Is the negative binomial a possible way of solving this problem? I am unsure if the negative binomial is applicable for this problem?
Imagine a soccer team playing in a tournment with 5 games. The probability that they win (1/2), draw (1/6) , lose (1/3). At the end of the tournment the coach will lose his job if the team losses three consecutive matches. Find the probability that the coach losses his job. 
Now ideally at first i would use negative binomial because it is asking for three success, and hence it would be X~NB(3,1/3) P(X=5), so would this be the correct way to approach this problem? (the probabilitys remain constant throughout the tournament and are independente of other matches)
 A: I'm not sure how you'd use a negative binomial to solve it, but this is a way to solve it using transition matrices.
There are $4$ states to consider:  the last game was a win ($S_0$), a 1 game losing stream $(S_1)$, a 2 game losing streak $(S_2)$, and finally anytime in the past was a $3$ game losing streak ($S_3$).  Let $M$ be a zero index matrix, and let $M_{y, x}$ be the probability transitioning from state $S_y$ to state $S_x$.  Then
$$M = \begin{bmatrix}
2/3 & 1/3 & 0 & 0 \\
2/3 & 0 & 1/3 & 0 \\
2/3 & 0 & 0 & 1/3 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}$$
Computing $M^5{}_{y, x}$ computes the probability of going from state $S_y$ to state $S_x$ in $5$ games, so 
$$M^5{}_{0, 3} = \frac{1}{3^5} \begin{bmatrix}
152 & 52 & 18 & \boxed{21} \\
140 & 48 & 16 & 39 \\
104 & 36 & 12 & 91 \\
0 & 0 & 0 & 243 
\end{bmatrix}_{0, 3} = \frac{21}{3^5} \approx 8.6\%$$ 
is the chance he loses his job.

The eigen values of $M$ are pretty complicated, so it is unlikely that any kind of binomial expanion makes this easy in the general $n$ game case.
