The Infamous Let's Make a Deal Riddle
This famous riddle from Marilyn vos Savant's column "Ask
Marilyn," got a lot of media attention when it appeared
years ago. It became one of the most widely-debated probability
problems of the past decade. Why? Because so many logicians and
others thought she was wrong. They even wrote nasty letters to
editors of newspapers. In the end, they wrote apologies, because
Marilyn was right all along.
Let's Make a Deal
In the original riddle, you are to suppose that you are a
contestant on the old television game show "Let's Make a
Deal." The emcee, Monty Hall, has told you that behind one
of the large doors on the stage is a new car. The other doors
have either a small gift or a goat. You get to choose a door,
and keep whatever is behind it.
After you choose, Monty reveals what is behind one of the
doors you didn't choose. It isn't the car. The car, then,
is behind the door you chose or the other one. Mr. Hall gives
you the option of switching to the other one if you wish. The
question, then, is whether you should switch or not, or doesn't
it make any difference (assuming you want the car more than the
goat or small gift).
Note: Perhaps one reason people had difficulty with
this problem is that they weren't familiar with the show. Monty
Hall never opened the door with the car after a contestant
made her initial choice. In other words, if the car was
behind one of the two doors, he would open the other one, and
then give you the choice of switching. Most people would
assume this, even if they hadn't seen the show, but it is worth
The Solution to the Let's Make a Deal Riddle
To increase the probability that you'll get the car, you should
switch doors. This seems counter-intuitive to some, because it
seems that no matter what Monty has revealed, you are left with
two doors and no way to know which one has the car. The probability
would seem to be 50% for either. It isn't.
Think of it this way: If you went through this routine 300
times, you would choose a door without the car about 200 of those
times, right? (2 out of three doors have no car, so 200 out of
300 times - on average - you would choose a door without the
car) Now, each of those times Monty will eliminated the other
non-car door for you, meaning if you switch you will get the
Of course the other 100 times you will have chosen the right
door from the start, and will lose the car when you switch. However,
winning 200 out of 300 times is better than 50%, right? With
a probability of 2/3, or 66.7%, you should always switch.
There are other ways in which people have explained this riddle.
To me, the interesting thing about the riddle is that it is an
example of the relevance of knowing how to figure probabilities.
This is how the game show was really played, after all. I wonder
what the statistics are on how many contestants switched or stayed
with their first choice?