Monty Hall Problem

1 October 2010, 22:41

The Monty Hall problem (sometimes called Monty Hall Dilemma or Monty Hall Paradox) is one of my favorite math problems involving probability. I first learned about it long ago, when I just started to learn more in depth about probability in college. And like many other students I was not immediately convinced and satisfied with the solution given by my lecturer at that time. The counter-intuitive nature of the solution to the problem is what intrigued me the most. Let me try to explain for those of you not familiar with this problem.

Supposed that in a game show there are three doors, behind one of which is a prize (let say a car). There’s nothing behind the other two doors. You choose one of the doors, say door no 1. The host of the game, who knows which door hides the car, opens another door, say no 3, which is empty. You are then given a choice again whether you want to stick to your first choice of door no 1 or whether you want to switch to door no 2. Now the question is, is it better for you to switch? In other words, do you have higher probability of getting the car by switching?

Note that in the original problem, the other two doors hide a goat each instead of being empty. I modified it a little in case some people consider a goat to be a better prize than a car. At first glance it seems simple enough that after the host has opened door no 3, there’s a 50-50 chance that the car is behind door no 1 or door no 2. And so it doesn’t matter whether you switch or not. But, is that really the case?

Well, let’s try to look at the problem from another angle. When you first chose door no 1, there was a 1/3 chance that that car was behind it. But it also means that there was a 2/3 chance that the car was not behind it. In other words, there’s a 2/3 chance that the car was behind door 2 or door 3. Now when the host opens door no 3, which is empty, the probability remains the same (because the host always open the empty door, since he knows what are behind the doors). So your chance of getting the car is actually higher if you switch.

There are other ways of understanding the problem, which I’m not going to explain all here. I list down some references below if you want to read and learn further about this problem. The mathematical proof of the solution involves conditional probability theory and can be shown using Bayes’ Theorem.

You can also try to understand the problems by listing out all the possible outcomes when you switch as follows:

  • Outcome 1
    • The car: door no 1
    • Your first choice: door no 1
    • The host open door no 3
    • You switch, and you get nothing
  • Outcome 2
    • The car: door no 2
    • Your first choice: door no 1
    • The host open door no 3
    • You switch, and you get the car
  • Outcome 3
    • The car: door no 3
    • Your first choice: door no 1
    • The host open door no 2
    • You switch, and you get the car

From the above we can see that 2 out of 3, you will get the car when you switch. I hope by now you can understand the problem and the solution better. And if you still need further proof or explanation, you can try some the following resources:

~ By Jimmy Sie




  1. You and your readers may enjoy working out the solution to this variation (which I have not seen published anywhere else so far). Imagine there are nine doors arranged in a three by three grid (like a tic-tac-toe game). Behind one door is the car; the other eight doors hide goats. You choose one door. Monty Hall then selects one row and one column, and opens those five doors. Of course he does so in a way that does not include either the door you picked or the door with the car. Should you switch to one of the three other unopened doors, and if so, which one? What are your odds of winning with each possible choice?

    Sam Wilson · 28 November 2010, 19:38 · #

  2. Thank you, Sam, for your contribution.

    Jimmy · 19 December 2010, 20:29 · #

  3. Here’s my attempt at solving Sam’s version (encoded in Rot13):

    V guvax lbh fubhyq fgneg ol cvpxvat gur zvqqyr qbbe. Vs gur pne vf va nal bs gur sbhe pbearef, Zbagl jvyy unir gb pubbfr gur bccbfvgr pbeare nf gur “vagrefrpgvba” bs gur ebj-naq-pbyhza. Ba gur bgure unaq, vs vg vf ba n fvqr, ur jvyy unir gb cvpx bar bs gur gjb “bccbfvgr pbearef”. Va nal pnfr, ol fgnegvat jvgu gur zvqqyr naq gura fjvgpuvat gb gur bccbfvgr bs Zbagl’f vagrefrpgvba qbbe, lbh jvyy vzcebir lbhe punaprf sebz 1/9 (ab jnl sbe vg gb or qvssrerag) gb 4/9 (V guvax). Urer’f n oernxqbja:

    (Ybpngvba bs pne • Bqqf bs orvat gurer • Vs gung’f jurer gur pne vf, gura ubj bsgra qb lbh jva ol guvf grpuavdhr?)
    (Pbeare • 4/9 • nyjnlf)
    (Fvqr • 4/9 • arire; lbh cvpxrq n pbeare naq vg jnf ba gur fvqr)
    (Zvqqyr • 1/9 • arire)

    Ubjrire, gurer znl or na rira orggre fgengrtl cbffvoyr.

    Lenoxus · 9 May 2012, 20:48 · #

  4. there’s a really good explanation of this with a diagram in the book “the curious incident of the dog in the nightime”

    _ · 17 May 2012, 00:25 · #