# DailyDirt: Pi Math

### from the urls-we-dig-up dept

National Pie Day is not actually March 14th (although it really should be, if only to make it more memorable). But here's to the number, not the delicious dessert.
If you'd like to read more awesome and interesting stuff, check out this unrelated (but not entirely random!) Techdirt post.

Filed Under: calculations, irrational, math, numbers, pi

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1. Anonymous Coward, 15 Feb 2013 @ 3:03am

### Re: Re:

your infinite sequence of non-repeating (RANDOM) numbers that does not contain a 9 would still work as long as your sequence also played by THE SAME RULES, and itself did not contain 9's.

sure, you could also when trying to 'make up' such a silly theory include extra things and call them what you like.

does not change the facts, and changing the rules still does not work.

and the fact is with a very very very long string of random numbers you will get all possible sequences by virtue of the fact that it is a very long non-repeating string of numbers.

we can all work out that the digits 0 thru 9 DO REPEAT and the statistics of any one digit being either 0,1,2,3,4,5,6,7,8,9 is 10%, it's going to be that set of numbers REPEATED AT RANDOM, and an even random distribution means that ANY 1 digit sequence WILL APPEAR in that random distribution with a specific statistical probability.

And you certainly do not need an infinite length of the original string to achieve it.

If you do happen to have that infinite string of numbers, then your sequence, no matter how long will appear within that string an infinite amount of times.

I have nothing further to prove to you, it is simply up to you to gain some basic understanding, sorry.

I am not responsible for your ignorance.

The solution to this problem is to use what's known as the Poisson approximation to the binomial, when the numbers are large. We can actually approximate the above formula as:

Odds(finding string of length k in N digits of pi) = 1 - 1/e(N*0.1d).

http://www.angio.net/pi/whynotpi.html

if you have positions that can only contain numbers 0 to 9 with a random distribution, you have a 10% chance of ANY ONE of those positions being filled with a particular number.

that's for a 1 digit size sequence of Decimal numbers (no cheating and removing some).

so you have a 100% chance of finding your 1 digit sequence in that 10 digit number.

(assume your sequence number is 2) for you to get a 100% chance of finding a 2 digit sized sequence you will need to ensure that every to has following it a 0 to 9 number, so you will need at least 10 2's in your random number, enough 2's for there to be one of every possible digit (0 thru 9).

you will also need 10 1's 3's ... up to 10 9's all followed by a 0 to 9 number.

to get a 3 number sequence you will have to get all the numbers you already have from the first two numbers and add a 0 to 9 after each of them, and so on, as long as they are random with enough of them you have 100% chance of matching your sequence.

With small sequence numbers the original number does not have to be very large. As the probability of any 1 digit being the right on is 10%. you can get away with a 10 digit number.

with a tiny size of 100 million digits you can get a 99.995 chance of matching a 7 digit number, and 100 million is tiny compared to infinity.

but as someone said earlier trying to omit a digit is just stupid..

I find it amazing you have so much trouble with very big and infinite.
And that you don't even need an infinite length to find ANY length of numbers within it, you can calculate (well I can) how many (or how long) a random number had to be until a string of ANY length can be found within a string of random numbers.

tell me how long your sting is and ill give you the number of random digits required before you have a 100% chance of that string occurring.

ALL FINITE NUMBERS TOO.. (and not leaving some out !!!)..

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