From the problem description on Wikipedia.
The director _randomly_ puts one prisoner's number in each closed drawer.
The director _randomly_ puts one prisoner's number in each closed drawer.
Consider, if you pick a completely random location then two numbers could end up in the same box so one prisoner's number implies there is some order.
Further, random does not mean even distribution. The logic is really based on a very specific kind of random distribution.
So, I have seen someone just move the top few cards when asked to shuffle the deck. This makes a small offset more likely than you might assume. Remember the odds should be evenly spread across 52! which means in all of human history each deck should be unique, but in practice people have played with a default deck. (http://www.philly.com/philly/news/new_jersey/20120904_A_C__s...)
Basically, even distributions are very rare in the real world.
aka, they can't all individually come up with the same random numbers without communicating. But, they could all independently come up with the same strategy if the all saw the same numbers on the outside of the boxes.
Before the first prisoner enters the room, the prisoners may discuss strategy—but may not communicate once the first prisoner enters to look in the drawers.
So they certainly could all use a pre-agreed random permutation to compensate for any insufficient or evil [non-]randomness introduced by the director.
If you assume something not explicitly as part of the puzzle it's not necessarily the correct solution.
The generic version of the scheme is, after looking in box N, look in box f(N) where f is any bijective function from [1, 100], and all the prisoners agree on the same choice of f. The version described on the wikipedia page corresponds to f(N) := N, but any of the other 100! such functions would do so long as they agree on the same one.
Picture them entering the room and the boxes are on a round filing cabinet. There is no obvious #1 to pick so some of the 100 would pick different #1's. Again, as a math problem it's reasonable to assume they can agree on an order, but it's not stated.
> Before the first prisoner enters the room, the prisoners may discuss strategy—but may not communicate once the first prisoner enters to look in the drawers
So they just write down a random permutation and give a copy to each prisoner while discussing strategy.
On the other hand, if they could communicate after getting a reasonable definition of the room and know they would not be moved then sure strict ordering is easy. But, nothing says they are given that.
Further, they need to pick random numbers without the warden deducing the scheme, but let's assume the can use a public key crypto to get around the need for privacy.
Since each number is in a physical draw in a physical room, and no two draws can occupy the exact same space, there is a strict ordering of the drawers given by each drawers distance from a fixed point in the room (say a specific corner agreed on beforehand) in each of the three orthogonal dimensions sideways (x), backwards (y), and up (z).
We say a drawer A is before a drawer B if A.x < B.x; or, if A.x = B.x then if A.y < B.y; or, if A.x = B.x and A.y = B.y then if A.z < B.z
If A.x = B.x and A.y = B.y and A.z = B.z then A = B.
This is a strict ordering that will number the drawers from 1-100, and can be determined beforehand.
Now the prisoners take that order, and shuffle it randomly, without telling the warden what they are doing.
So, while that works as a math problem, it may fail as an actual solution.