Monty Hall Problem

Say you come across a situation where you have to guess, and all cells have the same chance of containing a monster. If you have a minion or any item that flags or reveals a monster, this is a strategy you can do.

Say you have 2 monsters left in 3 hidden cells. Flag cell 1 which has a 2/3 chance of hiding a monster, then use your item to flag/reveal a monster. Let's say the monster was in cell 3. Cell 1 had a 2/3 chance of having a monster. Now there is 1 monster and 2 cells left. Since cell 1 had a 2/3 chance (and still does), the last cell, cell 2, only has a 1/3 chance of hiding a monster.

This unintuitive strategy was posed back in 1975 by a statistician, and is possible because your flagging of cell 1 constrained what your item could flag, likely forcing it to flag the last monster.

For a more general case:

Say you have X number of monsters and N number of cells. Flag 1 cell, then use your item to flag/reveal a monster. The cell you personally flagged had a X/N chance of having a monster. Now there are X-1 monsters and N-1 cells left, so the other N-2 cells equally have a [(X-1)-X/N]/(N-2) = [X/N]*[(N-1-N/X)]/(N-2)] chance, which is a lower chance because (N-1-N/X)]/(N-2) < 1. So you should guess one of these other cells.

This strategy should work even when the probability is not uniform among all hidden cells. In this case, you should flag the cell most likely to hide a monster, as that will make safer cells to become even safer to reveal.