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3 Doors and 2 Goats

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You are on a game show and there are three doors. The presenter tells you that behind one of doors there is a car and behind the other two are goats, if you pick the car you win it. After you have picked a door the presenter opens a different door with a goat behind it, he then gives you the chance to change what door you open, what should you do?

Hint: It is not 1/2 as you would first think.

The answer is 2/3, and here is the proof.

Call the doors x, y and z

Cx is the event that the car is behind door x

Cy is the event that the car is behind door y

Cz is the event that the car is behind door z

Hx is the presenter opening door x with the goat

Hy is the presenter opening door y with the goat

Hz is the presenter opening door z with the goat

Say you chose door x, the following formula shows the chance of getting the car.

P(Hz^Cy)+P(Hy^Cz)=

P(Cy).P(Hz|Cy)+P(Cz).P(Hy|Cz)=

(1/3.1)+(1/3.1)=

2/3

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Posted · Report post

You are on a game show and there are three doors. The presenter tells you that behind one of doors there is a car and behind the other two are goats, if you pick the car you win it. After you have picked a door the presenter opens a different door with a goat behind it, he then gives you the chance to change what door you open, what should you do?

Not sure what you mean by "what should you do?"

There's been another post on here (as I rememeber) that had a similar question and it was established that you have a 2/3 shot at getting it right, but are you asking whether changing doors will affect those odds?

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Posted · Report post

I mean what should you do so that you have the highest chance of getting the car?

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Posted · Report post

This is a classic puzzle, most recognizably called the Monty Hall problem.

It's been posted elsewhere in this discussion board.

Your chosen door has 1/3 chance of getting you a car.

The exposed goat door has a 0/3 chance of getting you a car.

The third door thus has a 2/3 chance of getting you a car.

You should swap.

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Posted · Report post

Hey Jamie,

That's a very interesting puzzle you have there...

And I can also provide you with an interesting twist...

Just visit the discussion board these guys mentioned...

Death Probability! Really Hard One!

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Posted · Report post

Welcome to the boards, jamie.

This riddle has already been posted and can be found here. Any further discussion can continue in that thread, so I'm going to close this one. Don't let that discourage you from posting other riddles, though.

If anyone wanting to post a riddle isn't sure if it's been discussed here before, just use the search feature for key words. For instance, for this riddle, a search for "goat" will come up positive. Thanks.

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Posted · Report post

Say you are given a heap of 50 coins, in which only one coin is real, while all other coins are fake. you can not recognize the fake or real coin by seeing or touching them.

When you grab a coin from the heap of 50 coins, there was only one out of 50 chances that you grabed right coin.

Suppose the host now removes 47 fake coins. Only two coins are left in the heap and one in your hand. Now what you will do?

Stick to the coin which you selected When there were 50 coins and there was only a very remote chance that the coin selected was real...?

OR

You will leave that and grabe another coin....?

Definitely when you do not know which coin is real, you shall like to leave the coin in hand which had more chances to have been fake, and grab a new coin which has now better chances to be real.

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