ForumsThe TavernThe Monty Hall Problem

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Programpro
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Programpro
562 posts
Nomad

Ok, I just saw the movie 21, and it still bugs me that some people don't understand this.

The Monty Hall problem involves a game show, in which there are three doors. Behind one door is a car, and behind the other two are worthless prizes. After you chose a door, the game show host reveals one door that has a worthless prize behind it. So, behind one of the two remaining doors is the car. He then offers you the opportunity to switch your choice...should you switch? I mean, there's a 50/50 chance, so it doesn't really matter, right? WRONG!!!

Okay, here's my reasoning. When you select an INCORRECT door (worthless prize), the guy has to reveal another door. He can't reveal your door, since you chose it, and he can't reveal the car. Thus, he eliminates the other incorrect door. This leaves you with your incorrect door, and the car, and switching in this case would get you the car.

So, if you choose an incorrect door, switching will get you the car. There are TWO incorrect doors, and only ONE correct door. that means that you have a 2/3 chance of getting an incorrect door, and if you get one, switching gives you the car.

So, if you switch, there's a 2 in 3 chance that you'll get the car.

Switching is the best answer. Feel free to dispute me and bring your own views to the table.

I just wanted to put my (and many others' opinion out there.

  • 5 Replies
ManUtd4life096
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ManUtd4life096
1,359 posts
Farmer

For your sake, I hope you didn't stay up all night thinking about that. But it's good reasoning.

Kanos
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Kanos
175 posts
Nomad

I think regardless its a 50/50 chance at getting the car.
Your not exactly picking 1 from 3 as they eliminate 1 option after your choice. Meaning you picking a car or a dud right from the start.

ManUtd4life096
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ManUtd4life096
1,359 posts
Farmer

Programpro, the thing is, you don't know if you picked a dud or not. You'll always have a 50/50 chance. The guy will show a Dud first, and then you're left with your door which could be the car, and the other door, which also could be the car. Your logic only works if you pick the dud.

StraightJaketMan
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StraightJaketMan
619 posts
Nomad

However, what would it mean if you choose the correct door at the time? No one really knows what is behind the other doors. No matter what the odds may be, you still can't be sure unless it's with 100% that you get the car, or, in your case, the wrong one. So, in reality, Switching may not be the best thing to do, without numbers backing you up, or a good enough reason to do so.

Yet, there are what we call "instincts," where that can alter your thoughts about what to do. So, it doesn't much rely on statistical data, or even luck, but the person who is playing at the time, and their mindset.

...well, if you want to think about the big picture, thre are a few things that could alter you're playing on the game. These would be:

1. Odds
2. Data (different from odds, by the fact that you have 100% certanty over 1 door, even if you're wrong. Also considered to be stubbornness)
3. Luck
4. Fate (differs from luck, through if you were meant to get the car, where as luck is a simple guess, with no idea at all about what you may have.)
5. The person (relates to Fate, since technically, everyone has their own fate)

So, all these factors, and some I might have not said, affect whether or not the person gets the car.

Thus is my reasoning for the Monty Hall Problem.

StraightJaketMan
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StraightJaketMan
619 posts
Nomad

well, I found this topic entertaining to think about, so I'll see what I can do to keep it alive. I still find it an interesting dillema, like most game shows out there.

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