Here we use generating functions to derive the number of successful sequences of length $2x$ containing $x$ consecutive equal characters, $x\in\{2,3,4,5\}$ over a $10$ character alphabet.
But before that we consider the special, simple case $x=1$.
Case $x=1$: Each word having length $2x=2$ is a successful sequence. So there are
\begin{align*}
10^2=\color{bue}{100}
\end{align*}
different words.
We continue by stating a generating function which counts words of a $10$ character alphabet with no consecutive equal characters at all.
These words are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.
The generating function $A(z)$ counting Smirnov words over a $10$ character alphabet is according to the reference
\begin{align*}
A(z)&=\left(1-\frac{10z}{1+z}\right)^{-1}
\end{align*}
The coefficient of $z^n$ of $A(z)$ gives the number of Smirnov words of length $n$.
Based upon $A(z)$ we generate all words over the $10$ character alphabet which contain for $x\in \{2,3,4,5\}$ no more than $x-1$ consecutive equal characters. This means each character can be replaced with one up to $x-1$ characters.
\begin{align*}
z\longrightarrow z+z^2+z^3+\cdots+z^{x-1}=\frac{z(1-z^{x-1})}{1-z}
\end{align*}
in the generating function $A(z)$.
We obtain this way a generating function $B(z)=A\left(\frac{z(1-z^{x-1})}{1-z}\right)$ with
\begin{align*}
B(z)&=\left(1-\frac{10\cdot \frac{z(1-z^{x-1})}{1-z}}{1+\frac{z(1-z^{x-1})}{1-z}}\right)^{-1}=\frac{1-z^x}{1-10z+9z^x}\\
\end{align*}
The coefficient of $z^{2x}$ of $B(z)$ gives the number of words which contain less than $x$ consecutive equal characters.
The number of all words over a $10$ character alphabet is given by
\begin{align*}
\sum_{n=0}^\infty (10z)^j=\frac{1}{1-10z}
\end{align*}
We conclude the generating function $C_x(z)$ with
\begin{align*}
C_x(z)=\frac{1}{1-10z}-\frac{1-z^x}{1-10z+9z^x}\qquad\qquad x\in\{2,3,4,5\}
\end{align*}
counts the words which contains runs of length $x$, i.e. words having at least $x$ consecutive equal characters.
The number of successful sequences of length $2x$ is
\begin{align*}
[z^{2x}]C_x(z)\qquad\qquad x\in\{2,3,4,5\}
\end{align*}
In the following we obtain the series expansions below with some help of Wolfram Alpha
Case x=2:
The number of successful sequences of length $2x=4$ is
\begin{align*}
[z^4]C_2(x)&=[z^4]\left(\frac{1}{1-10z}-\frac{1-z^2}{1-10z+9z^2}\right)\\
&=[z^4]\left(10z^2+190z^3+2710z^4+34390z^5+\cdots\right)\\
&=\color{blue}{2710}
\end{align*}
Case x=3:
The number of successful sequences of length $2x=6$ is
\begin{align*}
[z^6]C_3(x)&=[z^6]\left(\frac{1}{1-10z}-\frac{1-z^3}{1-10z+9z^3}\right)\\
&=[z^6]\left(10z^3+190z^4+2800z^5+36910z^6+\cdots\right)\\
&=\color{blue}{36910}
\end{align*}
Case x=4:
The number of successful sequences of length $2x=8$ is
\begin{align*}
[z^8]C_4(x)&=[z^8]\left(\frac{1}{1-10z}-\frac{1-z^4}{1-10z+9z^4}\right)\\
&=[z^8]\left(10z^4+190z^5+2800z^6+37000z^7+459910z^8+\cdots\right)\\
&=\color{blue}{459910}
\end{align*}
Case x=5:
The number of successful sequences of length $2x=10$ is
\begin{align*}
[z^{10}]C_5(x)&=[z^{10}]\left(\frac{1}{1-10z}-\frac{1-z^5}{1-10z+9z^5}\right)\\
&=[z^{10}]\left(10z^5+190z^6+2800z^7+37000z^8+460000z^9+5499910z^{10}+\cdots\right)\\
&=\color{blue}{5499910}
\end{align*}
Adding all the results we finally obtain
The number of successful sequences is
\begin{align*}
100+\sum_{j=2}^5[z^{2j}]C_j(x)&=100+2710+36910+459910+5499910\\
&=\color{blue}{5999540}
\end{align*}
