Probability question from the movie 21

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Discussion Overview

The discussion revolves around the "Monty Hall problem" as presented in the movie "21." Participants explore the logic behind the probability of winning when switching choices after one option is eliminated. The scope includes conceptual understanding and mathematical reasoning related to probability theory.

Discussion Character

  • Exploratory
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the logic of increasing chances by switching doors, questioning whether it is simply a 50/50 scenario.
  • Another participant identifies the scenario as the "Monty Hall problem" and provides a reference for further reading.
  • A later reply explains that understanding the problem became clearer with a diagram, suggesting that switching choices effectively reverses the odds, with a 2/3 chance of winning if the initial choice is a goat.
  • However, this same participant cautions that the assumption of always having the opportunity to switch is critical, implying that without this condition, the probability may not hold.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there is a mix of understanding and confusion regarding the problem's logic and the conditions under which the probabilities apply.

Contextual Notes

Participants highlight the importance of assumptions, particularly regarding the opportunity to switch, which may affect the validity of the 2/3 probability claim.

dratsab
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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.
 
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mathman said:
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!
 
dratsab said:
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.
 

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