Re: Random number generation and probabilities

Top-posting repaired. Response at bottom.

joyo wrote:
Randy Poe wrote:
Please don't top-post. Top-posting repaired.

joyo wrote:
Randy Poe wrote:
joyo wrote:
A set of numbers from 0 to 9 is generated randomly. When taken to
infinity, each number will output the same number of times.

Not phrased completely rigorously, but you are essentially
expressing the Law of Large Numbers.

if any n digit number(N) is extracted from this set of generated
numbers, then the probability of occurence that number would be (0.1)^n

If the mean of the output numbers is calculated, then it would be
sum(0....9)/10 = 4.5

Qn:

In a non infinite generation (but high amount) , if the mean is K, K
not equal to 4.5, would the probabilities of any n digit number
extracted(N), be clustered around (0.1)^n

Yes. The number of occurrences of an n-digit number
X in a large digit string is a random variable. It
follows a binomial distribution, which is closely
approximated by a normal (bell curve) distribution.
It clusters around (0.1)^n and has a spread which
depends on the length of the whole digit string.

and would the maximum/averagr
difference between prob(N) and (0.1)^n be a function of (k,4.5)?

I don't know what k is, but the distribution of possible
values of "# of occurrences of N" does depend on
the total number of trials.
K would be the average of the non-infinite set. 4.5 is the average of
the infinite set. (numbers generated 0 to 9

Upper and lower case are different. Note that you used
n and N to mean two different things, but tried to
use k and K interchangably.


let N be n digit number extracted from a non infinite (but large)
group of randomly generated numbers. (0 to 9)

Is prob(N)- (0.1)^n = f(K,4.5)



The question is confused by the fact that you are
misusing the word "probability". You are talking about
*one* trial of let's say M digits. Each digit occurs
a fixed number of times, not necessarily M/10.
Any n-digit number N occurs some number of times.

You want to know if the NUMBER OF OCCURRENCES
of N in M digits can be deduced from K.

The answer is no. Here's a simple argument:
Suppose all digits occur equally often. Then K=4.5.
Pick some 3-digit number, say 123. It occurs some
nonzero number of times (approximately M/1000).

Now suppose the only digits that occur are 4 and 5,
equally often. Then K is still equal to 4.5, but
123 never occurs.

what is top post?

Top-posting is posting a reply above the thing you
are replying to. It confuses things. It is a standard in
e-mail, but discouraged in most groups on Usenet.

- Randy

.

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