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### Question on Probability of Repeating Digits

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Date: 04/18/2001 at 02:21:07
From: Sanjayan Vinayagamoorthy
Subject: Question regarding Probability of Repeating Digits

Hello,

"Probability of Repeating Digits," archived at

http://mathforum.org/dr.math/problems/probability.repeat.digits.html

You said that the probability must be 1 that you can find any number
with that string in it. I don't understand how.

Here is my reasoning.  A number in one base is equivalent to any other
base. Thus if I limit myself to base 7 and generate all possible
strings of irrational numbers, none of them will have the digit 9, 8,
or 7. Also, I can get a one-to-one mapping to each of these numbers to
the real numbers in base 10. Thus isn't there an infinite number (an
infinity equal to the infinity of real numbers) that don't have the
string 987 three times over? This means that I should get a
probability of one of finding a number without the string 987 in it.

My question is, where is my argument flawed?

Regards,
Sanjay
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Date: 05/02/2001 at 17:48:12
From: Doctor Ken
Subject: Re: Question regarding Probability of Repeating Digits

Hi Sanjay,

Here's the basic flaw: just knowing that one infinite set A is
contained within another infinite set B doesn't tell you much about
the relative sizes of A and B, except that A can't be larger than B.
But they could be the same size, or A could be half as large as B, or
the same size as B, or "of zero size" within B.

For example: let B be the set of all real numbers between zero and
one.  The measure of this set is 1.  Then consider the following three
sets:

A1 = all members of B except 0.173
A2 = all irrational members of B
A3 = all members of B less than 0.5
A4 = all rational members of B

By the standard measure of sets of real numbers, A1 and A2 have
measure 1, A3 has measure 0.5, and A4 has measure 0 (because it is the
complement of A2). All four sets are infinite, but if you choose a
number at random between 0 and 1, the probabilities of choosing a
member of A1, A2, A3, or A4 are 1, 1, 0.5, and 0, respectively.

Note that the complement of A1 is only a single number, but the
complement of A2 is an infinite set, and that this fact doesn't affect
the final probability.

Now, in your line of reasoning you have shown that there are an
infinite number of decimal expansions that do not contain 7's, 8's, or
9's. This part of your argument is valid. But it does not follow that
this set has nonzero measure within the larger set.

To see it another way, note that there are also an infinite number of
decimal expansions containing ONLY the digits 7, 8, and 9 (by
correspondence with base-3 digits, if you like). And then there are
also an infinite number of decimals containing at least one digit from
0-6 and at least one from 7-9. Together with your set, these three
sets are mutually exclusive, and contain all possible decimals. But
the fact that each of them is an infinite set doesn't tell you
anything about their relative measures - it turns out that your 0-6
set and my 7-9 set both have measure zero, and the other set has all
the remaining measure.

- Doctor Ken, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Probability

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