Probability Question from Movie 21 - Help Needed

[cookiecrunch]

Hello all,


I stumbled upon a basic probability type question from the movie 21 and was hoping someone could provide me with an explanation. The question is as follows:

If i have 3 doors and behind one of the doors is a car and behind the other 2 doors is nothing and i have to choose one of the doors. In this case it doesn't really matter which door i choose because the probability of a car being behind any door is equal that is 33.33%, so say i choose door 2. Now given that i chose door 2 what if another door was added and i had an option to either leave door 2 as my choice or pick another one the answer was to pick another one to account for variability and the fact that picking between door 1,3 and 4 will increase my chances of winning.. I don't understand how the probability of 1,3 and 4 is higher given that i picked door 2 and after which another door that is door 4 was added..


Any help will be greatly appreciated!


Thanks

 

[DaveC426913]

This is a variant on the well-known and highly contentious Monty Hall Riddle. Probably easier if you read up on this first and then adapt it to your riddle.

Some very intelligent and highly respected people have weighed in on this - trouble is, different very intelligent and highly respected people have come to opposite conclusions!

Search PF for a host of posts on that very subject.

 

[Dale]

This variant doesn't work, I think. You know that door 4 does not have the car. You have gained no new information.

 

[plymouth]

It has something to do with the host's atitude towards the contestants of the tv show I believe and there are two solutions as a result ("friendly host" and "tricky host").

 

[cookiecrunch]

This variant doesn't work, I think. You know that door 4 does not have the car. You have gained no new information.

Well, practically speaking, we don't know that for sure that the car is not behind door 4 just because the host made it an option...i mean he could simply be toying with your mind..i think most people would think that given the fact that he opened the door after "I" made my choice that the car is probably not behind that door.

Actually i just listened to the question part of the movie again and it turns out that the person first chooses door 1, the car is actually behind door 2 and only the host knows that..and he opens another door that is door 4 after i make my choice..and apparently the probability of switching my choice given that another door has opened for me to choose from is better...Does this help?

 

[robert Ihnot]

DaveC426913: This is a variant on the well-known and highly contentious Monty Hall Riddle. Probably easier if you read up on this first and then adapt it to your riddle. Some very intelligent and highly respected people have weighed in on this - trouble is, different very intelligent and highly respected people have come to opposite conclusions!

Hold your horses! Didn't Marilyn vos Savant handle that problem? You can easily prove it to yourself: Let you computer generate pairs of random integers 1-3. The first integer represent your choice of door, the second the correct answer.

In the case, that you are correct, the two numbers match, and you will win. This will happen about 1/3 of the time. BUT IF YOU SWITCH, well the only number that now can be shown now with yours, if you were originally wrong, IS THE CORRECT ONE! This will be the case about 2/3 of the time.

Marilyn said that she suggested this type of tests to many people and no one failed to report that switching was better.

HOWEVER, there can be a question if we assume the announcer has some leeway in what he is going to do once you chose, but I don't see that as part of the original problem.