Jigsaw puzzle puzzle

[bobman0330]

Say you have a partially completed jigsaw puzzle with a large rectangular hole of substantial size in the middle. All the puzzle pieces are squarish with either a prong or a hole on each side. The way the puzzle has played out, there are just 3 or 4 each of 4-hole and 4-prong pieces. There are a lot of 3-prong and 3-hole pieces, and there are even more 2-prong pieces, with all but one of them having their prongs and holes on opposite sides, not adjacent.

If anyone needs clarification of that, let me know. This is not the ideal format for presenting such a visual puzzle.

Now, the space in the top right corner has to border two known pieces. One of them is showing a prong, one a hole. That means the correct piece will have to be either:
A. a 3-prong piece (orientable 1 way)
B. a 3-hole piece (orientable 1 way)
C. a 2-prong piece (orientable 2 ways)

For the moment, just assume that there are equal numbers of each kind of piece. Does the increased number of possible orientations of type C pieces make it more (twice?) as likely that the correct piece is a two-prong piece?

[LetYouDown]

There are no designs/etc. on these pieces, I assume? Assuming not, this seems like it's really just a matter of perspective. What defines the odds that it's "correct"?

I'd postulate that it's either 100% likely to be the correct piece or 0% likely, but I'd probably get shot =).

[RiverTheNuts]

I think its a pretty simple "yes" that being orientable in 2 ways makes them count as two different pieces as far as plugging into the corner is concerned...

(2 x #2-prongers)/total pieces that could fit = chance a 2 prong piece goes tjere

[bobman0330]

That's what I thought at first, but consider this situation:

There are three similar corners, all of which require a piece that is entirely bright orange. Conveniently for this hypothetical, you happen to have 3 bright orange pieces, 1 1-prong, 1 2-prong, and 1 3-prong.

Possible arrangements: (parentheses indicate possible orientations of the 2-prong)
1 2(1) 3
1 2(2) 3
1 3 2(1)
1 3 2(2)
2(1) 1 3
2(1) 3 1
2(2) 1 3
2(2) 3 1
3 2(1) 1
3 2(2) 1
3 1 2(1)
3 1 2(2)

For a total of 12 possibilities. In 4 of those possibilities, the 2-prong will be in the first position. This suggests that the probability of the two-prong being in any given space is 1/3. Doesn't it?