Even Marbles Everywhere I

Fill the top-left $7 \times 7$ block freely. Is the rest of the board then forced?

The last square of each of the first $7$ rows is forced. So is the last square of each of the first $7$ columns. What about the bottom-right corner?

Count the good boards. There are $2^{64}$ boards in total, all equally likely. We show that the good boards correspond exactly to the $2^{49}$ ways of filling the top-left $7 \times 7$ block.

Step 1: Fill the $7 \times 7$ block any way you like.

Step 2: Finish the first $7$ rows. In each of rows $1$–$7$, the last square (column $8$) is forced: white if the row has an odd number of whites so far, black if even.

Step 3: Finish the first $7$ columns. In the same way, the bottom square of each of columns $1$–$7$ is forced.

Step 4: The corner. Only the bottom-right square is left, and it must fix both row $8$ and column $8$ at once. That works only if row $8$ and column $8$ currently have the same parity. They do:

• Column $8$ so far has one forced square per row $1$–$7$, each matching its row's parity in the block. So its number of whites has the same parity as the total number of whites in the $7 \times 7$ block.
• Row $8$ so far has one forced square per column $1$–$7$, so by the same argument it also matches the parity of the block's total.

Both have the same parity, so one choice of corner fixes both: white if both are odd, black if both are even.

So every $7 \times 7$ block extends to exactly one good board, and every good board comes from its own top-left block. There are $2^{49}$ good boards.

Probability.

Answer: $0.00003$.