# how do i find my broadband phone number?

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You should then see a phone number in the panel that appears.

The chance of a particular 10 digit number coming up with 10 digits chosen is obviously 1/(10^10). Now, The chance of it coming up in 11 digits is just this but in 2 different ways, isn’t it? It doesn’t matter what the ‘spare’ digit is, so we forget any probabilities associated with that. So for 11 we have 2/(10^10). Then, for 12 digits, we have 3 ways of it coming up, with a 1/(10^10) chance of each, so 3/(10^10), etc…

Following this on, we have an (n – 9)/(10^10) chance of it coming up in n digits (where n β₯ 10 obviously).

To make this chance equal to 1, we get

(n – 9)/(10^10) = 1

n – 9 = 10^10

n = 10^10 + 9 = 1000000009 digits.

This should be the same for any string of numbers, since each digit in the long string is independent of the last, every digit has a 1/10 chance of being every particular number, hence every 10 digit number has an equal chance of coming up. Hence E_x = E_y = 10^10 + 9.

I have tried to find holes in this, but the more I think about it the more it seems this simple. Am I being naive? If someone is willing to point out a flaw, I’d welcome the clarification π

ADDED – ah I see, it’s not quitre as clear cut; I suspected as much :/ My result for the number E_x should be correct I think, in light of your link; it’s only when we have repeated numbers we have to recalculate. Oh well, back to the drawing board π

ADDED AGAIN: Ah, after looking at your link, and following the logic there, isn’t the answer

10^10 + 10^9 + 10^8 + … + 10^2 + 10, or

E_y = 1111111110 – since all numbers in the chain y are the same, we have an expected time of 10^10 to get to ten 9s, but then an expected time of 10^9 since we can ‘overlap’ the last and first eight 9s, plus 10^7 etc. etc.?

… hence E_x > E_y, which we expect since it would take longer to find that completely random string than it would to find the string of 9s; the number of possible ‘overlaps’ makes it much more probable (in the same way that the repeated ABRA in ABRACADABRA makes it more likely to find than 11 different letters in a string.)

This is probably counter-intuitive to what people think (the common view is that the lottery is likely to come up with an ordered sequence than a random one, where this is not true); but then, each ‘choice’ is not independent and anyway probability can be counter-intuitive sometimes π

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