How to win bets and make enemies with a simple card trick:

First, take 21 playing cards.

Fan out the cards, facing your sucker, and tell them to mentally choose one of the cards.

Shuffle the deck, and start dealing the cards out one at a time, face up, into three piles. Instruct your sucker to watch for their card and remember which of the three piles it lands in.

Gather up the cards, one pile at a time. Make sure you place the pile indicated by the sucker in the middle, between the other two piles.

Repeat this process two more times (a total of three times.)

At this point, oddly enough, the sucker's card is the tenth card from the top of the deck. Perhaps someone else can explain why this is so. I only know that it is so.

Now, here's the twist that will win your bet, or get you killed, or get a laugh, depending on how far you're willing to push it.

Deal the cards out one at a time, turning them over. Watch for that tenth card. That is the sucker's card. When you come to it, keep right on going and deal out a few more cards. Then stop, pause with the next card in your hand, and say:

"I bet you (some amount of money, preferably \$0.00) that the next card I turn over is your card."

At this point, the sucker has already seen his card go by. He knows the next card can't possibly be his. So naturally, he'll take the bet, it's a lock.

Then, you put the card that's in your hand back on top of the deck, sift through the pile of cards you already dealt, take the sucker's card out and turn it over, face down.

And now is a good time to run for cover.

Of course I have given only one way to locate the sucker's card. There are a million different ways to do that, of course.

I had a proof but here, but it was wildly wrong.

So here's the right proof! I'm so proud of myself (core pats self on back)

It utilizes the power of modulus.. ooooohh :)

Take, say, position 20:

```1st part
-------
(20-7)=13

2nd part
------------
(13%3)     =1, the second pile
floor(13/3)=4, the position in the second pile
4+1*7=11

(11-0)=11

3rd part
----------------------
(11%3)     =2, the third pile
floor(11/3)=3, the position in the pile
3+2*7=17

(17-7)=10
```

Tada! Rinse and repeat, for every other starting position. You'll see it works out nicely.

Interestingly enough, the fact that this works out so nicely mathematically is completely irrelevant - you could pick the most erratic behaviour you like, and no matter how painful the math was, the proof will be trivial - just prove it for each possibility. When there are a finite, very small number of possibilities, things become really, really easy.

*note: there is still an error hidden in this proof. Find it and win a bon bon. (even with the error, the proof is still valid. Whenever modulus is involved, you have a lot of leeway...)

Here is a more general mathematical proof because I am annoying and pedantic

Consider our pack of cards, numbered sequentially and starting from 0 partly because it makes the maths simpler but mostly to confuse people.

Let the card that our sucker chose be at position x.

Now, we deal the cards into 3 piles and then recombine them into a single deck. Since we are dealing into three piles and there are seven cards in each pile, it can be seen that the new position of our card in the deck y is given by

y = 7( x mod 3 ) + ( x div 3 )

where the first term gives the base position of the pile that the card is in and the second term gives the position within that pile.

We also perform an extra operation however, which is to take the pile with our card in it and make it the middle pile. In other words if the card is now in the first pile then we want to add 7 to its position, If its in the middle then we want to leave it where it is and if its in the last pile than we want to subtract 7 from its position. i.e. the position is now given by

y = 7( x mod 3) + ( x div 3) - 7( x mod 3 - 1 )

which simplifies nicely to

y = 7 + ( x div 3 )

In the trick we perform this operation 3 times, so after iterating we get

y = 7 + ( 7+ (7+ x div 3) div 3) div 3)
= 7 + 2 + 1 + ((x div 3) div 3) div 3)
= 10 + ((( x div 3 ) div 3) div 3 )

In our pack, x can only range between 0 and 20 so the second term always evaluates to 0

voila.
Too many people know that trick -- and that's a GOOD thing!

The sucker card location will definitely win beer and make enemies! If this is your primary method of obtaining drinks/sex/whatever, you will be disappointed to hear that particular trick is becoming very widely known (to the extent that I can guarantee to see it at least once at every performance. Although you can use this to your advantage.

If anyone ever approaches you in a bar to take this bet (and are not particularly interested in making any friends) go along with it. Pick your card, watch the trick as if you have never seen it before -- even smirk to yourself when the poor trickster passes by your card.

When the bet is laid down to you, insist that the trickster could not possibly turn your card over next (if you think he will pay, raise the stakes!). Once the negotiation is finished, you will need to do three things simultaneously. Reach out to shake his hand with your right hand (looking him square in the eye) and say "You've got yourself a bet" At the same time, with your left hand, turn the entire face-up pile of discarded cards (yours included) face down and give them a good mix, scattering them across the table. Thus rendering him incapable of "turning your card over next"

That will teach him to try and con you!

Log in or register to write something here or to contact authors.