Let's Make A Deal!
Monty Hall leads you up on stage and presents you with three closed doors. He announces that behind one of the doors is a wonderful prize, and behind the other two doors is a miserable copy of the home game. He offers to give you the prize behind the door of your choosing. You reason correctly that there is an equal chance of the prize being behind each door, so you pick Door #1 at random.
In an effort to make things interesting, Monty announces that of the remaining two doors, he will randomly pick one of the doors that does not have the glittering prize and open it. He then proceeds to do exactly that, and throws Door #2 wide open revealing one of the two consulation prizes.
Monty then offers you a choice. You may choose to stay with Door #1, or you may now change your mind and switch to Door #3. What should you do to maximize your chance of winning the non-cardboard prize, and why? Come up with your answer before reading on.
You should always switch. That's what they want you to do. Reverse-reverse psychology. Just kidding. But you should still switch. This phenomenon is hard to grasp since it seems unintuitive, so I'm offering three explanations. Try each one until you get it.
1) Logic. Before Monty opens the door, there is a 2/3 chance that the prize is behind either door 2 or 3. Why should that change when Monty opens one of them? If Monty said to you, one of the doors, 2 or 3 has a glorified paperweight behind it, what information is gained? Remember, Monty said he would open a non-grand-prize door. We already know that at least one of Doors #2 and #3 contains an unspectacular prize. We do not need to scrap the information from our original analysis. As a result the 2/3 chance is now attributed entirely to Door #3.
2) Reason. Doors #2 and #3 undergo a process to which Door #1 is immune. This process involves randomly opening one of the doors that does not have the grand prize. That means that the remaining door of the two has survived a random elimination. It is more likely that the door survived because it has the prize.
3) Math. Say you pick Door #1, and your strategy is to switch doors. Now before Monty opens a door, there are three possibilities which are all equally likely:
Prize is behind Door #1: (1/3)
Prize is behind Door #2: (1/3)
Prize is behind Door #3: (1/3)
So if you switch, 1/3 of the time the prize is behind Door #1 and when you switch you will have a 0% chance of winning. However, 1/3 of the time it's behind Door #2, and when you switch you will have a 100% chance of winning. Also 1/3 of the time the prize is behind Door #3, and when you switch you will also have a 100% chance of winnning. So by the Law of Total Probability:
[(1/3) * 0%] + [(1/3) * 100%] + [(1/3) * 100%] = 2/3
So, if your strategy is to switch, you will have a 2/3 chance of winning.
This is a classic probability problem, there are versions with 4 doors too. The following site has links to some great pages that explain the problem in more detail and even have door choosing simulation scripts: http://www.letsmakeadeal.com/problem.htm
In an effort to make things interesting, Monty announces that of the remaining two doors, he will randomly pick one of the doors that does not have the glittering prize and open it. He then proceeds to do exactly that, and throws Door #2 wide open revealing one of the two consulation prizes.
Monty then offers you a choice. You may choose to stay with Door #1, or you may now change your mind and switch to Door #3. What should you do to maximize your chance of winning the non-cardboard prize, and why? Come up with your answer before reading on.
You should always switch. That's what they want you to do. Reverse-reverse psychology. Just kidding. But you should still switch. This phenomenon is hard to grasp since it seems unintuitive, so I'm offering three explanations. Try each one until you get it.
1) Logic. Before Monty opens the door, there is a 2/3 chance that the prize is behind either door 2 or 3. Why should that change when Monty opens one of them? If Monty said to you, one of the doors, 2 or 3 has a glorified paperweight behind it, what information is gained? Remember, Monty said he would open a non-grand-prize door. We already know that at least one of Doors #2 and #3 contains an unspectacular prize. We do not need to scrap the information from our original analysis. As a result the 2/3 chance is now attributed entirely to Door #3.
2) Reason. Doors #2 and #3 undergo a process to which Door #1 is immune. This process involves randomly opening one of the doors that does not have the grand prize. That means that the remaining door of the two has survived a random elimination. It is more likely that the door survived because it has the prize.
3) Math. Say you pick Door #1, and your strategy is to switch doors. Now before Monty opens a door, there are three possibilities which are all equally likely:
Prize is behind Door #1: (1/3)
Prize is behind Door #2: (1/3)
Prize is behind Door #3: (1/3)
So if you switch, 1/3 of the time the prize is behind Door #1 and when you switch you will have a 0% chance of winning. However, 1/3 of the time it's behind Door #2, and when you switch you will have a 100% chance of winning. Also 1/3 of the time the prize is behind Door #3, and when you switch you will also have a 100% chance of winnning. So by the Law of Total Probability:
[(1/3) * 0%] + [(1/3) * 100%] + [(1/3) * 100%] = 2/3
So, if your strategy is to switch, you will have a 2/3 chance of winning.
This is a classic probability problem, there are versions with 4 doors too. The following site has links to some great pages that explain the problem in more detail and even have door choosing simulation scripts: http://www.letsmakeadeal.com/problem.htm
posted by flowbeus at 4:41 PM
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