The answer depends on the probabilities of the initial state.
You can work it out from basic axioms of probability. It's helpful as well as insightful to use matrix notation, but you only need to know two things about matrices: how to multiply them and the fact that multiplication is associative. Specifically, when $\mathbf{x}$ is a vector and $\mathbb{A},\mathbb{B}$ are matrices, then $$(\mathbf{x}\mathbb{A})\mathbb{B} = \mathbf{x}(\mathbb A \mathbb B)$$ whenever either expression is defined.
Let the initial state correspond to time $t=0.$ At any time $t\ge 0$ let the unconditional probability of state $j$ be $p_t(j),$ forming a row vector $\mathbf{p}_t$ with one component for each state in the set of all states $S.$ At time $t\in\{0,1,2,\ldots\}$ the answer in terms of the transition matrix
$$\mathbb{P} = (p_{ij})$$ is
$$\mathbf{p}_t = \mathbf{p}_0\mathbb{P}^t.$$
These matrix powers are defined recursively as
$$\mathbb{P}^0 = \mathbb{I}_S$$
(the identity matrix on the states $S$) and
$$\mathbb{P}^t = \mathbb{P}^{t-1}\mathbb{P}$$
for $t=1,2,3,\ldots.$
Because $\mathbb{P}^0$ is defined to be the identity matrix, the formula is true at $t=0.$ Suppose inductively that this formula is still correct at some arbitrary time $t-1.$ We need only show it will then be correct at time $t.$
To this end, decompose the event $\mathcal{P}_{t}(j):$ "the system is in state $j$ at time $t$" into a union of events of the form $\mathcal{P}_t(i\to j):$ "the system made a transition from state $i$ to state $j$ at time $t$" where $i$ ranges over all the states.
Being a Markov chain means these transitions are independent of the probabilities of their starting states, and independence only means the probabilities multiply:
$$\Pr(\mathcal{P}_t(i\to j)) = p_{t-1}(i)\, p_{ij}.$$
Because these transitions all start from different states, they have nothing in common. Thus, it is axiomatic that their probabilities add, giving
$$p_t(j) = \Pr(\mathcal{P}_{t}(j)) = \sum_{i\in S} \Pr(\mathcal{P}_{t}(i\to j)) = \sum_{i\in S} p_{t-1}(i)\, p_{ij}.$$
But that sum merely expresses the law of matrix multiplication when computing $\mathbf{p}_{t-1}$ times $\mathbb{P},$ and so can be written in vector form as
$$\mathbf{p}_t = \mathbf{p}_{t-1}\,\mathbb{P} = \left(\mathbf{p}_{0}\,\mathbb{P}^{t-1}\right)\mathbb{P} = \mathbf{p}_{0}\left(\mathbb{P}^{t-1}\mathbb{P}\right) = \mathbf{p}_{0}\,\mathbb{P}^t$$
as claimed, QED.
For example, suppose the transition matrix is $$\mathbb{P} = \pmatrix{\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 1}.$$
Then you can work out that
$$\mathbb{P}^t = 2^{-t}\pmatrix{1 & t & 2^t-t-1 \\ 0 & 1 & 2^t-1 \\ 0 & 0 & 2^t}.$$
If, say, all states are equally likely initially, that means $$\mathbf{p}_0 = \pmatrix{\frac{1}{3}, & \frac{1}{3}, & \frac{1}{3} }$$
and therefore the chances at time $t$ are
$$\mathbf{p}_t = 2^{-t} \pmatrix{\frac{1}{3}, & \frac{1}{3}, & \frac{1}{3} } \pmatrix{1 & t & 2^t-t-1 \\ 0 & 1 & 2^t-1 \\ 0 & 0 & 2^t} = \pmatrix{\frac{1}{3\,2^t}, & \frac{1+t}{3\,2^t}, & 1 - \frac{2+t}{3\,2^t} }.$$
Wikipedia has an example in its article on Markov chains with links to an entire article on Markov chain examples.
