# Using Markov chain instead of total probability

This question was given to me as a review for an upcoming exam:

If a baseball team wins a game, they have a 40% chance of winning the next game due to getting overconfident. If they lose the previous came, there is a 70% they will win the next game. Assume the first game that there is a 50% of winning. What is the probability the team wins the 4th game?

Is there a way to do this without using total probability?

So far I added all of the 8 scenarios that result in a win on the 4th game to get an answer, but I'm thinking there is a way to do this using matrix multiplication instead of total probability.

Yes, you are right. If state $$1$$ is win and state $$2$$ is loss then the transition matrix is $$P=\begin{bmatrix}.4&.6\\.7&.3\end{bmatrix}$$ The probabilities after $$4$$ games are given by $$\begin{bmatrix}w\\l\end{bmatrix}=P^4\begin{bmatrix}.5\\.5\end{bmatrix}$$ where $$w$$ and $$l$$ are the respective probabilities that the fourth game is a win or a loss.
If you look at the matrix multiplication, you will see that this involves exactly the same calculations as the calculations you've already done. You could short-cut it though by squaring $$P$$ twice. This would cut out one matrix multiplication.