# The Monty Hall Problem and Card Counting

In the movie 21, Professor Mickey Rosa poses an interesting puzzle to his class – The Monty Hall Problem.  Imagine you’re on a game show and the host presents to you three doors.  Behind two of the doors are goats.  Behind one of the doors is a car. Choose the correct door and you win the car.  Let’s say you choose Door #1.  The host, who knows what’s behind all the doors, opens  Door #3  and reveals a goat.  The host then offers you the choice of staying with Door #1  or switching to Door #2.   Should you switch doors at this point?  To answer this question you must first determine the probability of choosing the car and the probability of choosing a goat.

At first blush it may seem that it’s basically a coin flip.  After all there are two doors left, one of which will reveal the car.   Either door would be a 50/50 proposition, right?  Well,  as counter intuitive as it may seem, if you switch doors  you would  double your odds of winning.  Many people emphatically believe  that after Door #3  is eliminated, the odds of wining and losing are equal.    This would be true if there were only 2 doors to choose from at the start of the game.  But, the game began with three doors which means that when you selected Door #1, you had a 1 in 3  chance of winning the car and a 2 in 3  chance of getting a goat.  If you stick with your original pick of Door #1, those odds of winning and losing remain the same.

However, if you choose to switch doors, your odds of winning increase dramatically.  The key to this problem is that the host knows where the car is and will only open a losing door.  If you switch doors, you will win if the car is behind Door #2 or Door #3 because the host will always open a door that has a goat behind it.  You essentially have two shots at winning.   By switching, you  would only lose if the car happens to be behind Door #1.

In summary, if you stay with Door #1  you have a straight 1 out 3  chance of winning the car and a 2 out 3  chance of claiming a goat.  If you switch doors you will win if the car is behind Door #2 (1/3)  or Door #3  (1/3).  Adding these two probabilities yields a 2/3 chance of winning and a 1/3 chance of losing.  The Monty Hall Problem illustrates an important principle of card counting.   Card counters use information gained from the cards as they are dealt to update the constantly changing odds of the game.  In order to earn a long term profit, a good card counter must capitalize on this  by betting proportionally larger amounts relative to player advantage – in the same way Ben Campbell correctly chose to switch doors after determining that it would increase his odds of winning.

1. Steve says:

Good stuff! Like a lot of people I was stumped when I watched 21 as to why you should switch doors. I didn’t realize how important door 3 was.

2. Mew says:

Hey Fisher!
I was wondering why the odds to win by switching isn’t 1/2 rather than 2/3. The odds are 1/3 for each door at the beginning. Once the host reveals the goat in Door #3, the odds should change from 1/3 to 1/2 only for Door #2. This is true because when Door #1 was chosen there were 3 possible outcomes. Opposed to this, there are only 2 outcomes for choosing Door #2. Am I doing this correctly? Is this why switching would theoretically raise the odds?

Thank you for your time!

Mew

3. Anonymous says:

Read it again and you will understand:
However, if you choose to switch doors, your odds of winning increase dramatically. The key to this problem is that the host knows where the car is and will only open a losing door. If you switch doors, you will win if the car is behind Door #2 or Door #3 because the host will always open a door that has a goat behind it. You essentially have two shots at winning. By switching, you would only lose if the car happens to be behind Door #1.
In summary, if you stay with Door #1 you have a straight 1 out 3 chance of winning the car and a 2 out 3 chance of claiming a goat. If you switch doors you will win if the car is behind Door #2 (1/3) or Door #3 (1/3). Adding these two probabilities yields a 2/3 chance of winning and a 1/3 chance of losing.

4. Li says:

Read it again and you will understand:
However, if you choose to switch doors, your odds of winning increase dramatically. The key to this problem is that the host knows where the car is and will only open a losing door. If you switch doors, you will win if the car is behind Door #2 or Door #3 because the host will always open a door that has a goat behind it. You essentially have two shots at winning. By switching, you would only lose if the car happens to be behind Door #1.

In summary, if you stay with Door #1 you have a straight 1 out 3 chance of winning the car and a 2 out 3 chance of claiming a goat. If you switch doors you will win if the car is behind Door #2 (1/3) or Door #3 (1/3). Adding these two probabilities yields a 2/3 chance of winning and a 1/3 chance of losing.

5. read it again and you will understand:
However, if you choose to switch doors, your odds of winning increase dramatically. The key to this problem is that the host knows where the car is and will only open a losing door. If you switch doors, you will win if the car is behind Door #2 or Door #3 because the host will always open a door that has a goat behind it. You essentially have two shots at winning. By switching, you would only lose if the car happens to be behind Door #1.

In summary, if you stay with Door #1 you have a straight 1 out 3 chance of winning the car and a 2 out 3 chance of claiming a goat. If you switch doors you will win if the car is behind Door #2 (1/3) or Door #3 (1/3). Adding these two probabilities yields a 2/3 chance of winning and a 1/3 chance of losing.