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Similar Phone Numbers


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/   
    
Associated Topics:
High School Probability

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