Mathematicians shocked(?) to find pattern in prime numbers There is an interesting recent article "Mathematicians shocked to find pattern in "random" prime numbers" in New Scientist. (Don't you love math titles in the popular press? Compare to the source paper's Unexpected Biases in the Distribution of Consecutive Primes.)
To summarize, let $p,q$ be consecutive primes of form $a\pmod {10}$ and $b\pmod {10}$, respectively. In the paper by K. Soundararajan and R. Lemke Oliver, here is the number $N$ (in million units) of such pairs for the first hundred million primes modulo $10$,
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
&a&b&\color{blue}N&&a&b&\color{blue}N&&a&b&\color{blue}N&&a&b&\color{blue}N\\
\hline
&1&3&7.43&&3&7&7.04&&7&9&7.43&&9&1&7.99\\
&1&7&7.50&&3&9&7.50&&7&1&6.37&&9&3&6.37\\
&1&9&5.44&&3&1&6.01&&7&3&6.76&&9&7&6.01\\
&1&1&\color{brown}{4.62}&&3&3&\color{brown}{4.44}&&7&7&\color{brown}{4.44}&&9&9&\color{brown}{4.62}\\
\hline
\text{Total}& & &24.99&& & &24.99&& & &25.00&& & &24.99\\
\hline
\end{array}$$
As expected, each class $a$ has a total of $25$ million primes (after rounding). The "shocking" thing, according to the article, is that if the primes were truly random, then it is reasonable to expect that each subclass will have $\color{blue}{N=25/4 = 6.25}$. As the present data shows, this is apparently not the case.
Argument: The disparity seems to make sense. For example, let $p=11$, so $a=1$ . Since $p,q$ are consecutive primes, then, of course, subsequent numbers are not chosen at random. Wouldn't it be more likely the next prime will end in the "closer" $3$ or $7$ such as $q=13$ or $q=17$, rather than looping back to the same end digit, like $q=31$? (I've taken the liberty of re-arranging the table to reflect this.)  
However, what is surprising is the article concludes, and I quote, "...as the primes stretch to infinity, they do eventually shake off the pattern and give the random distribution mathematicians are used to expecting."
Question: What is an effective way to counter the argument given above and come up with the same conclusion as in the article? (Will all the $N$ eventually approach $N\to 6.25$, with the unit suitably adjusted?) Or is the conclusion based on a conjecture and may not be true?
P.S: A more enlightening popular article "Mathematicians Discover Prime Conspiracy". (It turns out the same argument is mentioned there, but with a subtle way to address it.)
 A: If I have read the New Scientist article correctly, the so-called "discrepancy" is: If a prime ends in 1 (as in the first class), the observed probability that the next prime also end in 1 is not 1/4. This observation can be explained by elementary probability with the assumption that the classes are indeed truly random. I was also troubled by this and I asked it in the math overflow forum: https://mathoverflow.net/questions/234753/article-in-the-new-scientist-on-last-number-of-prime-number. 
Let me restate the argument. Let's write the sequence of all numbers ending with 1, 3, 7, 9 (beginning with 7): 7, 9, 11, 13, 17, 19,... and flag each number with a probability of $p$. Let's denote $q=1−p$. Now if a number ending in 1 has been flagged, the probability that the next number being flagged ends in 1 can easy be computed: $\sum_{k=0}^\infty q^{3+4k}p=q^3p\frac{1}{1-q^4}$. That's not 25%! In order to make that more intuitive, suppose that p is close to 1 that is we flag each number with high probability (but nevertheless randomly). If we have flagged a number ending in 1, the probability that the next number being flagged ends in 1 is very small, because we can expect that at least one of the three following numbers (ending in 3, 7, 9) will be flagged (recall that we have expected p being close to 1). 
Now this model is oversimplificated. The probability that a random number $n$ is prime can be evaluated as $1/ln(n)$ (not as a constant $p$) by the prime counting function. If we know that the number ends in $1, 3, 7, 9$; this probability becomes $\frac{10}{4}\frac{1}{ln(n)}$ (assuming the classes are random). Because the sequence $q^{3+4k}p$ tends to zero rapidly for $k\rightarrow\infty$, if a number $n$ ending in 1 has been flagged, the probability that the next number being flagged also ends in 1 can be evaluated as $q_n^3p_n\frac{1}{1-q_n^4}$ with $p_n=\frac{10}{4}\frac{1}{ln(n)}$ and $q_n=1-p_n$. 
For $n=100\cdot10^6$, we find 19.8% which is not much different from the number cited in the article (take in mind the simplification in my argument, also the article seem to make the experiment for a random number between 1 and $10\cdot10^6$ not just a number which is approximately equal to $10\cdot10^6$). Moreover as $n\rightarrow\infty$, $p_n\rightarrow 0$ and we can check that $\lim_{p\rightarrow 0}q^3p\frac{1}{1-q^4}=\frac{1}{4}$ which seems to explain that the "discrepancy" vanishes when we take a longer sequence.
There are a lot of mysteries concerning prime numbers but it seems that the one pointed by the New Scientist is nothing more than misconception on elementary probability. 
