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### Neighboring Cards

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Date: 02/18/97 at 13:58:35
From: Christopher K.
Subject: Card probability

Hello,

I have a problem. I found that if you take a deck of cards, name any
two cards, and then look for them in the deck, they will be by each
other or only one card away from each other. I tried to find the
probability of this occurring and have asked a lot of people about it,
but I can't find the answer. So far, when trying to simulate this, I
have only failed once.
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Date: 02/27/97 at 21:10:09
From: Doctor Ken
Subject: Re: Card probability

Hi Chris -

That's some pretty good luck!

Here are some pointers when trying to find the probability.  When
computing ANY probability, you must remember the probability rule of
gold: probability is the number of favorable possibilities divided by
the total number of possibilities.

Often it's a good idea to break the problem into pieces.  In your
case, we can break the problem into two pieces: the probability that
the two cards are next to each other and the probability that they are
separated by a single card.  We can compute the two pieces separately,
and then add the probabilities together to find the final probability.

In your case, the total number of possibilities (for both pieces of
the problem) is the number of different arrangements of a deck of
cards.  So in order to calculate the probability that the conditions
you named are satisfied, we have to find this number.  So let's do it.
There are 52 cards in a deck.  If we have them in a certain order,
then there are 52 choices for the first card, 51 for the second, 50
for the third, and so on.  The total number of different arrangements
is thus 52 * 51 * 50 * ... * 2 * 1, which you can write as 52!
(pronounced 52 factorial).  So that's the denominator in our
probability fraction.

The harder part is to find the numerator.  But it's not too bad
either.  Let's concentrate on the first piece first.  Given two cards,
how many different arrangements of the deck have those two cards next
to each other?  Well, let's start by writing a few down.  If the two
cards are called A and B, then we could have these arrangements (I'm
using x to just mean any old card here):

A B x x x ..... x
B A x x x ..... x
x A B x x ..... x
x B A x x ..... x
and so on.

Since there are 50 x's in each line, each line actually represents
lots of different arrangements; you could order the cards that fill in
the x's in lots of different ways.  How many?  Well, 50 * 49 * 48 *
... * 2 * 1, which is 50!.  Now how many lines are there?  There are
51 different places we could put the "A B" pair (convince yourself of
that), and for each place there are two lines (one for "A B" and one
for "B A"), so in total there are 2 * 51 * 50! different arrangements
that have A and B next to each other.

So what's the probability?  Well, we divide:

2 * 51 * 50!        2        1
Probability = --------------  =  ----  =  ----
52!            52       26

So that's the first piece of the problem.  I'll leave it to you to
figure out the second piece, the probability of having A and B
separated by a single card.  Once you have that, you can add the two
probabilities together to find the total probability.  Good luck!

-Doctor Ken,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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