Probability that a bit is $1$ in an infinite binary sequence where $000$ must be followed by $1$ Probability of "1" and "0" are 1/2 for one position, except three "0" in a row makes next position must is "1", says "0001" is allowed while "0000" is prohibited. So what's probability of "1" in the whole sequence with infinite length?
Probability of "1" is defined as number of "1" divided by number of "1" and "0" in the sequence, what's the limitation when the sequence is infinite?
 A: We can turn this around, and ask "What is the average length of a sequence with $n\, 1$'s?"
Such a sequence can be seen as a sequence of "blocks" of the form $1$, $01$, $001$, and $0001$, with probabilities $\frac12,\frac14,\frac18,\frac18$ respectively. So the expected length of a block is $\frac12\cdot 1+\frac14\cdot 2+\frac18\cdot 3+\frac18\cdot 4=\frac{15}{8}$.
Hence the expected length of a sequence containing $n\,1$'s (ignoring end effects which become negligible as $n\to\infty$) is $\frac{15}{8}n$. So the probability of a $1$ tends to $\frac{8}{15}$ as $n\to\infty$.
A: Imagine Markov chain with 4 states which model arbitrary position in the string, far from beginning:


*

*1: position is $1$;

*2: position is $0$, previous position is $1$

*3: position is $0$, previous positions are $10$

*4: position is $0$, previous positions are $100$
$$\require{enclose}
\begin{array}{ccccccccc}   
\enclose{circle}{^2\ 0} & \xrightarrow{\tfrac12} & \enclose{circle}{^3\ 0} & \xrightarrow{\tfrac12} & \enclose{circle}{^4\ 0}\\\
& _{\tfrac12}\searrow\nwarrow^{\tfrac12} & \tfrac12\downarrow & \swarrow\ 1\\\  
 && \enclose{circle}{^1\ 1}\\
&&\circlearrowright\tfrac12\\
\end{array}  
$$
Evidently $p_2=p_1/2$, $p_3=p_2/2$, $p_4=p_3/2$, and
$$p_1+p_2+p_3+p_4=1$$
which gives
$$p_1=\frac{8}{15}$$
A: To clarify a possible ambiguity:  I think there are competing notions of the probability involved here (hence the incompatible results people are reporting).  The way I interpreted the question, I count all the good sequences of length $n$, where a "good" sequence is one that does not contain $0000$.  Call that number $A_n$.  Now count all the $1's$ in all those sequences, calling that number $T_n$. I then compute (or estimate) $\lim_{n\to \infty} \frac {T_n}{n\times A_n}$.  This may well not be what was intended, but it was how I interpreted the question.
To stress:  generating a good sequence bit by bit, forcing a $1$ after $000$, does not give the same result.  Indeed, I'd argue that this was a biased way of producing good sequences.  Consider, say, sequences of length $4$.  There are $15$ good sequences of length $4$ (as $0000$ is the only bad one) so a uniform measure would give each probability $\frac 1{15}$.  But the bit generator method gives $0001$ probability $\frac 18$, twice the probability of any other good sequence.  As that particular sequence is "poor in $1's$" this method underestimates the probability of getting a $1$ (from the unbiased viewpoint).
As to the unbiased probability:  Recursive methods give us some sense of the answer.
Let $A_n$ be the number of "good" series of length $n$.  Then $A_1=2,A_2=4,A_3=8, A_4=15$ and we have the recursion $$A_n=A_{n-1}+A_{n-2}+A_{n-3}+A_{n-4}$$ since any good sequence of length at least $4$ ends in one of $1,10, 100, 1000$
Now let $T_n$ be the total number of $1's$ in all the good sequences of length $n$.  Then $T_1=1, T_2=4, T_3=12, T_4=32$ and we have the recursion $$T_n=A_n+T_{n-1}+T_{n-2}+T_{n-3}+T_{n-4}$$
This is easily computed and the ratio $\frac {T_n}{n\times A_n}$ appears to approach a limiting value of about $.566$
To compute this analytically, we note that the recursion for $A_n$ has the characteristic polynomial $$x^4=x^3+x^2+x+1$$ which is a "Pisot" polynomial, in that one real root has norm greater than $1$ and all the other roots have norms less than $1$. The associated Pisot number is $\alpha \approx 1.9276$  and it follows that $A_n\sim C\alpha^n$ for some constant $C$.  It follows from this that, for large $n$, we have the ratios $$\frac {A_{n-i}}{A_n}\to \frac 1{\alpha^i}$$.  So the expected length of the "ending block" of a randomly chosen good sequence of length $n$ is $$\sum_{i=1}^4 \frac i{\alpha^i}\approx 1.7657$$
which implies that the desired answer is approximately $$\frac 1{1.7657}=.56634$$ in harmony with the recursion. 
