Sevens Are Everywhere

Probability Level 4

The number $7$ appears everywhere. There are $7$ days in a week, $7$ wonders of the ancient world, and $7$ continents on the planet earth.

If a positive integer less than or equal to $n$ is chosen randomly, the probability that it contains a $7$ in it's decimal expansion is $p_n.$ As $n$ grows large, what does $p_n$ approach?

The answer is 1.

1 solution

Brian Charlesworth
Aug 24, 2015

It's easier to first look at the percentage of integers that do not contain any $7$ 's and then take the complement.

For $n$ -digit positive integers, $n \ge 1,$ without $7$ 's allowed we have $8$ choices for the first digit and $9$ choices for (any) subsequent digits. So for the $10^{n} - 1$ positive integers with $n$ or fewer digits,

$8 + 8*9 + 8*9*9 + ..... + 8*9^{n-1} = 8*\dfrac{9^{n} - 1}{9 - 1} = 9^{n} - 1$

of them contain no $7$ 's. Thus the desired percentage is

$A = 1 - \lim_{n \rightarrow \infty} \dfrac{9^{n} - 1}{10^{n} - 1}.$

But this last limit goes to $\left(\dfrac{9}{10}\right)^{n}$ and thus $0$ as $n \rightarrow \infty,$ so $A$ goes to $\boxed{1}$ in the limit.

What this means is that, from a probability standpoint, "almost all" positive integers contain at least one $7,$ (the same of which can be said for any of the digits $1$ through $9$ .) Yet another weird result thanks to our friend $\infty.$ :)

Question: Can the same be said for any finite string of digits? What about any infinite string of digits?

A very nice and simple solution to a problem which at first,looked daunting!Thank you sir!+1

Adarsh Kumar - 5 years, 9 months ago

Thanks, Adarsh. Counting the complement is so often the best approach with problems like these. :)

Brian Charlesworth - 5 years, 9 months ago

Hi Brian! There's no problem with the first 1-9 integers having only one 7 as digit. The problem goes to the subsequent digits, how about 70,71,72, 701, 702,703 and so on or integers starting from digit # 7? The subsequent integers don't just contain 1-integer having atleast one 7 of its digits.

Rene Dela Cerna - 5 years, 9 months ago

Yes, it can be tricky to count the integers with at least one $7$ directly, which is why I chose to count the complement, i.e., those numbers that contain no $7$ 's. It makes for a much cleaner approach. Counting directly would require one to use inclusion/exclusion so as to avoid double-counting.

Brian Charlesworth - 5 years, 9 months ago

i too came up with the same result. :)

Halbert Joshi - 5 years, 9 months ago

Sevens Are Everywhere

The answer is 1.

1 solution

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