Probability question from the movie 21

[dratsab]

It has been over a year since I saw the movie, so I apologize if I'm not descriptive enough. If you have seen the movie, then you know that scene where Kevin Spacey says there are 3 doors, and the main character picks one, then the other is eliminated, and so the main character increases his chances of picking the right door, by then switching his choice. I didn't understand the logic in this, wouldn't it just be 50/50? According to the movie, he increased his chances of over 50%, even though it's been narrowed down to two doors. Totally confused me.

 

[mathman]

It sounds like you are asking about the famous "Monte Hall problem", although your description looks a little garbled. Reference:

http://en.wikipedia.org/wiki/Monty_Hall_problem

 

[dratsab]

It sounds like you are asking about the famous "Monte Hall problem", although your description looks a little garbled. Reference:

http://en.wikipedia.org/wiki/Monty_Hall_problem

That's it, I didn't understand how it worked until I saw the diagram. Now it makes perfect sense to me, what it is basically doing is reversing your odds, because by picking the car you will lose by switching, so you want to pick a goat, which has a 2/3 chance of happening... then switching. If you plan to switch from the start, then you would want to pick a goat, so the other would be eliminated. Thank you!

 

[Mensanator]

That's it, I didn't understand how it worked until I saw the diagram. Now it makes perfect sense to me, what it is basically doing is reversing your odds, because by picking the car you will lose by switching, so you want to pick a goat, which has a 2/3 chance of happening... then switching. If you plan to switch from the start, then you would want to pick a goat, so the other would be eliminated. Thank you!

But don't forget - you are making the assumption that you will ALWAYS be given the opportunity to switch. If not, then "win 2/3 of the time by switching" does not necessarily apply.