# Probability of Repeated Identical Numerals

1. Jun 17, 2007

### John14:6

I am attempting to determine what the statistical probability is of a repeated numeral occurring within a larger number. Specifically, the base number I am referencing has 8 digits. How likely would it be to have the same 3 numerals (such as 444) occur in the middle of the number (such as a number of 1944482)? This pattern would only occur 10 times (000, 111, 222, 333, 444, 555, 666, 777, 888, 999)--the digits to the left and to the right of the center 3 numerals could be any digits in any order.

Any help would be greatly appreciated.

2. Jun 18, 2007

### chroot

Staff Emeritus
You need to clarify the question more. Are the patterns RRRXXXXX, XRRRXXXX, XXRRRXXX, XXXRRRXX, XXXXRRRX, XXXXXRRR, where X is any digit and R is a repeated digit, all allowed?

Also, if this is homework, you must post your attempts at a solution, per PF guidelines.

- Warren

3. Jun 18, 2007

### John14:6

The only pattern allowed is XXRRRXX, and X can be any digit, including 0 (e.g., 0155586). Actually, I'm looking for the probability of just one particular set (555) occurring and not any others (i.e., 2377786, 6233358).

Although it has been years since my stats class in graduate school, the initial thought is that the probability would be 1 in 9,999,999. But the number must have 8 digits (no less); also, it can have 0 as a first digit (such as 0655532). Consequently, I realize my initial process is flawed.

Thanks for helping. Although this seems like homework (to me, too), it has to do with a nonprofit group I'm working with that wants to use this statistic in some materials it's developing.

4. Jun 18, 2007

### chroot

Staff Emeritus
Well, this is the second time you've repeated it.... but XXRRRXX doesn't have eight digits; it has only seven. As a result, I'm pretty thoroughly confused about what you actually want.

- Warren

5. Jun 18, 2007

### John14:6

Warren,

There is no excuse that I have not yet learned to count! Yes, sir, you are correct--it is seven digits and not eight. My sincere apologies!

It has been an extremely long day--but no excuses for incompetence!

Again, so sorry for all the confusion ... you are a good man for being so patient!

6. Jun 18, 2007

### chroot

Staff Emeritus
So, can you tell me again what you're looking for? Is it XXX555XX? or XX555XXX? Or either? Are numbers like 55555555 allowed, or must the repeated number appear nowhere else but in the center?

- Warren

7. Jun 18, 2007

### John14:6

OK, it is XX555XX, which is a total of 7 (yes, it is 7!) digits. And yes, the digit 5 can also appear as X. More specifically, X can be any digit in any order from 0 through 9, which means the 7 digits could start with 0. The only constant is 555 in the center of the 7 digits.

Thank you. Warren, you are a very patient person!

8. Jun 18, 2007

### chroot

Staff Emeritus
Well, if the position of the digits is constant (i.e. XX555XX), and the other digits are uncorrelated, then the problem is significantly reduced in complexity. You can simply ignore the X digits, and just ask:

What's the probability of a three-digit random number equalling 555?

Since there are 1,000 three-digit numbers, the chance of this is simply 1/1,000.

- Warren

9. Jun 18, 2007

### John14:6

With the question "What's the probability of a three-digit random number equalling 555?" does it make a difference that this three-digit number is embedded in a seven-digit number? I would think that it is not only a three-digit number equalling 555, but also that it is both preceded and followed by 2 random numbers. In other words, the 7-digit number could not be X555XXX nor XXX555X, but rather XX555XX. Furthermore, the three-digit number cannot be embedded in any more or less than a 7-digit number (e.g., it could not be XXX555XXX nor X555X).

Again, you are very patient!

10. Jun 18, 2007

### chroot

Staff Emeritus

Now ask: what's the probability of prepending a valid two-digit number? Since any two-digit number is acceptable, as you said, this probability is exactly one.

Now ask: what's the probability of appending a valid two-digit number? Since any two-digit number is acceptable, as you said, this probability is also exactly one.

The product of these probilities is:

$1 \cdot \frac{1}{1000} \cdot 1 = \frac{1}{1000}$

- Warren

11. Jun 18, 2007

### HallsofIvy

Staff Emeritus
Unless, of course, you want none of the X's to be a 5!

12. Jun 18, 2007

### chroot

Staff Emeritus
He verified this in post #7.

- Warren

13. Jun 18, 2007

### John14:6

Yes, I agree that the probability of both prepending and appending any valid 2 digits is one. However, the other factor here is the position of the 3-digit number of 555: it must occupy the third, fourth, and fifth position in the 7-digit number. Therefore, although prepending/appending any random digits still results in a probability of one, the position of these digits can only be first, second, sixth, and seventh. It seems to me that this positioning--as well as limiting the total number of digits to 7--would affect the probability. Thanks again for your followup.

14. Jun 18, 2007

### chroot

Staff Emeritus
Well, I'm afraid you're incorrect. The probability of the numbers you're talking about, XX555XX, where X can be any digit, is 1/1000. I've explained it several times, in several different ways, and cannot be any more straightforward.

If you do not believe my answer is correct, here's proof -- a simple Python program to measure the probability by looking at a million randomly-generated seven-digit numbers:

Code (Text):

import random

N = 1000000
hits = 0

for i in xrange(N):
number = random.randint(0, 10**7-1)
if str(number)[2:5] == '555':
hits = hits + 1

print "Hit probability: ", float(hits) / float(N)

When it is run:

\$ python prob2.py
Hit probability: 0.00102

The answer is one in a thousand. There is no room for disagreement.

- Warren

15. Jun 18, 2007

### nrqed

Another way to show that Chroot is right is the following: There are 10^7 possible 7 digits numbers (assuming that 0 is valid even as the very first digit). Of these 10 million numbers, how many are of the form XX555XX? Well, the answer is simply 10^4, right? Therefore the probability is 10^4/10^7 = one in a thousand.

Because numbers starting with a zero are acceptable, the answer does not depend on where the 5's are placed. The probability is the same to get a number of the form X5XX55X for example, or 5XXXX55.

16. Aug 6, 2007

### pinwheel111

Here's an easier way to think about it. If zero's are allowed to be one or both of the first two numbers, then there are 100,000 possible outcomes of the form aaxxxaa. Since there are 10,000,000 possible outcomes, then your probability is 100,000/10,000,000, or 1/100. Finding out that there are 100,000 possible outcomes is simply a matter of counting. 00xxx00, 00xxx01, 00xxx02, and so on. You can remove the x's and realize that you're just counting to 9999 (99xxx99) beginning with 0000. That's 10,000 numbers. Multiply this by 10 for all the possibilities of x (0-9). 100,000. Divide that by the 10,000,000 numbers from 0000000 to 9999999 and you get .01. For the probability of any one particular value of x, simply omit multiplication by 10, and you get 1/1000.

Last edited: Aug 6, 2007