You have an $m$ by $n$ rectangular grid. Each cell may contain a single bomb or a number (or neither, but not both). If a cell has no bomb and is adjacent (horizontally, vertically or diagonally) to atleast 1 bomb-containing cell, then it must have a number. The number (between $1$ and $8$) is the number of cells adjacent to it that contain bombs. This is how you must set up a grid for the opponent.
All cells with numbers are visible to the opponent. All cells with bombs and all empty cells look the same, these are the ones that need to be probed. The opponent picks up one cell at a time. If he picks up a bomb, he loses. If he picks up every blank cell, but no bomb, he wins.
Is it possible to set up a grid, where the opponent cannot use logic alone to win, and will have to use guesswork at some point? If so, please give the smallest* possible grid setup.
Twist: Instead of a 2D grid, if we have a 3D grid ($l$ by $m$ by $n$). All rules same as above, except that now a number could be anything between $1$ and $26$. What is the smallest* unsolvable grid setup?
*smallest in terms of area/volume of grid, not number of bombs