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### Mathematical Card Trick

```Date: 04/07/2002 at 21:56:39
From: Jason Demler
Subject: Mathematical Card Trick

What is the math or probability behind this? Can you please explain?
I have asked my college statistics professors and they do not know
how it works. Here it is:

Take a deck of 52 cards. Shuffle or do whatever to mix them up. Flip
over the top card and count from that number to King. (Ex. I flipped
over a 9, so I would count 9, 10, Jack, Queen and King. I then
flipped over a 5, 6, 7, 8, 9, 10, Jack, Queen and King.)

I do this until I can no longer count to a King. Those excess cards
become "dead cards." Sometimes you may be able to go exactly to a
king. You then flip over 3 piles so that they are now facing down.
The piles that were not flipped over become dead cards and are added
to your pile, if you have some, of dead cards. Next you flip over the
top card of 2 of the 3 piles. Add those numbers up and add 10 to them.
(Ex. I flipped over a 5 and a Queen (12). These add up to 17 and then
I add 10 to that so my total is 27.

Now, you remove 27 cards from your dead card pile. The remaining
number of cards in your dead card pile will be the number on the card
on the 3 pile that is still lying face down.

This works every time. It seems as though it is just random but there
have been looking for the answer to this for about two years. I am
stuck telling everyone that I just know that it works and that there
is math behind it, but I don't know how it works.

Thank you very much,
Jason Demler
```

```
Date: 04/08/2002 at 16:03:16
From: Doctor Peterson
Subject: Re: Mathematical Card Trick

Hi, Jason.

I'm surprised no one could figure this out; it's really pretty simple.
Maybe they expected something involving statistics. You just have to
keep track of how many cards are in each pile.

I assume that when you say you "flip over the top card and count from
that number to King," you mean that you put that card and those
following face up in a new pile while you count. If that card is A,
then you will count essentially from A through 13, putting 14-A cards
in that pile. Once you have done that with as many piles as you can,
you then reduce it to three piles; since the card order is arbitrary,
we can assume "without loss of generality" that you just dealt out
three piles, and what is left is dead. So you have this situation
after turning those three piles face down (which puts on top the card
that determined the number of cards in the pile):

Pile 1: 14-A cards, card A on top
Pile 2: 14-B cards, card B on top
Pile 3: 14-C cards, card C on top
Dead pile: 52-(14-A)-(14-B)-(14-C) cards = 10+A+B+C cards

Suppose you choose the first two piles. Add 10 to the cards on top,
and you have 10+A+B. Take that many cards from the "dead" pile, and
you are left with (10+A+B+C)-(10+A+B) = C cards. That's the answer.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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