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Comments: "The Monty Hall Problem"

http://www.techinterview.org/Puzzles/fog0000000045.html

Reading this problem, common sense tells you that you have 50/50 odds, regardless of your first selection. So when I saw your solution, I was intrigued. On the surface it looks correct, so I wanted to know why it was that my common sense failed me.

So, aha! I think I've found a flaw in your solution.

You start off correctly showing the three possible scenarios:

          door 1      door 2      door 3
case 1      $$          goat        goat
case 2      goat          $$          goat
case 3      goat        goat          $$

However, where the logic falls down is that once Monty has selected to show you a goat, there are now FOUR possible situations, vaguely related to your original three cases.

For the sake of illustration, let's assume you picked door 3 (it works out identically, regardless of which door you select, as you'll see):

In both case #1 and #2, Monty has a pre-described path he must follow; that is, to show you the _other_ door with a goat, since your door also has a goat.

However, in case #3, Monty has _two_ paths he may elect to take: he can open either door. Case #3 has become #3a and #3b, as such:

          door 1      door 2      door 3
case 1      $$        <goat>      [goat]
case 2    <goat>        $$        [goat]
case 3a    goat        <goat>        [$$]
case 3b    <goat>        goat        [$$]

[] = my door choice, <> is what Monty showed me.

Now you can see that your chances really are 50/50.

To sum up: it doesn't matter which door you selected first: Monty has at least one goat to show you. That means you're left with one good and one bad door, and from a statistical point of view, the past is irrelevant: it doesn't matter whether you selected the good door first or the bad door first. You're left with a 50/50 decision between good door and bad door.

Brad Wilson
Saturday, November 17, 2001

By the way, in response to Michael's suggestion, I google'd for "The Monty Hall Problem" and had a bunch of hits. The result is that, on average, people agree with the solution posted to TechInterview, and not mine.

Since this is supposed to be a statistics problem, there's nothing better than a nice little bit of statistics to prove what's going on. Make the run size large enough, and the statistics won't lie to you.

I wrote a C# console program that uses a runsize of at least one million. It performs the exact steps outlined in the original problem, namely:

- A door is randomly selected as "correct"
- A door is randomly selected as "contestant choice"
- If contestant's choice is wrong, then Monty MUST by definition show the OTHER wrong door (no choice for Month); if the contestant's choice is right, then I randomly select a door from the remaining two doors

I ran the test in two ways; in one, the contestant always switches, and in the other, the contestant never switches.

Both runs yield identical results: virtually 50% of the time the contestant picks the right door.

If you'd like the C# code, please contact me at http://www.quality.nu/contact.aspx and I'll e-mail it to you (it should be trivial to convert to Java or C++ if you want to run it yourself and don't have the .NET Runtime & command line compilers).

Brad

Brad Wilson
Saturday, November 17, 2001

D'oh!

Don't you know that ... just when you think you have the thing solved, you find something stupid. :-p

In my C# code, I called the Next method on the Random object, like so:

int CorrectChoice = rand.Next(1,3);

Of course, the simple docs for Next() docs weren't clear that the upper bound is never reached... yielding integers between 1 and 2. I wasn't printing out every choice, since I was running millions of iterations. I happened to find it out indirectly.

Turns out the original math (66 2/3% success when always switching doors) was indeed correct. My bad!

Brad Wilson
Saturday, November 17, 2001

You are correct that there are 4 cases, but 3a and 3b each occur with probability 1/6. 

Larry Rosenstein
Tuesday, November 20, 2001

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