Simple markov chain: passing cards around a circle Consider a circle of 5 people in which one person initially has $2$ cards in his hand. He passes each card to either the person on his left or his right (for each card, 50% chance left, 50% chance right). 
At each step, if someone holds a card, he passes it to the right or the left with equal probability. What is the expected number of steps to have someone holding both cards in his hands (as in the beginning configuration).

Seems like a pretty simple probability problem. At any point in time, there are there possible states: A: both cards held by the same person
B: the two cards held by two different people who are not adjacent
C: the two cards held by adjacent people
We can easily write the probabilities for transitioning from one state to another. I could write it as a matrix, but I don't know how to in latex so I'll just write it like this:
$A\to A: 1/2$
$A\to B: 1/2$
$A\to C: 0$
$B\to A: 1/4$
$B\to B: 1/2$
$B\to C: 1/4$
$C\to A: 0$
$C\to B: 1/4$
$C\to C: 3/4$
Define $a$ as the expected number of steps to reach state $A$ beginning at state $A$, $b$  the expected number of steps to reach state $A$ beginning at state $B$, $c$ the expected number of steps to reach state $A$ beginning at state $C$.
We thus have a simple system of equations:
$a=(1/2) \cdot 1+(1/2)\cdot (b+1)$
$b=(1/4)\cdot (1)+(1/2)\cdot (b+1)+(1/4)\cdot (c+1)$
$c=(1/4)\cdot (b+1)+(3/4)\cdot (c+1)$
which gives $(a,b,c)=(5,8,12)$.
Apparently this does not give the right answer, but I don't see the error.
Edit: even after fixing the mistake pointed in the answer below, the answer is still wrong.
A: [edit] Actually you have the right transition probabilities but B is non-adjacent, C adjacent. 
There is a problem with your equations below, as you should not have the $(a+1)$ terms (if you move to $A$ you have finished). So it should be
$a=\frac12\times 1+\frac12(b+1)$
and
$b=\frac14\times 1+\frac12(b+1)+\frac14(c+1)$.
