We consider a regular pentagon which the vertices are numbered from $1$ to $5$ in the direction of clockwise. Initially (i.e. at time $0$), two ladybugs are placed at the vertices $1$ and $3$. At each next moment, each of ladybugs moves independently of the other, to one of two adjacent corners with probability $½$ each two. How long on average does it take for the ladybugs meet at the same vertex?
I think we have to consider the number of edges between the two ladybugs.
I have to prepare for an exam for the course of Stochastic Processes, and a part of the material bothers me enormously; essentially it is a problem resolution by conditioning method. Does someone could solve my problem and clearly explain each step (with schematisation if possible)? I tried, but I can not go very far in the problem. It should also say that my first course probability goes back more than three years; this may be part of why I have trouble on that kind of question.
The game has three states $d\in\{0,1,2\}$, whereby $d$ denotes the shortest edge distance of the two bugs. I assume that the two bugs may use the same edge at the same time without disturbing each other.
Denote by $E_d$ the expected number of additional moves when the bugs are in state $d$. Obviously $E_0=0$. When the bugs are in state $1$ they certainly will make a move. With probability ${3\over4}$ this move will end in state $1$ (necessitating $E_1$ further moves), and with probability ${1\over4}$ this move will end in state $2$ (necessitating $E_2$ further moves). Arguing similarly when the bugs are in state $2$ one arrives at the system of equations $$\eqalign{E_0&=0\cr E_1&=1+{3\over4}E_1+{1\over4}E_2 \cr E_2&=1+{1\over4}E_0+{1\over4}E_1+{1\over2}E_2\ .\cr}$$ This leads to $E_1=12$, $E_2=8$. It follows that the average length of the game is $8$.