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Monty Hall problem (en.wikipedia.org)

submitted 1 week ago by [email protected] to c/[email protected]

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[–] [email protected] -2 points 1 week ago* (last edited 1 week ago) (1 children)

At start you have 1/3 chance if you randomly pick a door.

When you remove 1 door, you have 1/2 which is better.

But I'm with you, it's bizarre that it works 🤷🏼‍♀️!

Edit: For the non math curious downvoters: it is exactly how it works, you start out with a 1/3 choice (~33% win chance). If you randomly chose again after the door opening (the door conceniently being empty) you have 1/2, which is ... Better. So it is better to change. You can try it out with 10 or a houndred doors, the result is actually the same, you just get stuck on the "forced" change of doors because there are so few to chose from.

[–] [email protected] 7 points 1 week ago (1 children)

This is not how it works, this way you wouldn't improve your chances by switching doors.

[–] [email protected] 1 points 1 week ago* (last edited 1 week ago) (1 children)

Well explain how it works then. Because you know, it does work.

[–] [email protected] 4 points 1 week ago (1 children)

Ok, assume you pick door number one. There's three possibilities; the car is either behind door number one, two or three. Now:

First scenario, the car is behind door number one. Monty opens door two or three, you switch doors and don't win a car.
Second scenario, the car is behind door number two. Monty opens door three, you switch doors to number two and win a car.
Third scenario, the car is behind door number three. Monty opens door two, you switch doors to number three and win a car.

So two out of three times, you've won a car by switching doors. So you have a 2/3 chance of winnin by switching, or a 1/3 chance by not switching.

[–] [email protected] 2 points 1 week ago

Very good explanation, thanks!

I didn't provide the exact numbers, but stated that because the odds are now better for a random selection, is the reason for switching. Am I that bad at explaining 😭