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### Similar Phone Numbers

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Date: 10/21/96 at 11:41:49
From: Don Opper
Subject: Probability

Dear Dr. Math,

My brother and I have phone numbers with exactly the same numbers
but in a different order.  He thinks this is some sort of miracle.
I think it's fairly common.  What is the probability that two seven
digit numbers will contain the same numbers but in a different order?
How is this figured out?

Thanks,
D. K. Opper
```

```
Date: 10/22/96 at 12:10:12
From: Doctor Tom
Subject: Re: Probability

Hi Don,

This problem is probably a lot more complicated than you think.  For
example, the 3-digit prefixes are localized, so if you and your
brother live near each other, the odds that you share a prefix are
better, and that would improve the odds of a match.  Similarly, if you
live apart but in the same area code, it's likely the prefixes will
not match.

Some prefixes were illegal until recently - those with a "0" or "1"
as the second digit were reserved for area codes. No prefix begins
with "1" or "0", etc. I'll bet that given current religious
superstitions, a lot of places will not use the "666" area code.

I can state a mathematical problem that can be solved, but given the
statements above, it's unlikely that it will represent the true
situation.

We could solve this: Choose a 7 digit number uniformly from the range
0000000 through 9999999. Now choose a second. What is the
probability that the rearranged digits of the first are the same as
those of the second?

Even this gets messy. For example, if the first number I choose is
1234567, then any reordering of those seven digits would make a match.
And there are 7*6*5*4*3*2*1 = 5040 such combinations. But if the
number is 1111111, there is no other match except for 1111111. This
could happen, I suppose, if you and your brother lived in different
area codes.

So to solve the problem, you've got to find the number of numbers with
no duplications, with one pair, with two pairs, with three pairs, with
one triple, with two triples, with a triple and a pair, with a triple
and two pairs, with a quad, with a quad and a pair, with a quad and a
triple, with a 5-match, with a 5-match and a pair, and with a 6-match.

Multiply each of these probabilities by the probabilities of getting a
match in each case - another ugly calculation.

What I'd do is write a little computer program to count them up since
working this out by hand would be extremely laborious.

To show you what I mean, let me just work it out for 2 digit numbers.

There are ten pairs:  00, 11, ..., 99, so there are 90 non-pairs.

The probability is 9/10 that you'll get a non-pair and 1/10 that
you'll get a pair.  If you get a pair, there's only 1 chance in a 100
of a match.  If you get a non-pair, there are 2 chances in a 100 of a
match (for example, if you get 23, it matches either 23 or 32).  So
the probability of a match is:

(1/10)*(1/100) + (9/10)*(2/100) = 19/1000  (Nineteen times in
a thousand.)

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

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