The Monty Hall Math Puzzle

By Daniel Miessler on July 6th, 2006: Tagged as Mathematics | Programming | Science

A few weeks ago a buddy of mine busted out the MHP on me. For those not familiar, it’s a now-famous mathematics puzzle named after the talk show host who hosted the show it’s associated with.

The idea is simple, and is covered in depth at Wikipedia, but for those looking for a simple rundown it’s basic concept is this: you’re given three doors and told one has a car behind it while the other two have goats. You are asked to pick a door. After your choice is made you’re then showed a goat behind one of the other two doors.

You are then asked if you want to switch to the remaining door or keep the one you chose initially.

The natural inclination is to think that since there are only two open doors remaining, and you already have one of them, that it’s precisely equal odds between them — and therefore not beneficial (or detremental) to your odds to change your choice.

That logic, however, is flawed.

It is, in fact, beneficial to change your choice. In reality, it takes you from a 33% chance to a 66% chance when you switch, or from 1/3 to 2/3.

If you doubt this go spend some time at Wikipedia. Anyway, I was so intruiged with the puzzle that I decided to write a bit of code to illustrate how this plays out in reality. There are many such programs out there already; I just wanted to do it for myself.

This code is rather crude, which it turns out is something of a signature of my little wannabe coding projects, but it was fun to write. I don’t have the time right now to write the decision process portion of the puzzle, and currently it just models the staying and leaving dynamic, but you get the idea.

(Here’s the code for anyone interested) And here are some results given four different sample sizes: daniel@marius mhp $ ./mhp.py 10 If you STAY you win 38 times. If you SWITCH you win 62 times. daniel@marius mhp $ ./mhp.py 1000 If you STAY you win 319 times. If you SWITCH you win 681 times.

daniel@marius mhp $ ./mhp.py 10000 If you STAY you win 3360 times. If you SWITCH you win 6640 times. 40

daniel@marius mhp $ ./mhp.py 100000 If you STAY you win 33319 times. If you SWITCH you win 66681 times. The results are quite clear: 1/3 vs. 2/3. If you make a choice and stay with it you win 1/3 of the time; if you switch from that initial choice you win 2/3 of the time.

And that's with completely random (as much as pseudo can be "completely") placement of both the car and the contestant's choice.

Very interesting indeed. It is, in fact, better to switch.

  • Carl M

    Well, this is indeed a classic example of our faulty intuition when it comes to estimating probabilities. I have always been puzzled however by the fact that people create simulations to get a better “understanding” of what is going on.

    While I’m all for simulations in many situations (and in any case it’s perhaps a good exercise to create them), a simulation is not needed to analyze this particular puzzle because there are only 9 possible situations. The prize can be behind any of Door 1, Door 2, and Door 3, and we can select any of Door 1, Door 2, and Door 3. There are nine possible combinations.

    Though we could analyze all 9, let’s just look at three. Suppose that the prize is behind Door 1. If we initially select Door 1 and then switch to the unshown door (whichever it is – in this case the host could reveal EITHER Door 2 or Door 3), we will lose. On the other hand, if we select Door 2 then the host MUST reveal Door 3 (the host will not reveal the prize). So the unshown door is Door 1 (the PRIZE). Switching to that door will cause us to WIN. A similar situation holds if we initially select Door 3. So, in 2 of the 3 cases, switching wins. It’s as simple as that.

    Simple it may be, but intuitive it is not. That is why this is a standard example in my Elementary Statistics classes.

    PS If you want another nifty example of a counterintuitive probability result (related in fact to error correction in computers), let me know. :D

  • Carl M

    Well, this is indeed a classic example of our faulty intuition when it comes to estimating probabilities. I have always been puzzled however by the fact that people create simulations to get a better “understanding” of what is going on.

    While I’m all for simulations in many situations (and in any case it’s perhaps a good exercise to create them), a simulation is not needed to analyze this particular puzzle because there are only 9 possible situations. The prize can be behind any of Door 1, Door 2, and Door 3, and we can select any of Door 1, Door 2, and Door 3. There are nine possible combinations.

    Though we could analyze all 9, let’s just look at three. Suppose that the prize is behind Door 1. If we initially select Door 1 and then switch to the unshown door (whichever it is – in this case the host could reveal EITHER Door 2 or Door 3), we will lose. On the other hand, if we select Door 2 then the host MUST reveal Door 3 (the host will not reveal the prize). So the unshown door is Door 1 (the PRIZE). Switching to that door will cause us to WIN. A similar situation holds if we initially select Door 3. So, in 2 of the 3 cases, switching wins. It’s as simple as that.

    Simple it may be, but intuitive it is not. That is why this is a standard example in my Elementary Statistics classes.

    PS If you want another nifty example of a counterintuitive probability result (related in fact to error correction in computers), let me know. :D

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