Probability Question

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.

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.

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

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

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Originally Posted by 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

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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

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

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

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

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

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

Forgot about the 21 film. This is an old mathematical problem that goes back to the 18th or 19th century, but has been called in our times "The Monty Hall Problem" -- loosely based on the Monty Hall game show "Let's Make a Deal" where contestants could keep a known prize or exchange it for an unknown prize hidden behind a door.

And indeed, the correct answer is to switch cards. Sticking with your original card gives you a 1:3 chance of winning the money, while switching cards give you a 2:3 chance (you double your chances by switching!).

I heard about the problem many years ago in a column by Marilyn Vos Savant, the person with the highest IQ in the world. She stated that switching would increase your chances and it stirred a firestorm. PHDs and Professors from all over the world wrote her to say she was crazy, but eventually she convinced most that it is better to switch. And if you don't believe it still, play the game yourself with a friend 100 times once with each strategy. You will see on average you will win around 33 times by staying with your original card, and around 67 times when your switch.

Here are some of the letters that Ms. Vos Savant received when she first posited this problem in her column (in her column, the problem was written with 3 doors instead of 3 cards, the prize was a new car and the other 2 doors had goats behind them):

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Since you seem to enjoy coming straight to the point, I'll do the same. You blew it! Let me explain. If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2. As a professional mathematician, I'm very concerned with the general public's lack of mathematical skills. Please help by confessing your error and in the future being more careful.

Robert Sachs, Ph.D.
George Mason University

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You blew it, and you blew it big! Since you seem to have difficulty grasping the basic principle at work here, I'll explain. After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the odds are the same. There is enough mathematical illiteracy in this country, and we don't need the world's highest IQ propagating more. Shame!

Scott Smith, Ph.D.
University of Florida

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Your answer to the question is in error. But if it is any consolation, many of my academic colleagues have also been stumped by this problem.

Barry Pasternack, Ph.D.
California Faculty Association

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You're in error, but Albert Einstein earned a dearer place in the hearts of people after he admitted his errors.

Frank Rose, Ph.D.
University of Michigan

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I have been a faithful reader of your column, and I have not, until now, had any reason to doubt you. However, in this matter (for which I do have expertise), your answer is clearly at odds with the truth.

James Rauff, Ph.D.
Millikin University

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May I suggest that you obtain and refer to a standard textbook on probability before you try to answer a question of this type again?

Charles Reid, Ph.D.
University of Florida

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You are utterly incorrect about the game show question, and I hope this controversy will call some public attention to the serious national crisis in mathematical education. If you can admit your error, you will have contributed constructively towards the solution of a deplorable situation. How many irate mathematicians are needed to get you to change your mind?

E. Ray Bobo, Ph.D.
Georgetown University

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I am in shock that after being corrected by at least three mathematicians, you still do not see your mistake.

Kent Ford
Dickinson State University

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You made a mistake, but look at the positive side. If all those Ph.D.'s were wrong, the country would be in some very serious trouble.

Everett Harman, Ph.D.
U.S. Army Research Institute

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I highly doubt if u do it yourself you'll end up getting results of 33 or 67. I know thats the probabilty but probability doesnt make it so. Same way people play roulette and see it land on red like 10 times in a row and so people decide to bet on black because its due. When in truth the last 10 spins have no effect on the probability or a red or black coming on this spin.

Ah, and the solution!

Well, think of it this way. When the dealer has the 3 cards laid out on the table, you have a 1:3 chance of picking the Ace. However, after you pick your card and BEFORE it is revealed, we can separate our cards into 2 stacks -- stack ONE with your 1 card in it and stack TWO with the 2 cards you didn't pick. Obviously, stack ONE has a 1:3 chance of having the ACE and stack TWO has a 2:3 chance of having the ACE, since they have one and two cards respectively. But the dealer knows what the cards are, and he always turns over one of the cards from stack TWO and it is always a deuce.

Before the dealer does this, we have stack ONE with card A with a 1:3 chance, and stack TWO with cards B&C with a 2:3 chance.

Stack ONE: Card A 1:3
Stack TWO: Card B 1:3
Card C 1:3

after the dealer reveals one of the cards from stack TWO to be a deuce (for example, he flips over card C to reveal a deuce) there is still a 1:3 chance that the Ace was in Stack ONE and a 2:3 chance it was in stack TWO. THAT HAS NOT CHANGED. What has changed is we now know that card C is not the Ace, and therefore card B has a 2:3 chance of being the Ace while card A stays with its 1:3 odds.

Stack ONE: Card A 1:3
Stack TWO: Card B 2:3
Card C 0:3

For an explanation by Ms. Vos Savant herself, click the following link.....

Marilyn vos Savant | The Game Show Problem

What was her favorite drug? So if you have 3 people each pick a card and 1 guy is told he lost. Now all of a sudden you might as well give congrats to your other opponent case he's gonna win 66% of the time... 420

Spoken like a PhD!

The logic is sound If you think of it like in groups. You pick your card and then the other 2 are put into one group. Your 1 card has 1/3 chance so the other 2 as a group have a 2/3 chance. If one of the cards in the group is shown not the be the card you are looking for the group still has a 2/3 chance of being the card you are looking for.