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# Probability of picking a natural digit

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Are all the natural numbers written on slips of paper and put into a hat?

The probability is 1.

Agree.

If the problem is modified to: pick a random 'n' digit number (leading zeros are permitted). What is the probability that it has a digit '1'? The answer is: (10n - 9n)/10n. Now as, 'n' tends to infinity, the ratio approaches '1'.

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If the problem is modified to: pick a random 'n' digit number (leading zeros are permitted). What is the probability that it has a digit '1'? The answer is: (10n - 9n)/10n. Now as, 'n' tends to infinity, the ratio approaches '1'.

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It's also the probability that a "2" is present ... and so on.

It's not a surprising result, once one realizes that most natural numbers are greater than Graham's number.

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It's also the probability that a "2" is present ... and so on.

It's not a surprising result, once one realizes that most natural numbers are greater than Graham's number.

I read up a bit about Graham's number - but can you explain the connection to the puzzle here?

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Sure. Graham's number has so many digitis, it's a virtual certainty that it, and every greater number, contains at least one of each of the digits 0 1 2 3 4 5 6 7 8 and 0. But even though Graham's number is almost indescribably large, there are infinitely more larger natural numbers than smaller ones.

Therefore the probability requested in the OP is 1.

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karthickgururaj,

bonanova is correct that the probability is so close to 1 that it is extre-e-e-e-mely improbably that you would pick a number that did not contained the digit 1 from the set of natural numbers where the upper bound was Graham's number. What isn't true is that this probability is 1. Though extremely small, the probability is not infinitesmally small which is a requirement to correctly declare it to be exactly equal to 1. I believe bonanova was just trying to make the point that it is so close to one that you would not have lived long enough, even if you were born with the Big Bang and began your count then, to count the number of 9's past the decimal point in the probability when it finally was a different digit, thus it may as well be treated as 1.

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@DejMar Thank you for clarifying. That's what I meant to imply by "virtual certainty."

Also, in your post, change "upper bound" to "lower bound."

Strange to use lower bound and Graham's number in the same sentence. ##### Share on other sites

• 0 Right! Lower bound. And to further clairfy for others, the probability that is "virtually certain", as bonanova just stated in the previous posts, applies to using Graham's number as the lower bound. For the set of natural numbers, the probability is so infinitesimally distant from 1, that it is 1.

Edited by DejMar

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