# One-step transition probabilities for a markov chain?

Imagine m balls being exchanged between two adjacent chambers (left and right) according to the following rules. At each time step, one of the m balls is randomly selected and moved to the opposite chamber, i.e., if the selected ball is currently in the right chamber, it will be moved to the left one, and vice versa. Let $X_n$ be the number of balls in the left chamber after the nth exchange. For m=3 I want to find all the one step transition probabilities. I know the state space will be {0,1,2,3} and that I am looking for Probabilities, when it goes from 0->1, 1->0, 2->1, 1->2, 3->2, 2->3. I am struggling with how to account for the fact that the balls can have different starting positions? For example going from 1->0 you can either pick the one ball in the left chamber and move it, or pick one of the two balls in the right chamber and move it to the left making it a 1->2 transition, so what would the probability for something like that look like?

• "one of the $m$ balls is randomly selected" ... "going from 1-> 0" can be done only the following way: you select the ball in the left box. If you take a ball from the right box then you go from 1->2. – zoli Apr 27 '17 at 15:47

$\{0,1,2,3\}$ is the set of states, that is, the set of the possible numbers of balls in the left chamber. Assume that the system is in state $i$ ($i=0,1,2,3$). The probability that the sytem goes to state $i-1$ is $\frac i3$ because this is the probability that one selects a ball from the left box. The probability that the system goes to state $i+1$ is $\frac{3-i}3$ because this is the probability that one selects a ball from the right box.
For example, if the system is in state $1$ then there is only two possible transitions, as shown below
The system can go to state $2$ (with probability $\frac23$) or to state $0$ (with probability $\frac13$).
$$P= \begin{bmatrix} 0&1&0&0\\ \frac13&0&\frac23&0\\ 0&\frac23&0&\frac13\\ 0&0&1&0 \end{bmatrix}.$$