Riddles

[Firemeboy]

You are on a game show and are in the final round.  The host shows you three doors and tells you that behind two of the doors are boxes filled with bananas.  Behind the other door is a box filled with $10,000.  

You are told to pick on of the doors and decide to pick door #2.  After you pick your door, the host, as he always does, picks another door and opens it up.  In this case, he opens door number 3.  You see behind it a box filled with bananas.  You haven't seen what is behind door # one, nor do you know what is behind door #2.

The host asks you, "Would you like to switch doors?"

So, the question is, is it in your best interest to switch?  If so, why?  If not, why not

[Legion]

WOW, would not consider this a riddle, but ok.  OK well it really makes no difference whether you switch or not, you have a 50% chance of getting it right.

[Firemeboy]

That is a good guess, but incorrect...  

[Legion]

I am really puzzled by this because unless you know more about the game or the game show it really shouldn't matter if you switch your guess or not.

[Skar]

I suspect that it has to do with the fact that when you made the original choice you only had a 33.3333- % chance of getting it right.  So your original choice has only a 33.333- % chance being right.  But if you switch you have a 50% chance of that being right.

So sticking gives you 33.333-%  Switching gives you 50%

Am I right?

[The Holy Saint, Grand High Poobah, Master of Monkeys, Ehlers]

That would be a misrepresentation of statistics though.

Originally, you had a 33.3% chance of success
You made a choice.
One was revealed, not your choice, and that one had a 66.6% chance of not being the door you wanted. It wasn't. THat leaves you with two choices left

That means EITHER choice you make has a 50% chance of being correct. YOu are making a new decision. There are two options. One is right, one is wrong. Sure, switching has a 50% chance of being right, but staying where you are has a 50% chance of being right too. Thus switching does *not* give you an advantage. It's not dumber, but it's not smarter either.

[Firemeboy]

Skar, you are correct.  

[The Holy Saint, Grand High Poobah, Master of Monkeys, Ehlers]

except that the question was "is it in your best interest" and, as I have shown, it is not. It's not AGAINST your interests, it's just not necessary.

[Archon]

SE's right.

[Firemeboy]

Look at it a different way.  Let's say there are 10 doors, and you pick one.  You had a 1 in 10 chance that the door you first picked is right.  Now the hose opens 8 doors, revealing 8 boxes of bananas, and two closed doors (the one you picked, and the last door).

Which door should you pick then?  Sure, both doors now have a 1 in 2 chance of being right, but your originaly pick had a 1 in 10, while the other door has a true 1 in 2 chance.

You would be a fool not to switch to the other door, just as in this puzzle you have a better chance if you switch to the other door.

[Archon]

That is completely not true. Both doors have a 50% chance to be the right door. If you have ten doors and you eliminate one, then they all have an 11% chance. Eliminate another and they all have a 12.5% chance. And so on, until the remaining two have a 50% chance. You lose.

[The Jade Knight]

Archon's right.  The probability shifts every time something changes.

The original probability was for 1 of 3, or 1 of 10, not 1 of 2, which is a different equation.  Whenever something changes (ratio), so does the probability.

[Firemeboy]

The original probability was for 1 of 3, or 1 of 10, not 1 of 2, which is a different equation.  Whenever something changes (ratio), so does the probability.

No, think about it.   When you first picked that 1 in 10, it had a one in ten chance.  The fact that the rest are eliminated doesn't mean anything.   What if I held up a deck of cards and asked you to pick the jack of diamonds.  You picked one card (1 in 52 chance that you got it right, since you don't know where the jack is), then I (knowing where the jack is) turned over 50 cards, all of which are NOT the jack of diamonds.  There are now two cards left. When you picked the first card, you had a 1 in 52 chance that it was the jack of diamonds.  This last card now has a 1 in 2 chance.  You would be a fool not to switch cards.

[Archon]

No Firemeboy, that isn't the way it works. At the beginning, before any cards had been eliminated, the probability to be the jack of diamonds was the same for each card (1/52). When one card is eliminated, all of the cards still have the same probability of being the jack of spades, only now their probability is 1/51. If you continue eliminating cards, the denominator will be reduced each time, but the cards will still all have the same probability of being the jack of diamonds. The last two cards still have a 50% chance each. Think about it, the jack of diamonds is definitely one of the two cards. That means that together, the cards have a 100% probability. If there were only two cards, and one had a 50% chance, and the other had less than a 50% chance, where would the remaining percentage be?

[Firemeboy]

Yes, but think about it.  I know where the Jack of diamonds is.  You don't.  You pick 1 in 52 cards.  That card has a 1 in 52 chance.  I, knowing where the card is, remove the other 50 cards and leave one card left.  You can't say that the original card has the same chance of being the Jack of Diamonds as the 1 card that is left over...  

Here is an article describing how it works.  

http://www.straightdope.com/classics/a3_189.html