Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 9C / Solution

Order the states as sandwich $N$, Hall $H$, and own cooking $C$. The transition matrix, with current state indexing the row, isThe 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 polynomialThe three distinct eigenvalues imply that the sequence has the formusing 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 $C=3/2$. ConsequentlyThe 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.