## A simple puzzle and a lesson in probability

At the risk of sounding immodest, let me say I consider myself above average in mathematics. But when, sometime back, I heard of the Monty Hall problem, I was forced to reconsider the assessment of my skills. The problem kept me confused for a day. Then I forgot about it. I was reminded of the problem now because I am now reading Sam Harris’ *The Moral Landscape *(Let me give you some envy saying I bought this for $1.5 at a local used goods store), where the author writes about this problem to illustrate how people tend to *feel* they are right, even when it is clearly proven to them they are wrong. Since he was so confident of the answer, I was intrigued and looked up the problem on wikipedia.

Wikipedia:

The Monty Hall problem is a probability puzzle loosely based on the American television game show

Let’s Make a Dealand named after the show’s original host, Monty Hall.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 [but the door is not opened], 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?

The puzzle is a very simple one. The twist lies in the fact that in your first attempt, you have 3 doors to choose from, while the second time you are allowed to choose, you have only two doors. It seems reasonable to argue that since you had already chosen one, and because you have only two options in the second attempt, there is really no point in switching, as either door will have a probability of 1/2 of having a car behind it.

But the correct answer is that you should always switch. One way to understand that answer is to think as follows: Imagine, that the first choice you made was indeed the door with the car. The probability for that is 1/3. The probability that the car is in one of the other two doors is 2/3. So, if you do not switch, you will get the car only in 1/3rd of your attempts, but if you switch, you will get the car in 2/3rds of your attempts. Therefore you should always switch.

Though I understood this, I was still not very satisfied. Let me make my confusion clear. The events in the current scenario are as follows:

- I choose a door.
- The host then opens a door with a goat.
- I get the option to switch, which effectively means I can choose to open one of the two remaining doors.

I wondered how different this is from the slightly different scenario given below:

- The host opens a door containing a goat.
- The host then allows me to choose one of the remaining two doors.

I felt that these two scenarios were the same and so the answer did not really convince me. Then it finally dawned. Step 1 in the first scenario and step 2 in the second scenario are not the same. They are different because, in Step2 of Scenario 1, the host chooses the door he wants to open from only two doors. While in step 1 of the second scenario, the host chooses the door from all the three doors. Therein lies the answer to my confusion.

When the host chooses from two doors, the possibilities before him are as follows (assuming the player chose Door 1):

- Both Door 2 and Door 3 have goats in which case he will open any one randomly.
- Door 2 has a car, in which case he opens Door 3
- Door 3 has a car, in whcih case he opens Door 2

So, in two of the three possibilities, he is correctly telling us which door has the car. Only in the first possibility is he misleading us (so to say). So trusting him to guide us to the right door, means we will get it right 2 out of 3 times. And so we should always switch.

But Sam Harris says in his book:

Even when people understand conceptually why they should switch doors, they can’t shake their initial intuition that each door represents a 1/2 chance of success.

I do not share that experience. Once I saw what was wrong with my initial thought process, I clearly see that we should always switch. My initial conclusion that switching makes no different to my chances of getting a car, clearly seems to be wrong now after all that analysis. What do you think? Do you still feel, even after looking at the answer, that switching does not help?

I guess it is totally random. if the whole idea is to get the Car then any door will do good. They won’t allow me to open one after another . Switching works if they allow me to open all the three doors 🙂 Else it is immaterial to play on 2/3 🙂 Anyways i wrote this one just before i read yours . http://warpoffaith.blogspot.in/2012/08/truth-is-f-functioning.html . Am sure it involves the truth that no one else in that auditorium should be working out options for it will influence. If i work this in private without even the show presenter then no one will believe me for not opening the third door for failing with first two attempts. Good Luck 🙂

I understood the solution with 3 door problem since it was easy to analyse with less number of possible combinations. I am trying to understand the mathematical proof given in the wikipedia link.

Actually I was about to comment explaining my confusion. But in that process I understood the right solution.

Switching obviously helps.

Suppose you have to choose the Ace of Spades from a pack of cards.

After you choose your card, the dealer then goes through the pack, takes one card out, and show you the rest of the cards – none of which are the Ace of Spades.

Should you switch your selection to the card the dealer removed from the pack before he showed you the 50 cards that were not the Ace of Spades?