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3 Doors...1 car = 1/3 ??? I thought.


Pinch Paxton

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Don't worry about it if you haven't; it's very famous, but to understand it requires a lot of knowledge about quantum mechanics and special relativity. if you don't know about these then you won't understand it.

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  • 3 weeks later...
Yes that sounds perfectly correct, but mathematically your odds are supposed to improve if you change your mind. I shall post the maths later, but first I want to see if anyone can work it out. Supposedly, if you make a computer program that keeps changing its mind, you do actaully win more often. This completely confuses me.

 

 

Then how would Vegas ever get money? Gamblers Ruin...if you play enough you will lose.

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  • 4 weeks later...

This is another problem I got.

 

Monty Hall problem, tricked a shitload of statisticians back when it came out (70's or so I think).

 

Here is the reasoning. I don't know if there is a proof for it.

 

On your first pick, you have 1/3 chance of getting it right. That means that there is a 2/3 chance the car is in one of the other doors. Since the other goat is eliminated, there is now a 2/3 chance the car is in the remaining door.

 

So the best option is to switch.

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  • 1 month later...

We had this exact problem as a Problem of the Week in 8th grade math. Many people have repeated what I'm about to say, but I'll try to clarify. On your first guess, you have 3 choices: door 1, 2, or 3. You have an equal probability of picking each door, so the chances of getting the car on your first guess are 1/3. On your second guess, you are given the option to switch or stay. If you originally picked the car, you should stay. If you originally picked a goat, you should switch. Since there is a 1/3 chance that you originally picked the car and a 2/3 chance that you originally picked a goat, the probabiliy of winning the car is greater if you switch.

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  • 4 weeks later...
iglak-

 

You are artifically increasing the odds of winning by staying because you list four cases of door 3' date=' when the cases are not actually distinct- there are only two distinct cases at door three. What it should look like is this:

 

You pick 1. They eliminate 2. You stay on 1. You lose.

You pick 1. They eliminate 2. You switch to 3. You win.

You pick 2. They eliminate 1. You stay on 2. You lose.

You pick 2. They eliminate 1. You switch to 3. You win.

You pick 3. They eliminate 1 or 2. You switch to 1 or 2. You lose.

You pick 3. They eliminate 1 or 2. You stay on 3. You win.

 

 

The number of times you win by staying is 1/3. The number of times you win by switching is 2/3.[/quote']

 

Ding ding.

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me too, but, I am not really sure how people got to the stage that changing the door will increase your chances of winning!

 

Because it's true. Look at the post I quoted; there's no advanced concept there, it just lists every possible outcome, and the chance of you winning if you change is, indeed, twice that of you not.

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P.S. either way' date=' it's still 50:50 chances

the way you listed it shows 3 wins and 3 losses.

they way i posted it shows 4 wins and 4 losses[/quote']

 

2 wins for switch, 1 win for stay.

 

1 loss for switch, 2 losses for stay.

 

It's not rocket science.

 

Shouldn't it not say "1 or 2" but have seperate things for them?

 

No.

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But, then it depends which door you choose at the start. Let's say that they change the door number of the winning door to something else. Then, changing would be a wrong option. Changing would result in you getting 2/3 of the time wrong!

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Let's say that they change the door number of the winning door to something else.

 

Lets say that they remove the prize entirely! Then, change or not, you always lose! This is obviously a valid counterargument and I'm not just arbitrarily adding random irrelevencies to the argument at hand!

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Lets say that they remove the prize entirely!

 

That is too extreme and will cause bias in the probabilities. This cannot be used as a counter-arguement as it is unfair. You can only argue on something if it matches the rules given. In this case, there will be a prize behind one of the doors

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Lets try another tack. We've tried irrevocable proof, and you've ignored it.

 

How about this:

 

You have a 2/3 chance of choosing a 'losing' door.

 

If you change with a losing door, you ALWAYS win.

 

Therefore, if you change, you have a 2/3 chance of winning.

 

Or this:

 

If you decide to stay, the only way to win is by picking the winning door from the off. Therefore, the chance of winning by staying is 1/3. As the sum of the chances of the winning will be 1 (dur), the chance of winning by changing is 1 - 1/3, or 2/3.

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If you decide to stay, the only way to win is by picking the winning door from the off. Therefore, the chance of winning by staying is 1/3. As the sum of the chances of the winning will be 1 (dur), the chance of winning by changing is 1 - 1/3, or 2/3.

 

That explanation makes more sense to me. But what about the the fact that after one door has been eliminated, the chances of choosing the right door is 50%?

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