Dn's Card Trick

[SlightlyMad 0]

I agree ... I'm amazed with all of the math geeks you find hanging around poker rooms that Daniel had *not* seen this one before. There's some true comedy in "The Power of Logical Thinking" (Marilyn Vos Savant's book) when you read some of the responses from the Ph.D.s who think that Marilyn was barking up the wrong tree. If you happen to run into your local bookstore and want some good giggles, the comedy begins on the bottom of page 8 in the edition that is currently available.

[Jonbonwolf 0]

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

LOL! Nice first post.I wasnt "getting" that they could look each time they flipped a card. I was thinking it was a random discard. I get it now

[NoShowJones 0]

I just coded a program to simulate this also."Should have switched" means that you would have won the bet if you had switched cards when given the chance.Clearly, the odds are not still 1 in 3. Anyone who believes that is just silly.

[NoShowJones 0]

Some people have argued that using a greater number of cards illustrates the point. They say that if we used 52 cards, and I picked one, then it is very obvious that the Ace of spades is most likely in the 51 remaining cards. This is very true - but it is not the same as the original problem.If you let me choose a third of the deck, then you turn over a third of the deck.. the ace of spades still has a 50/50 chance of being in either my third or the other third.

[tunkpirate7 0]

If you let me choose a third of the deck, then you turn over a third of the deck.. the ace of spades still has a 50/50 chance of being in either my third or the other third.

Why would it be 50/50 - Look at it this way. You have the deck split into thirds and I take 2/3 of it and keep them face down, you take 1/3 and keep it facedown. Is it 50/50 that I have the ace of spades in my 2/3 of the deck? You already know that one of my thirds doesn't contain the ace of spades since there is only one ace. Once I look through both of my thirds I'm gonna flip up the one that doesn't have the ace of spades in it... Which doesn't even matter because you already knew that one of them didn't have it. -3 cards, 1 ace of spades. 2 of them are mine, 1 of them are yours. 1 of my cards DEFINITELY doesn't contain the ace of spades... so what difference does it make whether or not you see it?

[NoShowJones 0]

Why would it be 50/50 - Look at it this way. You have the deck split into thirds and I take 2/3 of it and keep them face down, you take 1/3 and keep it facedown. Is it 50/50 that I have the ace of spades in my 2/3 of the deck? You already know that one of my thirds doesn't contain the ace of spades since there is only one ace. Once I look through both of my thirds I'm gonna flip up the one that doesn't have the ace of spades in it... Which doesn't even matter because you already knew that one of them didn't have it.

So.. if we have 60 cards total, and only 1 is teh ace of spades. You take 40, and I take 20. There is a 2/3 chance you have the ace of spades, right? Of course!BUT, if you turn over 19 of your cards, and none of them is the ace of spades, there is no longer a 2/3 chance that your ONE down card is now the ace of spades.

[PedalMettle 0]

NoShowJones: where do you live? If you really think it's a 50/50 bet I'm willing to give you 6 to 5 I can pick the As by switching. We can do the bet as many times as you want until you are convinced or out of money. In fact, the more times the better! What you may be missing is that the person with the 2 aces looks at the cards and, if one is the As, turns up the other one. He is never going to turn up the As which is why it doesn't change the 2/3 chance that the As is one of those 2 cards. If he just flipped one at random without looking then you would be correct, it would be 50-50 in that case.If you still don't get it, read the wikipedia article I referenced above. If you STILL don't get it after that, then people will be lining up for your action.

[BadBeatBaKer 0]

I just coded a program to simulate this also."Should have switched" means that you would have won the bet if you had switched cards when given the chance.Clearly, the odds are not still 1 in 3. Anyone who believes that is just silly.

So did I, and when run 10000 times amazingly 66.66667% of the time you should switch.You are confusing your self by worrying if you can see one of the two cards. It clearly doesn't matter. The cards are set. There is no more randomization taking place, therefore the probability doesn't change. You know that one of the two you have is not the As, but with 2 out of three chances, there is a 50% chance the other one is the As. I can be certain that 2 out of 3 times I won't have it (averaged over a large number of trials).

[NoShowJones 0]

Ok. I have seen the light. My logic was wrong in the code I wrote to test the program (see.. you convince yourself that you're right, and you'll even code something to prove it). I corrected the code, and am now convinced that indeed you should switch. I'm stupid sometimes.

[LAS22 0]

I tried the trick last night with my wife and it worked 8 out of 10 times.

[AceyDeucy 0]

THE KEY FACT IN THE MONTY HALL PROBLEM:The host KNOWS he is opening a dead door. If the host is also opening a door at random the problem falls apart. It is the face that he knows the door he opens is no good that makes this problem go.Telling people this seems to get them to nod and agree.

[Balloon guy 158]

The guy WANTS you to change your pick, that's why he shows you the card. Don't switch, the odds don't matter. All that matters is that you picked the right card the first time and the conman wants you to switch. If you picked wrong he would show you the right card and take your money.What self interest would there be in a shillgame operator to give you second chance?Take your winnings and run.

[acesfull333 0]

The guy WANTS you to change your pick, that's why he shows you the card. Don't switch, the odds don't matter. All that matters is that you picked the right card the first time and the conman wants you to switch. If you picked wrong he would show you the right card and take your money.What self interest would there be in a shillgame operator to give you second chance?Take your winnings and run.

[youfred 0]

When you first pick you are more likely to have NOT chosen the Ace Of Spades.So switch.

Or say, "When you pick with the intention of always switching you are inverting your selection thereby selecting 2/3s of the deck instead of 1/3rd" If the game is presented without the looking and flipping of a card whohaa the switch is easy.

So switch.That aeroplane one just baffled me for a while!

I didn't like the linked explination very much... Planes move by accelerating air to push against unaccelerated air; Planes take off when they are moving fast enough through air that their wings generate lift.A conveyor belt can only effect a planes movement if the brakes are on. Would the belt moving in the same direction as the plane shorten it's time to reach take-off speed?In fact, all planes are taking off from conveyor belts, some going the same way, others opposite and many at angles (Runway 24L = 240degrees) because the earth is spinning! It is only the airspeed that matters, so they select the correct runway for the wind direction, not change the rotation of the earth.If you have six cards, three each of two suits, what are the odds of randomly drawing three of the same suit. Two will always match and the last is 3:1 against?

[Icec0o1 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

Haha and you thought you grasped the concept. Sorry, but you're wrong. When done this way, it would still be 1/3 more likely that the card from the 2 card pile from which he discarded is the Ace of spades. And as far as the children's thing, it's 50% chance the second one is a boy because the first one's irrelevant.

[PairTheBoard 0]

QUOTE(PairTheBoard @ Saturday, August 5th, 2006, 12:39 PM) 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

Haha and you thought you grasped the concept. Sorry, but you're wrong. When done this way, it would still be 1/3 more likely that the card from the 2 card pile from which he discarded is the Ace of spades. And as far as the children's thing, it's 50% chance the second one is a boy because the first one's irrelevant.

I was wondering if this gaffe would hook anyone. No Ice, you are wrong. Or else you're not understanding this setup. What makes the trick work is the fact that with the Group of 2 Cards, both cards are looked at and a non-spade is always revealed. That makes switching equivalent to choosing BOTH of the Group of 2 Cards from the beginning. However, in the Gaffed Version, Both Group of 2 Cards are NOT looked at. One is chosen at random to reveal. If BY CHANCE it's a non-space it LOOKS like the Trick is proceeding normally. If it Had Happened to be a spade Daniel would have had to reshuffle and do it over until it worked out the way he was trying to show it. As it was Daniel got lucky and turned over a non-spade.Here's the thing. The lucky non-spade card Daniel turned over could just as easily be considered grouped with the unknown Single Card he pulled aside as with the Unknown card he actually grouped it with. When done this way the remaining two cards - both unknown to everybody including Monty - are equally likely to be the spade.PairTheBoard

[jles96 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.

I work in a casino and did it on break with the dealers +350 for the night +1

[TheCorporation3 0]

LMAO....some people,