'Z-standardizing' data based on Poisson process Hello all this is my first post on Cross Validated, so please let me know if it is not in an acceptable form.  
I have been attempting to analyze a data set where I have a Bernoulli process that is generating a sequence of two outcomes with given probabilities. I am calculating the average length of runs of a given outcome  which I believe could be considered a Poisson process for example:
sample sequence= 0,0,0,1,1,0,0,0,0,1,1,1
mean run length=  3


This is a plot of some fake data that I generated to illustrate the distribution.
I am then attempting to quantify how the 'streakiness' of this sequence compares to a distribution of other sequences, and I would like the output to be in the form of a standard normal variable. More succinctly: given my sample sequence and the distribution above, how many 'standard deviations' more streaky is the sample sequence than the distribution?
So far I have tried using a Freeman-Tukey transformation e.g.: 
$r=sample \ run \ length$
$\mu=mean \ of \ distribution$
$X=\sqrt{r}+\sqrt{r + 1} - \sqrt{4*\mu+1}$
But this is providing some odd output.  For example if the run length in the sample is equal to the mean run length of the distribution I would assume X above should be 0 but it is not.  
My question is twofold.


*

*Am I correct in assuming that the run lengths above are actually the result of a Poisson process?

*If so, Is this the correct "standardizing transformation" for these type of data?

 A: Run lengths in a Bernoulli process will not be Poisson. 
The number of 1's until the next 0 is geometric; and the number of 0's to the next 1 is geometric (of the type starting from 0); consequently run-lengths of each type (0's and 1's) will be geometric (of the type starting from 1). 
The parameters of these two geometrics will not in general be equal (because $p$ is not always $\frac12$), so if you combine the two kinds of run-length together you'll have a "mixture" of geometrics. However, it's not quite a mixture in the normal sense because it always alternates between each type (i.e. they're sequentially dependent, not independent) - the number of runs of 0's and 1's cannot differ by more than 1. 
Nevertheless for very long sequences, we can get the distribution of run lengths readily enough:
$P(R=r) = \frac12[p^{r-1}.(1-p) + (1-p)^{r-1}.p]\,,\: r=1,2,...\: 0<p<1$
Consider simulating a set of runs with $p=0.2$:
 bern.2 <- rbinom(1000000,1,.2)
 t <- rowSums(table(rle(bern.2)))
 x <- as.numeric(names(t))
 r <- 1:max(x)
 pmf <- (.2^(r-1)*.8+.8^(r-1)*.2)/2
 plot(x,t/sum(t),log="y",xlim=c(0,50))
 lines(r,pmf,col=4)

which produces the following display of of observed log(proportions) at each run length, as well as logs of the above distribution (pmf):

It shows excellent agreement. 
[By contrast, the length of the first run in such a process will have a pmf of $p^r \cdot (1-p) + (1-p)^r\cdot p$.]

Even though the negative answer to the first question removes the need to address the second question, note that even in a Poisson, when $X=\mu$, the Freeman-Tukey transform of $X$ is not quite $0$. Note that $\sqrt{\mu}+\sqrt{\mu+1}\geq\sqrt{4\mu+1}$ for $x\geq 0$, with equality only at $x=0$.
