The short answer: You double your chances of picking the right door if you switch your choice.

Game ShowThe long answer
This is actually a very famous puzzle, known as The Monty Hall Paradox, and was the subject of fierce argument and debate among mathematicians after it was published in Parade magazine in 1990. The question was submitted by a reader to the magazine's resident genius, Marilyn Vos Savant. Her reply prompted hundreds of letters from scientists, mathematicians, schoolteachers and others disputing her answer and, in some cases, denouncing her as a fraud. This was because her answer was "counter-intuitive", and appeared to defy logic. Nevertheless, her answer was correct, and was subsequently confirmed in practical experiments.

Marilyn's reasoning went as follows: Your chance of selecting the right door is one in three; therefore the chance that you have not chosen the right door is two in three. By swapping your door for the one selected by the host, you double the chance that you have the right door.
Her opponents argued that the odds of any door being the right door are always one in three. Therefore the door which the host offers you is no more likely to be the correct door than the one you already have, and you might as well stay with your initial choice.
In response, Marilyn pointed out that if there were a hundred doors, instead of three, then the chance that you have the right door would be only one percent, and the chance that the right door is one of the others would be ninety-nine percent. Therefore your chance of swapping the wrong door for the right door would be ninety-nine, against one chance of changing the right door for the wrong door.
Marilyn's maths were compelling; and yet logic would seem to dictate that if there are two doors, and one of them holds a prize, the chances of choosing the right door should always be 50-50.
In fact, however, this is not the case. Impossible as it may seem, there is a 2/3 chance of the unselected door being the correct one, and only a 1/3 chance of the selected door being the correct one.

At the beginning, the contestant is faced with three choices, doors A, B, and C. A prize is hidden behind one of the doors, so there is a 1/3 chance that each of the doors holds the prize.
The contestant makes a choice, which changes the odds. Now there is a 1/3 chance that the chosen door holds the prize, and a 2/3 chance that one of the other two doors holds the prize. For example, if the contestant chooses A, there is a 1/3 chance that A holds the prize, and a 2/3 chance that either B or C holds the prize.
Next, one of the unselected doors is opened to reveal that it does not contain the prize. In our example, door B is opened and shown to not contain the prize. Now the contestant knows that the prize is behind either A or C, and you might think that each A and C has a 1/2 chance of containing the prize.
However, the revelation that B does not contain the prize has not changed the odds. There is still a 1/3 chance that A has the prize and a 2/3 chance that either B or C has the prize. But since it is now known that the odds of B having the prize is zero, all of the 2/3 chance that either B or C has the prize can be assigned to C, and the odds become 1/3 for A and 2/3 for C.
Thus, the contestant will win the prize 2/3 of the time by changing his selection, and only 1/3 of the time by not changing his selection.
If you still don't believe it, here is the explanation expressed in mathematical terms (courtesy of Richard W. Hamming):

The cases in the sample space are:

p(A,B) = 1/6 = p(A,C)
p(B,A) = (1-p)/3 = p(C,A)
p(B,C) = p/3 = p(C,B)

As a check, we note that the total probabilities add up to 1. The fact that the host opens the door with no prize eliminates p(B,A) and p(C,A) from our sample space. So the probability of winning is now:

P = {p(A,B) + p(A,C)}/{1 - p(B,A) - p(C,A)} = {1/3}/{1 - 2p)/3} = {1/3}/{1/3 + 2p/3} = 1/{1+2p}

So that:

For p=1 (host never reveals the prize) your P=1/3.
For p=1/2 (host randomly chooses), your P = 1/2.
For p=0 (host always reveals the prize if you haven't chosen it), your P = 1.

And if you still don't believe it after all that, you can verify it for yourself on a simulated version of the game show (created by the Dept. of Statistics, Univ. of South Carolina) , by clicking here.
You can also test it out with three playing cards and the help of a friend
Remove two face cards and a Queen from a pack of cards. Ask your friend to shuffle them and place them on a table, face down. Choose a card. Now have your friend play the game-show host by inspecting the undersides of the cards and revealing one of the face cards to you. Repeat this ten times staying with your original pick, then another ten times with the switch-to-the-other-card strategy. You will find that you pick the Queen twice as often when you switch from your original choice.

The strangefulness is terrific
To appreciate how truly weird all this is, consider the following. When you first make your choice, the odds that you have picked the right door are 1/3; and the car is just as likely to be behind the door you choose as it is to be behind either of the other two doors. Yet when one of the two doors concealing a goat is opened, the chances of the door you choose being the correct one change, and it is now twice as likely that the other unopened door conceals the car. Yet the car was just as likely to be behind the door you chose, at the beginning, as it was to be behind the door which is now twice as likely to conceal it. The car hasn't changed position - so how can it be more likely to be behind one door than another at the end of the game?
Here's another way of looking at it: suppose you are asked to pick one of three identical boxes, numbered 1, 2 and 3, and you are told that one of these boxes contains a valuable object - let's say a gold ring - and the other two boxes are empty. You pick a box - say box No3 - whereupon one of the remaining two boxes - say box No2 - is opened and shown to be empty. At this point, the chances of the box you chose being the one containing the ring become less than the chances of box No1 containing it. Yet at the beginning there was an equal chance of the ring being in box No1 and box No3.
Think about it. As far as you were concerned, you had a 50% chance of choosing the box containing the ring. Yet in reality no matter which box you picked, it would be only half as likely to contain the ring as the box you didn't pick! If you picked box A, box B would automatically become twice as likely to contain the ring. If you picked box B, box A would become twice as likely to contain the ring.
It sounds impossible - yet it can be proved quite easily.

Mind reading?
The Monty Hall Paradox works even when the boxes are imaginary!
Try this with a friend. Tell your friend that you are visualizing three boxes numbered 1, 2 and 3, and that one of these imaginary boxes contains a ring (you have to decide which box this is). Ask your friend to choose a box. Now tell your friend the number of an empty box (not the box they've chosen), and ask them: "Do you want to stick with the box you chose, or would you like to change your mind and go for the other box instead?"
If you try this a number of times, and keep a record of the number of times your friend guesses the right box, you'll find that he or she is twice as likely to make a correct guess if he or he switches boxes.
Yet these boxes don't actually exist! Which brings this piece of weirdness into the realms of mind reading.

Now here are a couple of questions for you to ponder (or, if you are the practical type, experiments for you to try):
Suppose someone shows you two boxes, marked A and B, and asks you to choose one.
Let's suppose you pick box A.
After you have made your choice you are told that there was originally a third box, box C, but this third box was thrown into the bin before you arrived, because it had been opened and found to be empty.
You are now given the option of changing your mind or sticking with box A.
The question is this: Does the elimination of a third box, whose existence you were not even aware of, reduce the chances of the box you picked being the correct one?
Remember, in this instance you have only been given a choice between two boxes.
The second question is this: What happens if, instead of one person being asked to choose a door (or a box), two people are asked to make this choice? In the three doors example above, a contestant in a game show is asked to choose one of three doors. But what if there are two contestants, and the second contestant is offered the same choice and picks a different door?
For example, let's suppose that contestant A chooses door No1, and contestant B chooses door No3. The host opens door No2 to reveal a goat. He offers contestant A the option of changing his or her mind and switching to door No3. He then offers contestant B the same option - ie, to switch from door No3 to door No1. Now, since we know that switching doors doubles the chances of being correct, this would mean that both contestants are more likely to choose the correct door if they switch their choice of doors - which is clearly impossible, since the car can only be behind one door at a time!
So how can this apparent contradiction be explained?

The Monty Hall Paradox is very strange, and the more one thinks about it, the more strange it seems; and it demonstrates that reality does not always conform to our notions of what is logical or possible.

Here are a few of the critical letters from scientists and mathematicians sent to Parade magazine:

Robert Sachs, Ph.D., George Mason University
There is enough mathematical illiteracy in this country, and we don't need the world's highest IQ propagating more. Shame!

Scott Smith, Ph.D., University of Florida
I am in shock that after being corrected by at least three mathematicians, you still do not see your mistake.

Kent Ford, Dickinson State University
I am sure you will receive many letters from high school and college students. Perhaps you should keep a few addresses for help with future columns.

W. Robert Smith, Ph.D., Georgia State University
You are utterly incorrect . . . How many irate mathematicians are needed to get you to change your mind?

E. Ray Bobo, Ph.D., Georgetown University
If all those Ph.D.'s were wrong, the country would be in very serious trouble.

They were all wrong!


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