The solution to part B follows from recognizing that each time the sequence is interrupted with the random selection of an opposing team fan, we are back to the exact same process but with fewer total fans left than when we started. This suggests a recursive approach:

Let  \(X_{a,m}\) represent the number of shuttle trips when we have \(a\) American and \(m\) Mexican fans and \(E_{a,m}\) be the expected number of shuttle trips so \(E_{a,m} = E(X_{a,m}) \). Also let \(A_{k,a,m}\) and \(M_{k,a,m}\) represent the event when \(k\) Americans or \(k\)  Mexicans are in the first shuttle. We can now write \(E_{a,m}\) conditioning on the number of fans in the first shuttle:

\begin{align*}
E_{a,m} &= \sum_{k = 1}^{a}E(X_{a,m}|A_{k,a,m} ) P(A_{k,a,m}) \ + \ \sum_{k = 1}^{m}E(X_{a,m}|M_{k,a,m} ) P(M_{k,a,m}) \\
\end{align*}

But if the first shuttle picks up \(k\) American fans, then this means that the situation is the same as having 1 shuttle trip more than when starting with \(a-k,m\) American and Mexican fans and similarly for \(k\) Mexican fans. i.e.

\begin{align*}
E(X_{a,m}|A_{k,a,m}) &= (1\ +\ E_{a-k,m}) \\
E(X_{a,m}|M_{k,a,m}) &= (1\ +\ E_{a,m-k})
\end{align*}

Finally, \(P(A_{k,a,m})\), the probability of getting \(k\) Americans in the first shuttle is just the probability that we pick \(k\) Americans and then a Mexican from \(a\) Americans and \(m\) Mexicans. i.e.

\begin{align*}
P(A_{k,a,m}) &= \frac{a}{a+m}\cdot \frac{a-1}{a+m-1}\cdot \cdot \cdot \frac{a-k+1}{a+m-k+1}\cdot \frac{m}{a+m-k} \\
P(A_{k,a,m}) &= \frac{m}{a+m-k} \prod_{i=0}^{k-1}\frac{a-i}{a+m-i} \\
& \textit{and similarly} \\
P(M_{k,a,m}) &= \frac{a}{a+m-k} \prod_{i=0}^{k-1}\frac{m-i}{a+m-i}
\end{align*}

Where the \(\prod\) sign refers to pi notation for the product of a sequence

Finally, the expected number of shuttle trips when we have only one set of fans is obviously 1. Hence our final recursive relations are:

\begin{align*}
E_{a,m} &= \sum_{k = 1}^{a}\left ((1\ +\ E_{a-k,m})\frac{m}{a+m-k} \prod_{i=0}^{k-1}\frac{a-i}{a+m-i} \right ) \ +\ \sum_{k = 1}^{m}\left ((1\ +\ E_{a,m-k})\frac{a}{a+m-k} \prod_{i=0}^{k-1}\frac{m-i}{a+m-i} \right ) \\
E_{y,0} &= E_{0,y} = 1 \qquad y \geq 1
\end{align*}

The equations above look more formidable than they really are. Writing a simple program to use the recursive relation gets us the required expected number of trips:

\begin{align*}
E_{11,7} &\approx \boldsymbol{{\color{Red} {9.55}}}
\end{align*}

Sample code to calculate this as well as compare with the results of several simulated runs of the shuttles is given below.