Re: Randomness




Matt Zellman wrote:
Matthias Klaey wrote:
"Matt Zellman" <matt.zellman@xxxxxxxxx> wrote:


mensanator@xxxxxxxxxxx wrote:
Matt Zellman wrote:
are the following string of digits random?

Not likely.


I suppose that was kind of a stupid question. Are they random by some
definition of random? and if so, what definition?


0101000000000000000000000000000010000100100000000110001000000000

0000110110111110001010100010011100110011011011011001100110100110

1101110011101111010000001101011100000111101111111101000110100110

0000000000000000000000000100000000010000000001000000000000001000

1010110100111010011010100010011001100010011010010001100100100110

1010100111100001010110111101000100001000010100101001000001011100

here's a better question. Could the preceding six strings of digits
conceivably be randomly generated?

Sure.

This question somehow reminds me of a theorem that I heard of, but
never actually seen: "For any given set of data there is a test such
that the data will pass the test" (or converse, "such that the data
will fail the test/such that the null hypothesis is rejected")
I believe it is attributed to Kolmogorov.
Does anyone know something more precise about this?

And Matt, by this theorem the answer to your firts question is "yes".
For the second: I have no idea. The best reference I can give you is
D.E.Knuth, The Art of Computer Programming, Volume 2, Chapter 3, Third
Edition, Addison-Wesley 1998, ISBN 0-201-89684-2. There you may learn
more about random sequences than you ever wanted to know :)

Greetings, Matthias Kläy

Thanks. The way the strings were generated is as follows:

I started with the first however many digits of pi (a number whose
digits are provably normal

They are?

), and applied a series of tests to it:

Oh, a series of tests.

I feel like the guy with the spitoon, I thought it was all one string.


the first sequence gives a 1 for every digit that is a 0 or 1

So eight out of every 10 digits will give 0. How is that random?
Any tests that doesn't produce the same number of 1's and 0's
certainly isn't going to be random. Even if it does, it will still
depend on how they are distributed.

Your coin is definitely biased.

the second sequence gives a 1 for every digit greater than or equal to
5

So half will be 1's and half will be 0's. It still won't be random,
but it'll be harder to show that.

the third sequence gives a 1 for every digit that is odd

Again, same number of 1's as 0's.

the fourth sequence gives a 1 for every digit that is the same as the
previous digit

So you'll get too many 0's. Not random.

the fifth sequence gives a 1 for every digit that is greater than the
previous digit

Tie goes to 0, so you'll have too many 0's. Also note that under
this rule, you can't have more than 9 consecutive 1's. This will
be an obvious giveaway of non-randomness when you make
the sequence big enough since, in a random distribution, all
sequence lengths occur eventually.

the sixth sequence gives a 1 for every digit that is a 3,4,5, or 6

Again, simply too many 0's. Obviously non-random.


The resulting sequences are not (necessarily) normal, but I think they
can still be described as "random" as long as there is some nonzero
chance that a digit could be either a zero or a one.

But you have to say up front what the probabilities are if they
aren't 1/2. So, no, the sequences with differing 1 and 0 counts
can't be described as random. And the fifth sequence can't be
described as random even if you give the probabilities.

.



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