To win the game you must find the two 'good' squares before the one 'bad' square, out of seven uncovered one at a time. Call the other four squares 'okay'.
Suppose it takes $k$ turns to until you win, where $k$ is some number between $2$ and $6$. Then to have won you must have selected $k-2$ 'okay' squares and $1$ 'good' square out of all the ways to select $k-1$ squares out of all $7$, and then have selected the second 'good' square out of the $8-k$ remaining squares.
So the probability of wining is: $${\begin{align} \mathsf P(W) & = \sum_{k=2}^6 \left(\frac {{4\choose k-2}{2\choose 1}}{{7\choose k-1}}\cdot\frac{1}{8-k}\right) \\\\[1ex] & = \sum_{k=2}^6\left(\frac{4!\,2!\,(k-1)!\,(8-k)!}{(k-2)!\,(6-k)!\,7!\,(8-k)}\right) \\\\[1ex] & = \frac{4!\,2!}{7!}\sum_{k=2}^6 (k-1)(7-k) \\\\[1ex] & = \frac{4!\,2!}{7!}\sum_{k=1}^5 k(6-k) \\\\[2ex] & \ddots \end{align}}$$
~~Simplify and~~ evaluate.