(here illustrated for $k=7$) and so detailed balance gives the invariant distribution as proportional to (1,2,2,2,2,2,2,2).
Unfortunately, the examiner had put 0 0 0 0 0 0 2 2 in the bottom row and so calculated the invariant distribution as proportional to (1,2,2,2,2,2,2,1). But if the runner starts with $k$ shoes at the front door, the probability that she has $k$ shoes there after the next two runs is 3/4, not 1/2.
This question was intended to be similar to Example Sheet 2, #11 (the professor carrying umbrellas between home and office.)
This year's course finished on 13 November 2011. I handed out the standard faculty feedback forms in the lecture on 8 Novembers. I obtained 63 forms back.
You can also give feedback using my equivalent on-line form. You may find it easier to tick the radio buttons and type into this form than to handwrite onto the paper form - and you have space to write interesting comments. 8 of these have been received so far.
This will be sent to my email anonymously. After reading the responses, I will forward printed copies to the Faculty Office.