The sample set \( \Omega \) can be defined in the following way:
\[ \Omega=\{1,3,5,6, 21,22,\dots,26,41,\dots,46\}.\]
The set of events \( \mathcal F \) consists of all possible subsets of \( \Omega \). The probability \( \mathbb P \) is defined as \( \mathbb P:\Omega\to[0,1] \) as
\[ \mathbb P(\{\omega\})=\left\{\begin{array}{ll}\frac16,&\mbox{ if }\omega \mbox{ is an one-digit number from }\Omega;\\
\frac1{36},&\mbox{ if }\omega\mbox{ is a two-digit number from }\Omega.\end{array}\right.\]
The random variable \( X \) is a function \( X:\Omega\to\mathbb N \) defined in the following way:
\[ X(\omega)=\left\{\begin{array}{ll}
-6,& \mbox{ if } \omega \mbox{ is one-digit number equal to 6},\\
8,& \mbox{ if } \omega \mbox{ is one-digit number odd},\\
7,&\mbox{ if }\omega \mbox{ is a two digit number with the second digit equal to 5},\\
4,&\mbox{ otherwise.}\end{array}\right.\]
The cumulative distribution function of the random variable \( X \) is the function \( F_X:\mathbb R\to[0,1] \) defined as \( F_X(t)=\mathbb P\left(X\leq t\right) \). Observe that the smallest amount of money that the player can earn is \( -6 \) dollars. Therefore \( F_X\left(-6\right)=\frac16 \) and \( F_X(t)=0 \) for \( t< -6 \). Moreover if \( -6\leq t< 4 \) than the event \( \{X\leq t\} \) means that the player earns exactly \( -6 \) dollars. The probability of this is \( \frac16 \). Thus \( F_X(t)=\frac{1}6 \) for \( -6\leq t< 4 \).

For \( t\in [4,7) \) the event \( \{X\leq t\} \) means that the player either won \( -6 \) or \( 4 \) dollars. The probability of this is
\[ F_X(t)=\mathbb P\left(X\leq t\right)=\mathbb P(X=-6)+\mathbb P(X=4)=\frac16+\frac{10}{36}=\frac{16}{36}.\]

Assume now that \( 7\leq t< 8 \). Then the event \( \{X\leq t\} \) satisfies
\[ \{X\leq t\}=\left\{X=-6\right\}\cup \left\{X=4\right\}\cup \left\{X=7\right\}.\]
Its probability is precisely \( F_X(t) \) and satisfies
\[ F_X(t)=\frac{3}{6}.\]

Finally, if \( t\geq 8 \) then the event \( \{X\leq t\} \) occurs always and as such has probability \( 1 \). Thus the cumulative distribution function \( F_X \) satisfies
\[ F_X(t)=\left\{\begin{array}{ll}
0,&\mbox{ if }t \in (-\infty,-6),\\
\frac{1}6,&\mbox{ if } t\in [-6,4),\\
\frac{16}{36},&\mbox{ if }t\in[4,7),\\
\frac{3}6,&\mbox{ if }t\in[7,8),\\
1,&\mbox{ if } t\in [8,+\infty).
\end{array}\right.\]

The graph of \(F_X\) is given below.

The expected value of \( X \) is
\begin{eqnarray*}
\mathbb E\left[X\right]&=&-6\cdot\mathbb P\left(X=-6\right)+4\cdot\mathbb P\left(X=4\right)\\
&&+7\cdot\mathbb P\left(X=7\right)+8\cdot\mathbb P\left(X=8\right)\\
&=&-6\cdot \frac{1}6+4\cdot \frac{10}{36} +7\cdot \frac{2}{36}+8\cdot \frac{3}{6}=\frac{162}{36}.
\end{eqnarray*}