What is the Expectations of all 3 ants meeting at same point?

Question

Say we have 3 ants in three corner's of triangle. What is the expectations that all 3 ants meeting together given that the ant moves in any direction. So by just seeing it I figured out that in 2 ways all three Ant can meet together in either in A ,B or C . But i think it is a two state discrete time markov chain problem.

Ant in A moving to B and then A with the Ant in B moving to C and then A and Ant in C moving to B and then A Ant in A moving to C and then A with the Ant in B moving to C and then A and Ant in C moving to B and then A

This will apply for all 3 ant meeting at B and C . But Is it right?

Answer 1

In section 3.3 of Problems and Snapshots from the World of Probability by Blom, Holst, and Sandell, the authors solve this problem. Here is their solution.

Three ants start at different corners of a triangle. At every time point, each ant independently jumps to a new corner, either to the left or to the right with probability 1/2 each. Eventually, all three ants will meet at one corner and the game is finished. How long on average until this happens?

Define "state 1" to be when all ants are at different corners, "state 2" when two ants are at one corner, and the third ant is at a different corner, and "state 3" when they are all together. The diagram below illustrates how we might move from state 1 to state 2.

We are interested in $M$, the number of steps needed to finish the game, starting in state 1. To solve this problem, we also need the random variable $N$, the number of steps needed to finish starting in state 2.

To begin, imagine starting in state 1. There are eight equally likely possibilities depending on how the ants jump. In two of the cases we remain in state 1, and in six cases we end up in state 2. Hence $$M=\cases{1+M^\prime&with probability $1/4$\cr 1+N&with probability $3/4$,}$$ where $M^\prime$ has the same distribution as $M$.

Now imagine starting in state 2. There are eight equally likely possibilities depending on how the ants jump. In one case all the ants go to the same corner and the game is finished. Otherwise, in five cases we stay in state 2, and in two cases we go to state 1.

Therefore $$N=\cases{1&with probability $1/8$\cr 1+ N^\prime&with probability $5/8$\cr 1+M&with probability $1/4$,}$$ where $N^\prime$ has the same distribution as $N$.

From these relationships, we find that the probability generating functions satisfy \begin{eqnarray*} G_M(s)&=&{s\over 4}\, G_M(s)+{3s\over 4}\,G_N(s),\\[5pt] G_N(s)&=&{s\over8}+{5s\over 8}\, G_N(s)+{s\over 4}\,G_M(s). \end{eqnarray*}

Solving this system of equations shows that $$G_M(s)={3s^2\over 32-28s-s^2}={3\over 32}\,s^2+{21\over 256}\,s^3+{153\over 2048}\,s^4+{1113\over 16384}\,s^5+\cdots.$$ Differentiating $G_M$ and setting $s=1$ gives $\mathbb{E}(M)=G_M^\prime(1)=12.$