Monty Hall Problem stumping intuition

In 1963, there used to be an extremely popular game show in the US Television named "Let's Make a Deal". It's host was Monty Hall. The game was pretty simple. Some people from the audience would be picked up to play a very simple sounding game of chance. They'd make deals with the host of the game show. They'd be offered something of value and given a choice to exchange it for a different item; the catch being, they wouldn't know what that other item was. It has a rich history of denials and a nasty reputation of being counter-intuitive. A popular form of this problem is as [link]:

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 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?

What do you think?

On a very rudimentary level, it seems that at each decision, it does not matter what has happened before. For example, when we are first told that the car may be behind any of the three doors, we would have an equal chance of landing up with a car, or either of the two goats. Mathematically, our chance of winning is \(\frac{1}{3}\). So far so good. Now, we make a choice, say, we choose the first door. The host goes behind, checks the doors, and then opens up a door which we did not choose, and also did not contain the car. Here, he does not tell us whether the car was indeed behind the door we chose, or behind the other one which is still unopened.

Now, we are faced with a choice, whether to switch or to stay. Here, things start to get interesting. He just opened a door which contained a goat. Now, either the car is behind the one we chose, or behind the other unopened door. Eventually, most lay-people argue that since the car can still be behind any of the two doors, it makes no difference whether one switches or stays. Do you agree?

Monty Hall Problem

(Image Credits)

Well, mathematics does not agree. It claims that given such a situation, it is always favorable to switch choices! But how? Here, we'll attempt a verbose explanation, mathematical albeit. But first, let Hollywood explain it to you.

There is one big catch that often eludes the eye in this case, leading to a wrong proposition in our head for the second argument. Let us walk through the game again. On choosing any door as our first bet, we have \(\frac{1}{3}\)odds of winning. That part is true. Also, note that the chances that we chose a wrong door is two-third. But, this divides our universe (Universal set \(U\) ) of 3 doors into two. In one of those universes, only one door exists, which may contain the car with \(\frac{1}{3}\)odds of winning (probability). The other universe has 2 doors, and hence \(\frac{2}{3}\)chances of being the prize holder. Right now, we do not have any additional information about the bet or the odds.

Now, the host of the game show visits the second universe and opens a door which absolutely does not contain the car. But, by doing so, he has supplied us with new information about that universe. Now, no one told us that the odds are still equally likely, we just assumed it because we were not paying attention to the conspiracy of the universe. What we are supposed to do is to build upon our previous knowledge and deduction.

We had deduced that there are two-third chances that the prize is in this universe containing 2 doors. We just didn't choose the second universe because there were two many doors and we had only one chance! But if we were a little less blinded by greed or nervousness or excitement or ignorance, we would want to make amends. Given that we know that we have most likely made a wrong choice in the first place (because who bets a \(1/3^{rd}\) odds of winning against \(2/3^{rd}\) ?), we'd rather change our choice if given another chance!

Now, the host gives us that chance by opening one of the other two doors and by virtue of that, has exclaimed that the way we were distributing the chances of success is no longer valid. Now, he has removed one of the two choices from the universe of \(2/3^{rd}\) chances. Thus, our \(2/3^{rd}\)chances must now be distributed as 0 chances of being in the door that was opened, and \(2/3\) in the door still closed.

Based on this reasoning, it would be a better strategy to switch choices. The main thing to notice is that equal likelihood of each door holds only when there were 3 doors. After one door has been revealed, nowhere is said (but erroneously presumed) that the likelihood of the prize being behind either of the two doors is equally likely. Of course, it is still a game of chance, because even if there was a chance of 0.000...1 percent of the car being behind the door in the smaller universe, it would still be possible.

This can also be proven mathematically, by a simple application of Bayes law. Let's see how.

Let's label the doors as A, B and C. Suppose, we place our bet on door A. So, \(P(A)\)is the probability of the car being behind A.

\(P(A) = \frac{1}{3}\)

Consequently, \(P(B\text{ or }C) = \frac{2}{3}\)

Now, let the host open door B. We now know that B does not contain the car for sure. But this move happened because we bet on A. So, it is conditional on our choice.

\(P(B|A\text{ is chosen}) = 0\) is the probability that B contains the car, given that we chose A (and is hence, 0).

Now, the probability of the car being in B or C can be written as the following by the Total Probability Theorem:

\(P(B\cup C) = P(B \cup C | A) + P(B\cup C|A') = \frac{2}{3}\)

Since, we have chosen A, the \(A'\)part becomes 0.

Thus, \(P(B\cup C) = P(B\cup C | A) = P(B | A) + P(C|A) = \frac{2}{3}\)

And since we know (because the host opened the door B) that the car is not behind B, the above equation gives the result as,

\(P(C|A) = \frac{2}{3}\). Thus, given that we chose A, the probability of the car to be behind C is \(2/3\). Thus, we would be at better odds if we switched our choice!

The problem and the solution would still stay the same even if there were 1000 doors to choose from, and on choosing 1, the host opened 998 of them. Stumped?

Here's a good explanation from Khan Academy.

Well, it seems that sometimes, it is much better to focus our attention on not making a wrong choice rather than making the right choice!