you win a goat. If you swap to door 3, you win the car.

Case 3. You pick door 3 concealing the car.  The game-show host opens either door 1 or door 2, it doesn’t matter which, revealing one of the goats. Door 3 conceals the car.  If you stick with your original choice, door 3, you win the car. If you swap to the other door, you win the goat.

Here’s a summary of what happens for these three cases.

As you can see, if you swap your first choice of door you have two-out-of-three chances of winning the car. If you don’t swap your choice, you have one-out-of-three chances of winning the car. Thus, it is always to your advantage to swap your choice of door once the location of one of the goats has been revealed.

The moral of this story is don’t trust to intuition, especially where probabilities are concerned. Look at the underlying mathematics. Mathematics cannot lie!
If this problem intrigues you, the Wikipedia article is quite good.

Okay, no more mathematics… for a while.

(^_^)