Statistic: Pattern with context dilemma

[pyfgcr]

Homework Statement

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

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

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

Homework Equations

The Attempt at a Solution

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

P(000)=a³=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)=a²*(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)=a³+c*a=0.552
P(011)=a*b+c*b+d*(b+c+d)=0.095
P(100)=c+(b+d)*a²=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.

 

[haruspex]

For any bit in the sequence, there are 9 possibilities:
- the '0' in a '0' pattern
- one of the two bits in 11 pattern
- one the 3 in a 100 pattern
- one the 3 in a 101 pattern
Can you calculate the relative frequencies of these 9 cases?
Having done that, suppose you pick a bit as the first of a triplet. For each of the 9 possibilities, can you calculate the probabilities that the next two bits are 00, 01, 10, 11?

 

[pyfgcr]

Oh, now I understand my mistake: I don't take relative frequency in account. I calculate again and now the sum is 1. Thank you very much.

 

[aralbrec]

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

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

P(000)=a³=0.512
P(011)=P(110)=a*b=0.08

This isn't right unless I have misunderstood.

000 could be generated by ("0" "0" "0") or ("100" "0")
011 could be generated by ("0" "11") or ("100" "11") or ("101" "100") or ("101" "11") or ...

 

[pyfgcr]

Yes, this was my mistake before, but I have calculated again with relative frequency by haruspex suggested and now the problem has been solve. The new result is 0.536, 0.134, 0.095, 0.085, 89/1500, 41/1500, 7/150, 1/60 of 000, 001, 010, 011, 100, 101, 110, 111 respectively