Interfering With PI

The note Cantor's Diagonal Proof discussed the application of 
Cantor's diagonal argument to the list of rational numbers

   1   1/2  2/1   1/3  3/1   1/4  2/3  3/2  4/1 ...

where we take k=1,2,3,... and for each value of k list all the 
fractions n/d  with  n + d = k  and  gcd(n,d)=1, in increasing 
order of n.  Taking the decimal representations of these numbers
and substituting a new digit for each number along the diagonal
we start out with

     [3]. 0 0 0 0 0 0 0 0 0...
      0 .[1]0 0 0 0 0 0 0 0...
      2 . 0[4]0 0 0 0 0 0 0...
      0 . 3 3[1]3 3 3 3 3 3...
      3 . 0 0 0[5]0 0 0 0 0...
      0 . 2 5 0 0[9]0 0 0 0...
      0 . 6 6 6 6 6[2]6 6 6...
      1 . 5 0 0 0 0 0[6]0 0...
      4 . 0 0 0 0 0 0 0[5]0...
              etc.

Thus our new number begins with the digits 3.14159265...  The 
question raised by the earlier post was

 Does anyone know at what point we would have to deviate 
 from the digits of pi as we continue down this list?

I've checked and found that we can continue to select the
digits of pi until reaching the fraction 4/13, which is the
83rd fraction on the list.  The 82nd digit (past the decimal 
point) of 4/13 = 0.307692 307692 307692 ...  is a 6, which
also happens to be the 82nd digit past the decimal point of
the number pi.  Here is the sequence of digits for pi and
the corresponding sequence of digits of the rational numbers:

PI:  3 1415926535 8979323846 2643383279 5028841971 6939937510
 Q:  1 5030060000 6003005060 0700001530 0204300600 0500380405

PI:    5820974944 5923078164 0628620899 86...
 Q:    6003006750 6018301000 6947503003 46...

It seems a little surprising that it takes this long to reach
a pair of matching digits.  Assuming the digit strings are
uncorrelated and that the digits of pi are "normal", I would
have expected this to be like a sequence of Poisson trials
with a probability of 1/10 for each trial.  The distribution
of waiting times for the 1st "success" in a Poisson process
with rate 1/10 is given by the gamma distribution

               f(t)  =  (1/10) e^(t/10)

which, not surprisingly, has an expected value of 10.  This
means the expected "waiting time" for the first match is 10
digits.  The probability of the first match occurring at or 
above the 82nd digit is e^(-82/10) = 1/3641.  I suppose this
unusually long waiting time is due mainly to the fact that
the first "0" in pi doesn't appear until the 32nd digit,
whereas the rational sequence Q has an excess of 0's, 
especially in the early numbers.  Still, even after the 0's
start to appear in pi at a more "normal" rate, they continue
to miss the 0's in Q for quite a while, and in fact the first
match is a 6, rather than a 0.

What is known about the asymptotic distribution of digits
in the Q sequence?  Do they approach a "normal" distribution?