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.

 

[blue_raver22]

It is understandable that this scene from the movie may be confusing to some viewers. The concept being portrayed is known as the "Monty Hall Problem" in the field of probability. It is named after the host of the game show "Let's Make a Deal", where a similar scenario was presented to contestants.

In this problem, the contestant is given the choice of three doors, behind one of which is a prize. The host then opens one of the remaining two doors to reveal that it does not contain the prize. The contestant is then given the option to switch their choice to the remaining unopened door or stick with their original choice.

The key to understanding why switching increases the chances of winning is to consider the initial probability of choosing the correct door. At the start, the contestant has a 1/3 chance of choosing the correct door. However, after one of the incorrect doors is eliminated, the remaining unopened door has a higher probability of containing the prize. This is because the eliminated door had a 1/3 chance of containing the prize, leaving a 2/3 chance that the remaining door contains the prize.

To put it simply, by switching their choice, the contestant is essentially betting on the fact that the host has intentionally opened a door that does not contain the prize. This increases their chances of winning from 1/3 to 2/3, making it a more favorable decision.

In conclusion, while it may seem counterintuitive at first, the probability of winning does increase when the contestant switches their choice in this scenario. This is a well-known concept in the field of probability and has been studied and proven by mathematicians.