coupon collecting around a ring A host and his $N$ guests are around a table. He has a present and passes it to his left or his right with prob 0.5. Each guest in turn does the same thing. What is the expected amount of time until each guest has been passed the present?
BTW, I know that the prob that any given guest will be the last to receive the present is ${1\over N}$. This was quite surprising to me...
 A: If you consider the question of $n$ people having touched the present and the present being at position $p$ in this group of $n$, then the expected number of turns before the present reaches either end of the group is $f(p,n)=(p-1)(n-p)$.  You can check this by seeing $f(1,n)=f(n,n)=0$ as the present is already at one end, and $f(p,n)=1+\frac{f(p-1,n)}{2}+ \frac{f(p+1,n)}{2}$ when $1 \lt p \lt n$.
If the present is at one end of a group of $n$ then the expected number of turns to move the present to a new person is $g(n)=1+\frac{p(2,n)+g(n)}{2}$ which gives $g(n)=n$.  So for example, at the start the host will pass to a new person in $1$ turn, and the present will then go to another new person in an expected $2$ turn. Each time it goes to a new person, the present is at an end of the group of those who have touched the present.
So for the present to have been touched by the host and $N$ guests (i.e. when all $N$ guests have been passed the present, though the host might not have been passed it), the expected number of turns is $1+2+3+\cdots+N = \dfrac{N(N+1)}{2}.$
