Probability Question

[Tiarra]

Here is an interesting question that could test your knowledge and understanding of probabilities. If you fail to get the correct answer the first time, or even after considering it for a long time, do not worry, you are in esteemed company. Most get the answer wrong when they first attempt the problem, so take time to think about it.

You are in a casino playing in a tournament and during the break, a dealer takes out 3 cards from the deck -- two Deuces and one Ace. He then mixes the 3 cards, places them in a row, and takes a look at them (without showing anyone else) so he knows which card is the Ace and which are the 2 deuces.

He then says he will give you $10,000 if you can choose the Ace. You point to one of the cards (it does not matter which one) but before flipping it over, the dealer flips over one of the other 2 cards that you did NOT choose showing a deuce. Remember that the dealer knows what the cards are, so he always shows you a deuce, never an Ace. The dealer then offers you the chance to stick with your original card, or switch to the card you did not pick originally and that the dealer did not reveal to be a deuce.

So for example, you choose card B, the dealer flips over card C showing a deuce and says you can stick with card B, or switch to card A.

What gives you the best chance of winning the $10,000?

1. It makes no difference if you switch cards. As there are 2 remaining cards (1 Ace and 1 Deuce) you have a 50% chance of drawing the Ace either way.
2. Stick with your original card. You have a much better chance with it than with the other card.
3. Switch cards. Your odds improve if you dump your original card for the other.
4. There is not enough information to answer the question.

[eberetta1]

I have heard of something like this before. The story went like Deal or no deal, choose 1 of 30 cases. You buzz thru all the cases until the last case. And they offer you to switch your case for the last other case.
The logic stated to take the case you do not own, because the odds that you chose the right case in the very beginning was only like 3%. I chose to stick with the original case, but the debator said it is right to change cases. So I don't know, whichever works and wins you the dough is the right answer to me, not the averages. I'll go with option 3.

[watte29]

I guess its answer 3, I think I heard about this, but can´t explain

[Prawney]

You are meant to change as your card has 1/3 chance of being an ace. Other card has 2/3 chance. Pretty sure this was in 21

[SendCookies]

You are meant to change as your card has 1/3 chance of being an ace. Other card has 2/3 chance. Pretty sure this was in 21

We are talking about probability and odds in the long run. You have better EV by switching cards. When first purposed w/ the question your odds where 1 out of 3. After the card is expose you now have 1 out of 2 being your new odds on switching. I don't know the math or reasons why mathematically but, overall ev is better by switching cards. Great ?. Now lets see if we get an answer. GL all

[abwil2]

in 3 card monty isnt the odd card taken out with slight of hand? LOl i havent heard this one before so ill be the odd ball and go with option number 2

[aina27]

like all the options can be poibles to win, I prefer the first intention is intuition, option 2

[wasgeht]

gotta be 3... but i dont get what that should tell us xD

[panthers63]

Probability hasn't changed at all. If you cant go with your choice I'd advise using the exit door.

[wagon596]

Yes, same theory was in the movie "21" so I'm not going to tell. Goodluck people.