# Re: Probabilities

*From*: jt64@xxxxxxxx*Date*: 3 Feb 2007 15:45:21 -0800

On 4 Feb, 00:25, j...@xxxxxxxx wrote:

On 3 Feb, 21:53, j...@xxxxxxxx wrote:

On 3 Feb, 18:39, "RNoster" <nos...@xxxxxxxxxx> wrote:

<j...@xxxxxxxx> wrote in message

news:1170456859.947502.206370@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Given 8 random bits

0 =1/2 50%

00 =1/4 25%

000 =1/8

0000 =1/16

00000 =1/32

000000 =1/64

0000000 =1/128

00000000 =1/256

that is 8 random bits out of an 8 bit word ?

how many words have just one 0 in it, or just a leading 0 ?

Well i do not think random material comes in word it is 8 bits chosen

with a PRNG from random source.

Maybe a bit unclear but it is leading or starting if you will of 8

bit

0XXXXXXX =1/2 50%

00XXXXXX =1/4 25%

000XXXXX =1/8

0000XXXX =1/16

00000XXX =1/32

000000XX =1/64

0000000X =1/128

00000000 =1/256

Given 16 random bits is the formula for two reocurrences of any 8 bit

string or the string above

when read out bytewise

1/256*1/256 ?

how long is the message string?

Given 16 random bits i would say it is 16 bit long.

how many distint patterns are you looking for?

Given two reoccurences of any 8 bit string that would be two

what are the word boundries?

Either you are talking above my head, or it isn't relevant information

answer is quite complex.

No it can not be with 16 bits i can use a PRNG to pick chunks of 16

bit from a random source and statistical check if my formula correct.

I will not do it though this is a *TRIVIAL* problem there must be a

formula and i think my is correct for 16 bits with two occurences of

an eight bit string.*when read out bytewise*

Jonas Thörnvall

When it comes to math i am sometimes a bit dent :)

Given 24 random bits is the formula for two reocurrences of any 8 bit

string or the string above when read out byte wise

2'*(1/256*1/256) ?

Ok this is most certainly up the walls :D

Jonas- Dölj citerad text -

- Visa citerad text -- Dölj citerad text -

- Visa citerad text -- Dölj citerad text -

- Visa citerad text -

Clarification: Given two reoccurences of any 8 bit string that would

be two identical 8 bit patterns within the 16 bit string when read out

bytewise- Dölj citerad text -

- Visa citerad text -

Sometimes the probability for it to happen is 1 but it still does not

happen and the string is still perfectly random. Yes yes how could

that be well in infinity the the probability goes to infinity. It

isn't that useful though, maybe that i where the boundaries comes in?

Well no my association goes stray again.

(I guess statistics have a hard time to deal with infinity so have

i...............)

And maybe Die HARD to ?

Jonas Thörnvall.

.

**Follow-Ups**:**Re: Probabilities***From:*jt64

**References**:**Probabilities***From:*jt64

**Re: Probabilities***From:*RNoster

**Re: Probabilities***From:*jt64

**Re: Probabilities***From:*jt64

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