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Are you up for a Brain Buster?

I stumbled across this logic/odds problem, and I admit it took me several googles and re-reads until I finally “got it”. It twisted my brain around enough times that I thought it worthy to post. And since Ph.D. mathemeticians are also often perplexed, it made me feel good when it finally sunk in.

It’s called the “Monty Hall Problem“, named after the old game show host of “Let’s Make a Deal”. Maybe you have heard of this. The problem centers around a choice of three closed doors, one of which has a car behind it, the other two have goats. You are asked to choose a door, so you pick door #1. Then Monty Hall opens up door #3 to reveal a goat, giving you another chance to revise your choice between the remaining closed doors #1 and #2.

(interestingly, I did a silly banner with this concept last year… ):

What should you do?! That is the problem. What are the remaining odds of the last two doors to contain a car vs. a goat?

I’m reading your mind, and you are currently thinking that CLEARLY since there are two left, and we know that one has a goat, and one has a car, that it MUST be 50/50, so it is truly a coin-toss as to which door to choose.

And you are wrong!

Its a really interesting problem, and I read a number of articles on it before I figured out how to put it in terms that make it understandable.

The answer?

The answer is, that you are 2/3 more likely to get the car if you SWITCH your answer, regardless of the initial choice.

WTF?

OK, so I googled this, and found few articles that did a good job explaining this. After much reading, which you may do on your own as well, I figured out that the keys to the “odds factor” in this problem are two-fold:

1) Your initial choice - The most important piece of this puzzle is to understand that your initial choice has a 2/3 chance of picking a GOAT. Let this sink in! When you FIRST PICK, while there are still THREE DOORS to choose from, you most likely get a goat.
2) Monty always eliminates a goat - Since our pal Monty Hall has to open a goat door, this “elimination” isn’t random, and doesn’t change your odds of having a goat!

So, through this process, and because of the odds of your first choice, you will be sitting on two closed doors, with a 2/3 probability that your existing choice is a goat. This means that 2/3 of the time, the other choice must be a car.

Another way it made sense to me is to look at it in reverse: The only way you can win by not switching is if your first choice is the car.

Here’s a quick diagram of the problem that makes it pretty plain:

Pretty cool, huh? It’s funny too, just read through some of the initial comments by Ph.D. professors on Marilyn vos Savant’s blog. These math types get all fired up about this problem. And it’s really pretty simple in the end… she doesn’t do a good job explaining it though…