First, I'll assume that the amount won is in fact $\$k$, not $\$n$, which I don't think would make a lot of sense.
Second, I'll assume that the player is trying to maximize her expected payoff and plays optimally. (You hadn't stated those assumptions.)
Third, to simplify things, I'll assume that $n$ is even; doing the same thing for $n$ odd is straightforward.
Let's work backwards. The expected payoff if she uses her third try is $(n+1)/2$, so she'll use it if she got less than that on the second try. The expected value of the second try is thus $((n/2+1+n)/2+(n+1)/2)/2=5n/8+1/2$, and she'll use it if she got less than that on the first try.
Thus, there's a $1/n$ probability for each of the numbers from $\lceil5n/8+1/2\rceil$ to $n$ from the first try, then there's a $\lfloor5n/8+1/2\rfloor/n^2$ probability for each of the numbers from $n/2+1$ to $n$ from the second try, and then there's a $\lfloor5n/8+1/2\rfloor/(2n^2)$ probability for each of the numbers from $1$ to $n$ from the third try. (These all add up.)
There's a slight complication in that sometimes (e.g. for odd $n$ if the second draw yields $(n+1)/2$) the two strategies have the same expected payoff but lead to different probabilities; you didn't specify what would hppen in such a case.