Why is my counting wrong?

A $3\times3$ square is partitioned into $9$ unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is the rotated $90^\circ$ clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability that the grid is now entirely black?

Firstly we observe that the middle square must be black. Then we realize that the middle "cross" is disparate from the corners, that is, the coloring of the cross does not affect the coloring of the corners after rotation.

Therefore we look at the minimum number of black squares required to fill in the cross and the corners separately.

To have the cross black, we must initially have either the middle column or the middle row filled in. To have the corners black, we must have one of the two diagonals filled in. Therefore we get the total number of possible colorings as $2 \times 2 \times 2^4$ where $2^4$ accounts for the other four squares which are irrelevant.

Thereby we get the probability as $\frac{2^6}{2^9} = \frac{1}{8}.$ But the answer is $\frac{49}{512}.$ Where did I go wrong?

• "To have the cross black, we must initially have either the middle column or the middle row filled in" this is wrong. They don't have to be filled in. For example, you can have black-white-black pattern in top most column, then the one it gets mapped to after rotation would just have to have whatever-black-whatever, then next one b-w-b and again next one whatever-b-whatever – user160738 Jan 30 '17 at 17:45
• You have overcounted. For instance, the case where the entire cross is black is in your argument counted both as "We choose the vertical to be black, and we choose the horizontal freely, and they happen to become black", and as "We choose the horizontal to be black, and we choose the vertical freely, and they happen to become black" – Arthur Jan 30 '17 at 17:46
• There are four ways of colouring the cross so that top and bottom are black and four ways so that left and right are black, BUT two of these ways (all the cross black) are the same so there are seven possibilities in total and not eight. – Mark Bennet Jan 30 '17 at 17:47
• @MarkBennet Wait so after considering the cross, how do I consider the corners to get the right answer? – Airdish Jan 30 '17 at 18:20
• @Airdish Can I suggest you try that yourself - think of it as a diagonal cross - but that will tell you whether you have understood. – Mark Bennet Jan 30 '17 at 18:30