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My phone is 2543663 (Posted on 2014-04-15) Difficulty: 2 of 5

I repeat rolling a fair die and writing the result down, - after a while I let the software to continue random generation of the integers 1 to 6 till the 1000th digit.

What is the probability for my 7-digit phone number being somewhere within the sequence?

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

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Two different exact answers | Comment 3 of 5 |
I couldn't help noticing that my "exact" answer has many more digits in fractional form than Charlie's "exact" answer.  Clearly, at least one of them is inexact.

Upon reflection, it seams that mine is the "inexact" one.  The problem, for instance, is that the probability that the first 7 digits contain the author's phone number is not independent of the probability that 2nd through 8th digits do.

The probability that the first one does not contain the number is 1-(1/6)^7 = 279935/279936 = .99999642775.

But if it does not contain the number, then it increases very very slightly the probability that digits 2 - 8 do, because it is not possible that digits 2-8 are 543663n.  In this case, the probability that digits 2-8 does not contain the number is 
(279935-6)/(279936-6) = .99999642767

Charlie:  Could I trouble you to calculate my original solution with some precision?  I am curious how close it is to the correct answer. 

  Posted by Steve Herman on 2014-04-15 18:29:07
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