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Probability of Repeated Identical Numerals 
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#1
Jun1707, 09:09 PM

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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
Jun1807, 01:43 AM

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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
Jun1807, 06:56 AM

P: 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
Jun1807, 01:53 PM

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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 


#5
Jun1807, 02:38 PM

P: 6

Warren,
There is no excuse that I have not yet learned to count! Yes, sir, you are correctit is seven digits and not eight. My sincere apologies! It has been an extremely long daybut no excuses for incompetence! Again, so sorry for all the confusion ... you are a good man for being so patient! 


#6
Jun1807, 02:40 PM

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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
Jun1807, 03:05 PM

P: 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
Jun1807, 03:11 PM

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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 threedigit random number equalling 555? Since there are 1,000 threedigit numbers, the chance of this is simply 1/1,000.  Warren 


#9
Jun1807, 03:33 PM

P: 6

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


#10
Jun1807, 03:37 PM

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Start with a threedigit number. Ask: what's the probability that that threedigit number is 555? The answer is: one in a thousand.
Now ask: what's the probability of prepending a valid twodigit number? Since any twodigit number is acceptable, as you said, this probability is exactly one. Now ask: what's the probability of appending a valid twodigit number? Since any twodigit number is acceptable, as you said, this probability is also exactly one. The product of these probilities is: [itex]1 \cdot \frac{1}{1000} \cdot 1 = \frac{1}{1000}[/itex]  Warren 


#11
Jun1807, 03:54 PM

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



#13
Jun1807, 05:08 PM

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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 3digit number of 555: it must occupy the third, fourth, and fifth position in the 7digit 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 positioningas well as limiting the total number of digits to 7would affect the probability. Thanks again for your followup.



#14
Jun1807, 05:19 PM

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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 randomlygenerated sevendigit numbers:
$ python prob2.py Hit probability: 0.00102 The answer is one in a thousand. There is no room for disagreement.  Warren 


#15
Jun1807, 06:50 PM

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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
Aug607, 11:27 AM

P: 7

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 (09). 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.



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