According to this site, the number of primes between 0000000000 and 9999999999 is 455052511, or about 4.55 percent of them. If you were to throw all 1010 of these integers into a hat and draw them out at random (with replacement), how many would you have to draw before being 99.99 percent sure that at least one of them was prime? Well, the probability of not getting a prime in one draw is 1 - 0.0455 = 0.9545. So the probability of getting no primes in n draws is 0.9545n. We want to draw enough times to reduce this probability to 0.0001, so we need to solve 0.9545n = 0.0001. This is the same as saying n log 0.9545 = log 0.0001, or n = log 0.0001 / log 0.9545 = 197.78.

What's the point of all this? It's simply to show that you don't need a million digits of e to find a 10-digit prime embedded in them. 207 (198 + 9) are probably (i.e. with 99.99% assurance) enough. This assumes, of course, that sets of consecutively overlapping 10-digit subsequences in the decimal expansion of e represent independent, random samples from the integers between 0000000000 and 9999999999. That may or may not be true.

In reply to Re: Caclulation of e by Dr. Mu
in thread Caclulation of e by doc

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