Given the following markov chain, where T1 is the start state, the labels are shown on the state( 'a' in this case) and p and 1-p are probabilities for that transition happening:
Now, for what values of p is it the case that there is greater than or equal to 0.5 chance of reaching state the labelled a, within a maximum of two steps, from the initial state T1?
So, say p=0.5 then, the transition $(T1 \rightarrow T2)$ will happen with a probability of $1-0.5=0.5$. But with the transitions $(T1 \rightarrow T1 \rightarrow T2)$ will happen with a probability of $p \times (1-p)$ i.e. $=0.5 \times (1-0.5) = 0.25$. So if p=0.5 this will be less than 0.5.
I can't get my head around finding the values for p where the overall probability of reaching label 'a' within a maximum of 2 steps is greater than 0.5. In PCTL this would be written as: $T1 \models P_{\geq 0.5} (F^{\leq 2} a)$
(Sorry for the poor drawing)
From what you have said, you need to solve $$(1-p)+p(1-p) \ge \frac12$$ subject to the usual condition for any probability that $0 \le p \le 1$.
This is at worst a quadratic which I will leave for you to solve.
Your example of $p=0.5$ has an error: the probability of reaching state $T_2$ in one step would indeed be $0.5$ and of reaching state $T_2$ exactly on the second step would be $0.25$. So the probability of reaching state $T_2$ in a maximum of two steps would be $0.75 \gt 0.5$.