You are on let's make a deal. There are three doors. Behind one door is a brand-new car behind the other two doors are goats. The host of the show knows exactly what's behind each of the doors. You pick your door. He then opens up one of the doors revealing a goat.
You are then asked whether you would like to stick with your door or switch to the other door that you did not choose.

Would it be to your advantage to switch doors?

Yes

If yes how much do you increase your odds by changing ?

I agree with FTI.

You increase your odds by 33%

Are you on the game theory chapter in your Econ book?

No correct answers yet.

**** the goat, marry the car, kill door 3

You increase your odds from 33% to 50%.

What if I find goat tasty, and consider all possibilities a win?

http://m.youtube.com/?#/watch?v=tvODuUMLLgM

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**** the goat, marry the car, kill door 3

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I'd stick with my door, and make some cabrito

One of you smart young whipper snappers explain the answer. You have one of the two doors. The car is behind one of the two. It would seem the odds are 50/50 you have the car.

nvm misread it

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**** the goat, marry the car, kill door 3

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ok no it doesn't increase your odds to change. Either way, you've picked one out of the two remaining doors.

If I recall correctly it's about a 17% increase over staying with your first choice.

When you first picked you had a 2/3 chance of picking a goat. Now that one goat has been revealed, you switch, because you were more likely at first to have picked a goat than you were to have picked the car with your first choice.

If you stay with your answer you have a 1/3 chance to have guessed correctly. If you switch, you will only be wrong 1/3 of the time (and correct 2/3 of the time).

The lady or the tiger?

Chicken Dance is correct. Initial odds are 1/3. The host has to pick between the two other doors and has to choose a door that contains a goat no matter what is behind the door you chose, which means we learn nothing about the initial door we chose that would change its odds of being correct. Those two doors have a cumulative 2/3s chance of having the car behind it. By guaranteeing that one of those two doors does not have the car behind it, there now becomes 2/3s chance that the unopened door you did not choose is the correct one, while you are still stuck at 1/3 odds on your initial door.

I agree with Chicken Dance and NYDA

Mensa

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A boat's a boat but the mystery box could be anything. It could even be a boat! You know how much we've wanted one of those. We'll take the box.

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Yep, Chicken Dance and the (NY)Doc have it. If you decide a priori to ALWAYS switch, the only way you lose is if you initially pick the car, which is 1/3 of the time. Switch!

Monty Hall problem explained

How big are the goat's tits?

Congratulations Chicken Dance

That question was tough.

Can't decide this without a Rule #1.

Of course if you wanted a goat instead ....

My life is complete.

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**** the goat, marry the car, kill door 3

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The only correct answer

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**** the goat, marry the car, kill door 3

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The only correct answer

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toucan just popped a major chub!

You removed a set from the equation. The Odds are now 50/50.

Thats like saying all 3 doors are open and you picked the car, your odds of winnng the car are still 1 in 3.

The orginal odds were 1 in 3 as there were 3 unknowns, but one of the unknowns is a given. You can't add or subtract data and have the equation remain the same.