## Probability of Repeated Identical Numerals

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.
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 Recognitions: Gold Member Science Advisor 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
 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.

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## Probability of Repeated Identical Numerals

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
 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!
 Recognitions: Gold Member Science Advisor 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
 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!
 Recognitions: Gold Member Science Advisor 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
 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!
 Recognitions: Gold Member Science Advisor Staff Emeritus Start with a three-digit number. Ask: what's the probability that that three-digit number is 555? The answer is: one in a thousand. 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
 Recognitions: Gold Member Science Advisor Staff Emeritus Unless, of course, you want none of the X's to be a 5!

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 Quote by HallsofIvy Unless, of course, you want none of the X's to be a 5!
He verified this in post #7.

- Warren
 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.
 Recognitions: Gold Member Science Advisor 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: 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

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