Will they fit?

There are 4 corner pieces, 32 edge pieces and 64 centre pieces. Two pieces fit with corners, three with edges and four with centres. So the probability is $\frac{4}{100}\times\frac{2}{99} + \frac{32}{100}\times\frac{3}{99} + \frac{64}{100}\times\frac{4}{99} = \frac{2}{55}$ , and the answer is $\boxed{57}$ .

There are $9$ adjacent pairs of pieces in each row and column of the puzzle. Thus there are $20\times9=180$ pairs of pieces which are adjacent, and so fit together,. There are $\binom{100}{2} = 4950$ pairs of pieces that can be chosen, so the desired probability is $\tfrac{180}{4950} = \tfrac{2}{55}$ , making the answer $\boxed{57}$ .

Imagine the puzzle is solved; and imagine each piece to be a dot. So, you get a 10x10 dotted matrix. You have to find the no. Of ways of choosing 2 adjacent dots. Keep 2 sticks vertically so that they cover any 2 adjacent columns of dots, and keep one stick horizontally on any row of dots( think of the letter H). The rule is: move the two sticks together, and the horizontal stick on any horizontal series of dots. Intersection of sticks give 2 horizontally adjacent dots.

No. Of ways to move 2 vertical sticks= 9. No. Of ways to move the other stick = 10. so, p=9 10= 90. You can choose 2 dots vertically as well. Repeat the same experiment by switching the orientation of sticks(like the letter H rotated 90 deg). Again q=90. Total no. Of ways of choosing adjacent dots = p+q=90+90=180=m(say). No. Of ways of choosing any 2 dots = 100C2=n(say). m/n= 2/55 (in coprime form). 2+55=57. *By this method, we can safely ignore the edge piece/corner piece assumptions, as asked in the question