Rough estimate:
Considering only the first three cards, the probability of three equal colours is approximately $\frac14$; the precise value $\frac{52\cdot 25\cdot 24}{52\cdot 51\cdot 50}=\frac 4{17}$ differs a bit from this because the colours are not independent. Fortunately, expected values are additive even in case of dependency. Therefore, we expect $50\cdot\frac 4{17}\approx 11.8$ instances of three consecutive cards of same colour.
Going by rule of thumb, we may estimate that this number is distributed with mean and variance both $=11.8$. So zero instances is about $3.4$ standard deviations too low - and that is very unlikely.
Less hand-wavy:
Let $f(n,k)$ be the number of ways to break a line $n$ same-colour cards into $k$ pieces of length $1$ or $2$.
We have the recursion
$$f(n,k)=f(n-1,k-1)+f(n-2,k-1) $$
and clearly, $f(n,k)=0$ if $k>n$ or $2k<n$. The recursion reminds of binomial coefficients, and indeed we find that
$$ f(n,k)={k\choose n-k}.$$
If our deck has $k$ blocks of red cards (each of length $1$ or $2$), then it can have
- $k-1$ blocks of black cards in-between the reds
- $k+1$ blocks of black cards with the red blocks in-between the black blocks
- $k$ blocks of black and the first block is red, the last black
- $k$ blocks of black and the first block is black, the last red
Hence the total way to have a 52 card deck without blocks longer than $2$ is
$$\begin{align} \sum_k f(26,k)\cdot& \left(f(26,k-1)+f(26,k+1)+2f(26,k)\right)
\\&=\sum_k {k\choose 26-k}\cdot \left({k-1\choose 27-k}+{k-1\choose 25-k}+2{k\choose 26-k}\right)\end{align}$$
where we need only $k=13, \ldots, 26$.
Evaluating this and dividing by $\frac{52!}{26!^2}$, we arrive at a probability of
$$ \frac{23246263956\cdot 26!^2}{52!}=\frac{1937188663}{41326544412342}\approx 0.000047.$$
Fixed prescribed red/black pattern
For a fixed prescribed red/black pattern the answer is much simpler: If we prescribe which cards are red and which are black, there are $26!^2$ "good" sequences out of $52!$ in total. We arrive at a probability of $\frac{26!^2}{52!}\approx 2\cdot 10^{-15}$.
Best Answer
The expected number of runs is 27.
Let $X_n$ be the color of the n'th card. For n<52, the n'th card is the end of a run if and only if $X_n\not=X_{n+1}$ and the last card in the pack is always the end of a run. So, the total number of runs is $$ N=\sum_{n=1}^{51}1_{\{X_n\not=X_{n+1}\}}+1. $$ The expected number of runs is $$ \mathbb{E}[N]=\sum_{n=1}^{51}\mathbb{P}(X_n\not=X_{n+1})+1. $$ Whatever colour the n'th card is, there are 51 remaining cards in the deck of which 26 of them are a different colour from $X_n$. So, $\mathbb{P}(X_n\not=X_{n+1})=26/51$, giving $$ \mathbb{E}[N]=51 (26/51) + 1=27. $$