Dn's Card Trick

[Mattnxtc 0]

The card trick he is using is called the Monty Hall game.I dont have the formula on me but using the bayes formula you can mathematically show why its always correct to switch

[PimpRock 0]

There was a massive debate about it after something similar was done on an American gameshow in the 1970s. Huge mathematical, statistical and philosophical debate from some of America's brightest minds which heaper scorn on a whole bunch of academics.It was quite interesting hearing it explained in lay person's (Daniel's) terms as the section in the book took a few reads. I think I read about it in the book called Freakonomics.

[PedalMettle 0]

The card trick he is using is called the Monty Hall game.

http://en.wikipedia.org/wiki/Monty_Hall_problemActually, DNs explanation was one of the better quickie explanations I have seen for this. It helps to see the 3 cards and think "2/3 chance the ace is here". The great thing about using it for gambling is that stupid people just can't understand it and will argue until they are blue in the face that it is 50-50.

[Goshkruse 0]

It's an old problem - commonly found in most entry level statistics courses. I've also seen it referenced - quite poorly - on the TV show "Numbers". When I tell it, I use the "Let's Make a Deal" scenerio. Choose a door, then give that person the opportunity to switch . . don't get zonked! The easiest example to explain it that I've found - is to take it to a larger number than 3. Suppose you did the same trick with all 52 cards in a deck, pick one, then I flip 50 of them over - leaving one turned down - would you switch now. Of course it becomes obvious to switch. There's still only a 1-52 chance you originally picked the correct card. The key is that the initial odds NEVER change. It still takes most people quite a while to rationalize the situation - and some people never get it. Like Daniel mentioned - no one I've come across analyzes the problem correctly the first time.

[Jam-Fly 8]

my friend told me a similar story, but it was the game show thing, and it took me like 3minutes and i proved that it is mathematically correct to switch. So does that mean I'm a maths genius (thats true btw, i really did)

[therrinn 0]

It was made famous because a bunch of dumb math and statistics professors didn't thinkt he problem through and criticised vos Savant when she made the problem famous - its really pretty trivial to prove if you take a moment and think about it.

[Zeatrix 0]

Man, DN should have used his reading skills instead. When the camera guy looked at the cards for the longest while and then put them face down and just seemed to pick one of the two cards to show, DN should have realized that the camera guy was picking between two none spade Aces

[cdipierr 0]

FYI, this is the same as the classic problem:Assume parents have two children. Putting aside twins and other abnormal factors, let's assume the chance of having a boy & girl are the same (ie. 50%).If I tell you that one of the children is a boy. What's the chance the other one is a boy as well?Same principle of conditional probability applies.

[LongLiveYorke 38]

FYI, this is the same as the classic problem:Assume parents have two children. Putting aside twins and other abnormal factors, let's assume the chance of having a boy & girl are the same (ie. 50%).If I tell you that one of the children is a boy. What's the chance the other one is a boy as well?Same principle of conditional probability applies.

GBBGBB1/3.

[qsoundrich 0]

3 Semesters of discrete math, and 2 of probability and I never understood this problem until Daniel just explained it. I remember reading about it in Ask Marilyn, but she just said it's true because she said so, never gave a reasonable explanation. Hopefully Daniel will retire from poker soon and teach all of us how to play. He seems to have a gift for communicating.

[Consciousone 0]

Anytime you make this bet against someone giving them 6-5 odds you will be getting +33% EV. Im going to try it on alot of people.

[ninjaunderwear 0]

Simple explanation:After you've randomly picked one card out of the three, you can seperate the cards into two groups, (1) the card you picked and (2) the other two cards you didn't pick. Group (1) has a 1/3 chance of containing the Ace of Spades. Group (2) has a 2/3 chance of containing the Ace of Spades. Revealing one card in Group (2) has no effect on either group's chance of containing the Ace of Spades.Ask Marilyn used an extreme example to further illustrate the point. If you used one million cards with one Ace of Spades among them, and asked somebody to choose one card out of there, you could again split the cards into two groups, (1) the card they picked and (2) the other 999,999 cards they didn't pick. Now, Group (1) has a 1/1,000,000 chance of containing the Ace of Spades and Group (2) has a 999,999/1,000,000 chance of containing the Ace of Spades. Now, if somebody revealed every card except one in Group (2), leaving behind only one card from each group, you'd obviously choose to switch to Group (2). As each non-Ace of Spades card is revealed, the odds of group (2) overall remain the same, but each individual card in group (2) is more likely to be the Ace of Spades. Continuing down until only one card remains in group (2), that single card now carries the entire 999,999/1,000,000 chance as the entire group.Switch for the win.

[Linque 0]

I ran into some trouble during my military service when I played around with this trick. The thing that really boggles me is how so many people can't comprehend the problem even after someone gives you and explanation.I guess each one of us approaches these kinds of problems differently and don't really see the match in 'real life' problems like this.That's a good explanation from Negreanu though. Funny that I should run into this trick on two such different occasions.

Ask Marilyn used an extreme example ... only one card remains in group (2), that single card now carries the entire 999,999/1,000,000 chance as the entire group.

Even after using this example, not all of the guys believed me.

[Leppo 0]

I just read this, The Monty Hall Problem in King Yao´s book Weighing the odds in Hold'em Poker and there is a VERY Crucial thing to explain that the person doing this "trick" to you is ALWAYS going to offer you a switch possibility, and that he sometimes does not just flip the card you have chosen. I recommend you all to read the book, it is a book for Limit Hold'em and has some really interesting chapters. Good Luck in the Main Daniel, and I´ll show you some real card tricks if I ever run into you.

[randy 0]

i have seen this problem many times before, i got it right away but it helps when a math student shows you

[hblask 1]

FYI, this is the same as the classic problem:Assume parents have two children. Putting aside twins and other abnormal factors, let's assume the chance of having a boy & girl are the same (ie. 50%).If I tell you that one of the children is a boy. What's the chance the other one is a boy as well?Same principle of conditional probability applies.

I think this falls into the "trick wording" category. If you say "the first child is a boy, what are the odds of having another boy?", it's 50/50. This is what most people will think you are asking. Assuming the two children already exist and one of them is a boy, the answer is different. Just as with the Monty Hall problem, you have to ask whether he alway shows a door and offers a switch. On the show he didn't.

3 Semesters of discrete math, and 2 of probability and I never understood this problem until Daniel just explained it. I remember reading about it in Ask Marilyn, but she just said it's true because she said so, never gave a reasonable explanation. Hopefully Daniel will retire from poker soon and teach all of us how to play. He seems to have a gift for communicating.

Maybe next he can explain the "airplane on treadmill" problem to those who refuse to see:http://www.straightdope.com/columns/060203.htmlhttp://www.straightdope.com/columns/060303.html

[Dane 0]

Two cards 2/3One card 1/3So 2/3 of the time it will be one of the two cards paired, showing a card is just to mess with your mind.There was a lot of fuss about some gameshow, that used this, but I can not remember the name.

[carmdog 0]

There is an analogous situation in the card game bridge called the Rule of Restricted Choice.The basic situation is: AJTxx opposite xxxx to lose only one trick. First you play small to the J, when it loses to the K or Q, the next trick you play small to the T. Since your opponent had a choice to win the K or Q (equal cards) it is about 2:1 more likely that the other opponent has the other high card.This is the same Bayes choice in the 3 Aces problem. In 2 cases, Daniel has to show the card that is not the Ace of Spades, and in one case, Daniel has a choice to eliminate either other Ace. He's laying you 6:5, but actually you're getting 1:2.

[Jonbonwolf 0]

Simple explanation:After you've randomly picked one card out of the three, you can seperate the cards into two groups, (1) the card you picked and (2) the other two cards you didn't pick. Group (1) has a 1/3 chance of containing the Ace of Spades. Group (2) has a 2/3 chance of containing the Ace of Spades. Revealing one card in Group (2) has no effect on either group's chance of containing the Ace of Spades.Ask Marilyn used an extreme example to further illustrate the point. If you used one million cards with one Ace of Spades among them, and asked somebody to choose one card out of there, you could again split the cards into two groups, (1) the card they picked and (2) the other 999,999 cards they didn't pick. Now, Group (1) has a 1/1,000,000 chance of containing the Ace of Spades and Group (2) has a 999,999/1,000,000 chance of containing the Ace of Spades. Now, if somebody revealed every card except one in Group (2), leaving behind only one card from each group, you'd obviously choose to switch to Group (2). As each non-Ace of Spades card is revealed, the odds of group (2) overall remain the same, but each individual card in group (2) is more likely to be the Ace of Spades. Continuing down until only one card remains in group (2), that single card now carries the entire 999,999/1,000,000 chance as the entire group.Switch for the win.

OK, so lets say we did flip all 999,999 and were down to the last card. What odds would you give me now. Would you give 10,000 to 1, If I agree to keep the card I picked at first? Ya I know 999,999 cards ago your choice was a HUGE FAV. But at this point would anyone take this bet?

[PairTheBoard 0]

Daniel actually made a technical gaffe at the end of his presentation when he tried to do the whole thing without the cameraman's assistance. He put out the 3 cards face down, pulled one aside, and then picked one of the other two at random to show, hoping it was a non-spade. Luckily it was a non-spade. He then concluded that the other card was twice as likely to be a spade as the one he originally pulled aside. Except NOW, that's no longer true. It's 50-50 when done this way.PairTheBoard

[ninjaunderwear 0]

OK, so lets say we did flip all 999,999 and were down to the last card. What odds would you give me now. Would you give 10,000 to 1, If I agree to keep the card I picked at first? Ya I know 999,999 cards ago your choice was a HUGE FAV. But at this point would anyone take this bet?

Where would you like to meet? Maybe we could do this in the poker room at the Wynn?

[Ecclesbury 0]

This is a weird problem. Once you get it, it seems so obvious that you have to switch, but until then people seem adamant that that it should be 50-50.A friend of mine, who is very intelligent, and studied Maths at University briefly, just would not accept that it wasn't 50-50. He even wrote a computer program that ran the scenario 1000 times and showed that it wasn't 50-50 but still wouldn't believe it! It was the increasing the numbers involved answer that finally swayed him, and now he feels a bit stupid!Anyway, there is a point to this post. The simplest, and often best, explanation that I have come up with (for non-mathematical viewers) is this:When you first pick you are more likely to have NOT chosen the Ace Of Spades.So switch.That aeroplane one just baffled me for a while!

[ballin4life 0]

The explanantion for people who still don't get it:You have a 1/3 chance of picking the correct card.2/3 of the time you guess wrong, and the other guy only has one card he can flip up, so 2/3 of the time the other card is right.

[greatwhite 0]

Am I the only person who doesn't get what's so complicated about this trick? The guy gets the ace of spades 2/3 times. The fact that he flips over one of those cards that's not the ace of spades doesn't change that. Since he has the ace of spades 2/3 times, he'll still have it 2/3 times once he looks at his cards and flips over one that's not the ace of spades. I was actually kind of dissapointed when he showed this.

[NOTHER 0]

What's up gambiens,Mind-boggling to me is the fact that the mentioned card trick was mind-boggling to Daniel and his friends.