= Solution
Order the states as sandwich $N$, Hall $H$, and own cooking $C$. The <transition matrix>, with current state indexing the row, is
$$
P=\begin{pmatrix}1/2&1/4&1/4\\1/3&1/3&1/3\\1&0&0\end{pmatrix}.
$$
The two probabilities $1/4$ in the first row are unconditional alternatives to the probability $1/2$ of returning. Let $a_n=(P^n)_{HH}$, so $n$ counts days after the initial Hall lunch. Expanding the determinant gives the <characteristic polynomial>
$$
\det(\lambda I-P)=\lambda(\lambda-1)(\lambda+1/6).
$$
The three distinct <eigenvalues> imply that the sequence has the form
$$
a_n=A+B(-1/6)^n+C0^n,
$$
using the specified $0^0=1$. Directly, $a_0=1$, $a_1=1/3$ and $a_2=7/36$. The last two equations give $A=3/14$ and $B=-5/7$; then $C=3/2$. Consequently
$$
\boxed{a_{60}=\frac3{14}-\frac5{7\,6^{60}}.}
$$
The limiting value is also the Hall entry of the <stationary distribution> $(4/7,3/14,3/14)$. The tiny finite-time correction is retained in the boxed exact probability. Sixty days later means sixty transitions, rather than the sixtieth day including the initial day.
Back to article page
