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.
.
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