Do similar pieces make a puzzle easier or harder? (idea)

Consider jigsaw puzzle P. If its pieces are identically-shaped, is the difficulty of the puzzle greater or smaller than with a puzzle where pieces can only be fit into some of the others?

Claim. Differently-shaped pieces make a puzzle easier to solve.

Proof.

Consider a box P, where every piece p_i has an exact match with either two (for corner pieces), three (edge pieces), or four (internal pieces) other pieces P_match,i = {p_m} (so 2 ≤ |P_match,i| ≤ 4). Consider also those pieces P_imp,i = {p_j}, for which an imperfect match exists with p_i. There is a tolerance t_i ∈ T associated with each p_i, where t_i > z for some lower bound z. This tolerance expresses the probability of a piece matching any other piece, and is a function of the shape of its teeth and concavities.

The tolerance for any p_i in a puzzle with identically-shaped pieces is the limit as t_i → 1, since any piece can fit into any other piece. The size of the set of pieces with which p_i fits (|P_fit,i| = |P_match,i| + |P_imp,i|) is directly proportional to the tolerance for that piece, or |P_fit,i| = x|1 - t_i|, for some x ∈ ℜ: the lower the tolerance, the less likely p_i is to fit with any p_j, and thus there are fewer pieces with which it can have an imperfect fit. Note that P_imp,i varies with t_i, but P_match,i does not. Thus, as t_i → 0, |P_imp,i| → 0. If |P_imp,i| = 0, then the only p_m which fit p_i are the 2-4 pieces which are the correct neighbors.

Thus, the more similarly-shaped the pieces of a puzzle are, the more possible matches there are for any given piece, and the harder it is to find its correct neighboring pieces. QED.

Now, you may have noticed that this is not the most elegant proof out there. So I challenge you to write me a proof worth of the Book.

1/15/08: We have a winner! DTal, as you can see, has responded. Of course, there's still space for a runner-up if you can write a better mathematical proof than mine.