During one of my Management Science classes at the university few weeks ago, our professor introduced the Monty Hall problem to the class and asked everyone what they think the right answer was. Since I was familiar with this problem, I asked him not to count my vote. But majority of the students got their answers wrong – which was obvious because otherwise this problem wouldn’t have been called a veridical paradox!
Here’s the statement of the famous Monty Hall problem (source):
Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to [switch and] pick door #2?” Is it to your advantage to switch your choice of doors?
Please think about this before reading further: should you switch or stay with your initial choice? Needless to say, you win if you pick the door that has a car behind it. There are no tricks or loopholes here, it has a purely logical – and alternatively, a probabilistic – solution.
Most people think that after the host opens a door (revealing a goat behind it), both of the remaining doors have equal chances (50/50) of having a car. Hence, it doesn’t matter if you switch or not, your chances of winning remain the same.
But the correct answer is – and this might surprise you if you’ve never heard of this problem before – you should switch. Switching will actually increase your chances from 1/3 to 2/3.
The simplest explanation (IMO) is: Let’s say you pick door A. The probability of having a car behind A is 1/3. Hence, the probability of not having car behind A is 2/3. Which also means that the probability of having a car in the other two doors combined – B and C – is 2/3. Now, the host, by exposing a goat behind one of those two doors, eliminates that door – let’s say B. So the probability of having a car behind the remaining door C is 2/3.
Hence, if you don’t switch, your probability of winning remains the same (1/3). But if you switch your probability of winning increases (2/3).
This wonderfully counter-intuitive answer was challenged by many intellectuals and mathematicians including Ph.D.’s when it was first published during the early 90’s, and it’s considered to be one of the best mathematical brain teasers. If you’re not convinced with the answer there’s a Wikipedia page (link) for you. And if that’s not enough, there’s an entire book written about it: The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brain Teaser (which I haven’t read yet.)
Addendum: There’s an interesting variant of this problem called Monty Fall or Ignorant Monty Problem (PDF link):
In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door?
The answer to this is: your probability of winning is the same (50/50) whether you stick or switch.