How does the Variable Change strategy work in the Monty Hall problem?

[ultimateguy]

I've seen this problem explained in the movie 21, as well as the show Numbers. I'll use the example given in 21.

You're on a gameshow, and you're shown 3 doors. Behind one of the doors is a new car, and behind the other 2 are goats. You pick door number 1.

The host then opens up door number 3, behind which a goat is revealed. The host then says "do you want to switch to door number 2, or stay with door number 1?"

Basically, it is said that it's in your best interest to change your answer to door number 2. When you first picked door 1, you had a 33.3% chance of getting the car, but once door 3 was revealed, door 2 now had a 66.6% chance of it being the car, because of variable change.

I have a BSc. in Physics, and this makes no sense to me whatsoever. Here's an argument: you initially pick door 2 in the exact same scenario. By the same logic, you should switch to door 1 after it's revealed that door 3 has a goat behind it. Using the same logic in both scenarios, you would be right in one and wrong in the other.

Regardless of what came before, you're still faced with 2 doors in which only one is correct... hence the odds are 50/50.

Comments? Corrections?

 

[NoMoreExams]

http://en.wikipedia.org/wiki/Monty_hall_problem remember what conditional probability is.

 

[ultimateguy]

Wow, I actually get it now. What a confusing problem! :yuck:

 

[sepmayr]

I've seen this problem explained in the movie 21, as well as the show Numbers. I'll use the example given in 21.

You're on a gameshow, and you're shown 3 doors. Behind one of the doors is a new car, and behind the other 2 are goats. You pick door number 1.

The host then opens up door number 3, behind which a goat is revealed. The host then says "do you want to switch to door number 2, or stay with door number 1?"

Basically, it is said that it's in your best interest to change your answer to door number 2. When you first picked door 1, you had a 33.3% chance of getting the car, but once door 3 was revealed, door 2 now had a 66.6% chance of it being the car, because of variable change.

I have a BSc. in Physics, and this makes no sense to me whatsoever. Here's an argument: you initially pick door 2 in the exact same scenario. By the same logic, you should switch to door 1 after it's revealed that door 3 has a goat behind it. Using the same logic in both scenarios, you would be right in one and wrong in the other.

Regardless of what came before, you're still faced with 2 doors in which only one is correct... hence the odds are 50/50.

Comments? Corrections?

Hi, I am Josef:
What you need to keep in mind is that the concept of "chance" is based on statistics. If you flip a coin 5 times, you might get "heads" five times. But if you flip it 1000 times, you will end up with somthing like 487 heads and 513 tales. The more often you flip it, the closer you'll get to 50/50. Now if you have a 1/3 chance and you try 1,000,000 times, you will be quite close to hitting it one third of the time. In any case, a single try does not play out the whole scenario. If you hit the "car" the first time. it does not have anything to do with luck, it simly means that at the time you made your choice the time was up for your pick to be right. One more thing: To make this a true scientific scenario, we need to control the variables: We either need to keep picking the same door while the location of the car randomly changes; or we can leave the car at the same door and randomly set the selection.
It is the subject of current cuttinge-edge theoretical science to find a (possible) sequence in which (let's say) heads and tails occur to mount up to a 50/50 chance.

 

[HallsofIvy]

Wow, I actually get it now. What a confusing problem! :yuck:

About 20 years ago I had an introductory course in Probability and Statistics and the textbook gave this problem as a homework problem in chapter 1!

 

[Tac-Tics]

About 20 years ago I had an introductory course in Probability and Statistics and the textbook gave this problem as a homework problem in chapter 1!

On the one hand, that's really cruel considering the number of mathematically educated people get it wrong (and are certain they're right).

As an introductory example for why we need a formal study of statistics, though, there is no better.

 

[sepmayr]

On the one hand, that's really cruel considering the number of mathematically educated people get it wrong (and are certain they're right).

As an introductory example for why we need a formal study of statistics, though, there is no better.

This is perhaps because in dayly life do not get thousands of shots at any given scenario and actually do not get the benifit of the odds. One shot, that's it!

 

[NoMoreExams]

This is perhaps because in dayly life do not get thousands of shots at any given scenario and actually do not get the benifit of the odds. One shot, that's it!

Def. disagree. The law of large numbers is used a LOT in finance, insurance, etc.

 

[HallsofIvy]

I've seen this problem explained in the movie 21, as well as the show Numbers. I'll use the example given in 21.

You're on a gameshow, and you're shown 3 doors. Behind one of the doors is a new car, and behind the other 2 are goats. You pick door number 1.

The host then opens up door number 3, behind which a goat is revealed. The host then says "do you want to switch to door number 2, or stay with door number 1?"

Basically, it is said that it's in your best interest to change your answer to door number 2. When you first picked door 1, you had a 33.3% chance of getting the car, but once door 3 was revealed, door 2 now had a 66.6% chance of it being the car, because of variable change.

I have a BSc. in Physics, and this makes no sense to me whatsoever. Here's an argument: you initially pick door 2 in the exact same scenario. By the same logic, you should switch to door 1 after it's revealed that door 3 has a goat behind it. Using the same logic in both scenarios, you would be right in one and wrong in the other.

No, you would be "right" to change in both cases because that would increase your odds. Saying the your odd increase does not mean you are certain to win!

Regardless of what came before, you're still faced with 2 doors in which only one is correct... hence the odds are 50/50.

Comments? Corrections?

The crucial point is that the host has additional knowledge: he know which of the doors has the car is behind and uses that information in choosing which door to open.

Let's say you choose door A initially. If the car is behind door A, which has a 1/3 a-priori probability, the host can open either of the doors B and C. Switching your choice to the remaining door means you "lose"-not switching you win. If the car is behind either B or C, (each of which has a 1/3 a-priori probability so 2/3 probability of either B or C) the host opens the other one. Switching you "win"- not switching you lose. That is, if you switch you have a 2/3 chance of winning, if you dont' switch 1/3.

By the way, although it is a bit more complicated, it is possible to show that if you assume the host does NOT know which door the prize is but just happens to open a door it is not behind, then it does not matter whether you switch or not.