The Monty Hall Problem gets its name from the host of the popular television game show Let’s Make a Deal, which originated in the United States in 1963 and has been produced in many countries.

The problem is as follows. Suppose you are on the show, and you are asked to choose one of three doors, numbered one, two and three. All you know is that behind one of the doors there is a car, and behind each of the other two there is a goat. Suppose you choose door one and then the host reveals that behind door three there is a goat and offers you the possibility of switching to door two. Should you switch? Does it make any difference to your chances of winning the car?

If you think that sticking with door one and switching to door two offer the same probability of winning, then you are probably in the majority. When the problem was popularised, several hundreds of people with PhDs refused to accept otherwise, including Paul Erdős, one of the most influential mathematicians in history! The reasoning would be that we know the car is behind door one or door two, and the two possibilities should be equally likely irrespective of the whole process.

However, the mathematics of probability theory tells us that the probability of winning is 1/3 if you stick with door one, and 2/3 if you switch to door 2. Loosely speaking, the initial 2/3 probability that the car is behind door two or behind door three shifts to door two once it is revealed that behind door three there is a goat.

This is a paradox of the veridical type, that is, a result that appears absurd. However, it becomes much clearer with many doors. It is intuitive that if there are 100 doors (and still one car), you choose door one, the host opens 98 of the unchosen doors that have a goat behind them, and you may switch to the remaining door, then you should switch.