(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An infinite-length binary text create from these patterns, with probability:

( For example: 0111001001010 ("0" "11" "100" "100" "101" "0") is a valid excerpt from the text)

- "0": 0.8
- "11": 0.1
- "100": 0.05
- "101": 0.05

I temporary call a,b,c,d for P("0"), P("11"), P("100"), P("101")

Problem: create a probability table for length-3 patterns (that is, find P(x): x="000","001","010",...,"111", x is excerpted at any point of the text)

2. Relevant equations

3. The attempt at a solution

Some of the patterns can be determined easily: 000, 011, 100, 101, 110 with

P(000)=a^{3}=0.512

P(011)=P(110)=a*b=0.08

However, there are no evidence about 001,010,111. So I think of context, e.g., 001 is create with 2 "0" and anything start with 1 ("11","110","101") or 2 first zero from "100" and final 1 from anything start with 1, hence:

P(001)=a^{2}*(b+c+d) + c*(b+c+d)=0.138

With this logic, we have:

P(010)=a*d+c*d+d*(b+c+d)=0.0525

P(111)=b*(b+c+d)+(b+d)*b=0.035

However, that logic must be applied for 000,011,100,101,110 as well:

P(000)=a^{3}+c*a=0.552

P(011)=a*b+c*b+d*(b+c+d)=0.095

P(100)=c+(b+d)*a^{2}=0.146

P(101)=d+(b+d)*a*(b+c+d)=0.11

P(110)=b*a=0.08

And after all, the sum of (1≤x≤8),P(x)=1.2085

I know that there is something wrong with taking the context to account, but I don't really know any other ways than considering the context.

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# Homework Help: Statistic: Pattern with context dilemma

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