I was issued a cellphone number that has my 6-digit birth date in it. Does anyone know how to calculate the odds of that happening? If there are 10 billion possible 10-digit phone numbers, and a 6-digit number can only occur 5 times within a 10-digit number, does that mean the odds are basically 1 in 2 billion? I know there aren't exactly 10 billion possible phone numbers, maybe around 8 billion, but I'm just trying to get a general idea.
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$\begingroup$There are $10^{10}$ possible $10$ digit phone numbers.
Your birthdate has six spots so if it begins is spot $a$ it will ocuppy spots $a,a+1,a+2,a+3,a+4,a+5$ and $a$ can be $1$ to $5$. So there are $5$ places for this to begin.
There are $4$ spots that are not in your birthdate. they can be anything. There are $10^4$ things they can be.
The probability is $\frac {5*10^4}{10^{10}} = \frac 5{10^6}$ or $1$ in $200,000$.
$\endgroup$ 3 $\begingroup$There will be two cases $$ CASE I. $$ When date is single digit because 10 digits of cell phone do not start with 0. If e.g if date of birth is 02 Aug 99 digits will be 020899 and 10 digit cell number will not start with 0. No. Of such dates will be $ 9*12 = 108 $ it can start from 2 nd place to 4th place hence number of ways are $ 3*9*{10}^3 = 27000 $
$$ CASE 2 $$ when date is 2 digit number then again 1st digit cannot be 0 hence number of ways will be from all substract when begin with 0 i.e $ = 4*{10}^ 4 -3 *9 * {10}^3 = 13*{10}^3 = 13000 $
Hence total number is 40000
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