mathmagician

Below are closed-form expressions for 1xn and 2xn cases

Technically, you're right. What i mean to ask though is how many times will he pick up a piece. Note: When a piece does not fit, he puts it back into the pile and may pick it up again immediately. If so, it still counts as picking up another piece. The solution I found is recursive. I have closed forms for m=1 and m=2. I'm still working on generalizing it.

I keep coming up with the same answer using the exact same math, but when I simulate it on the computer, it takes greater that 2.25 steps for r=1.5m (takes about 3.5 steps) and greater than 9 steps for r=3m (takes about 11.4 steps).

A blind old man loves jigsaw puzzles. He starts them by picking a piece at random and putting it down. He then picks another piece at random and tries to attach it to each side of the piece he started with (trying all four rotations and flipping it over) and repeats until the piece fits. Then he picks another piece at random and tries to attach it to the part of the puzzle he already has. He continues doing this until the puzzle is solved. What is the expected number of pieces he will try in order to solve an m by n puzzle?

there are three non-integer solutions, but as they require cube roots, I am to lazy to post them.