BMAD Posted October 17, 2014 Report Share Posted October 17, 2014 What is the probability that you can pick a natural number that contains the digit 1 in the set of natural numbers? Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted October 17, 2014 Report Share Posted October 17, 2014 Are all the natural numbers written on slips of paper and put into a hat? The probability is 1. Quote Link to comment Share on other sites More sharing options...

0 karthickgururaj Posted October 17, 2014 Report Share Posted October 17, 2014 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: (10^{n} - 9^{n})/10^{n}. Now as, 'n' tends to infinity, the ratio approaches '1'. 1 Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted October 18, 2014 Author Report Share Posted October 18, 2014 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: (10^{n} - 9^{n})/10^{n}. Now as, 'n' tends to infinity, the ratio approaches '1'. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted October 19, 2014 Report Share Posted October 19, 2014 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. Quote Link to comment Share on other sites More sharing options...

0 karthickgururaj Posted November 1, 2014 Report Share Posted November 1, 2014 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? Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted November 1, 2014 Report Share Posted November 1, 2014 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. Quote Link to comment Share on other sites More sharing options...

0 DejMar Posted November 1, 2014 Report Share Posted November 1, 2014 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. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted November 1, 2014 Report Share Posted November 1, 2014 @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. Quote Link to comment Share on other sites More sharing options...

0 DejMar Posted November 1, 2014 Report Share Posted November 1, 2014 (edited) 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 November 1, 2014 by DejMar Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted November 2, 2014 Report Share Posted November 2, 2014 Read more. Quote Link to comment Share on other sites More sharing options...

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What is the probability that you can pick a natural number that contains the digit 1 in the set of natural numbers?

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