and behind door number three…

You may have heard this scenario before:

You’re on a game show and the host has you pick one of three doors. You win whatever is behind your door. Behind one of the doors is a car. Behind the other two doors are goats. You choose one of the doors. The host then opens one of the doors you did not choose to reveal a goat, and gives you the opportunity to switch your choice. Will switching give you a better chance of winning?

The answer is that switching will give you a 2/3 probability of winning, so you should switch.

Having just finished a graduate level course in probability, I was thinking about this problem, and how to justify the answer. It turns out that it doesn’t take a graduate-level math education, it only takes some careful thinking.  I’ll try to explain.  (I will assume that the reader has some basic knowledge about probability.  For example:  the sum of all probabilities must equal 1.)

You choose a door, say door #2. The host then opens a door to reveal a goat. You know that there is a goat and a car left, and two doors haven’t been opened, so there’s a 50-50 chance of winning the car, right? Wrong. Probability is all about information. If you receive information about a random variable, that changes it’s probability. The host just gave you information about the door he opened.  What’s not obvious is that the host also gave you information about the other door.  Think about it this way:

You chose a door and have a 1/3 chance of winning the car.  Now forget this door and concentrate on the remaining two doors.  There is a 2/3 chance that the car is behind one of them.  Well guess what?  The host just told you which of the remaining doors the car could be behind!  The door that the host didn’t open now takes on this 2/3 probability.  It’s as if the host offered you a choice between two groups:  “If the car’s behind the first door you chose, I’ll let you have it.  Or you can switch and if the car is behind either of the other two doors, I’ll let you have it.”

If you like visuals, here you go:

The prizes behind the doors can be arranged like this:
Game show doors

For any door there is only one way a car can be behind it, but two ways a goat can be behind it.  This means you have a 1/3 chance of winning the car for any door that you pick.  Again, let’s say you pick door #2.

The host is now going to open one of the doors that you didn’t pick, so let’s concentrate on these.

Remaining Doors

Notice that there’s only a 1/3 probability that both remaining doors have goats behind them.

The host opens a door to reveal a goat.  Remember there’s only 1/3 probability that both doors have goats, so the probability that the other door has a goat is 1/3.  So there’s 2/3 probability that the car is behind the door that the host did not open.  Switch and you’ll double your original chances of winning!

If this doesn’t click right away, don’t feel bad.  This problem has stumped numerous PHDs and mathematicians.  Not because it’s difficult math, however.  This is an exercise in reasoning, and because of the way it’s presented it’s not immediately intuitive.  Like an optical illusion, it’s one of those tricks that fool us humans upon first glance.

If you are interested in where this originated, check out the following link:


A mathematician, hacker, fabricator, and all around do-it-yourselfer. I like building my own tools, learning new technologies, and breaking stuff.

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