Minesweeper thermodynamics

Summary

You know how sometimes you start a game of Minesweeper and immediately get stuck? Like maybe there are some cells that you know are mines, but there aren’t any places that are safe to click. In this example there are five different ways you could fill in the mines in the neighbouring cells. Note that there’s no cell which is safe in every possibility, so there’s nowhere we can safely click to get more information. So in order to plan our next click, it would be good to know how likely it is that each cell is safe. You might think that each of the five possibilities is equally likely, in which case the probability that a cell is safe would be the proportion of them in which it isn’t a mine: But it’s important to notice that the five possible arrangements have different numbers of mines. One has $5$ mines, three have $6$ and the last has $7$ (in addition to the $5$ mines we already found). Let’s say for example that we’re playing the ‘expert’ version of Minesweeper: a $30\times 16$ board with $99$ mines. That means that outside the solved region there are $444$ remaining cells and $94$ remaining mines. So for each possibility the total number of ways to arrange the mines in the rest of the board is one of the following.$$\binom{444}{94-5}=1.930\times 10^{95}$$ $$\binom{444}{94-6}=0.483\times 10^{95}$$ $$\binom{444}{94-7}=0.119\times 10^{95}$$ Different versions of Minesweeper randomise the mines slightly differently, but after your first click it’s a good approximation that every possibility is equally likely. So the possibility with only $5$ mines is about $16.2$ times as likely as the possibility with $7$ mines! This means we should calculate how safe each cell is by finding the proportion of the possibilities in which it’s safe, but weighted by the above numbers. That gives us the following probabilities: Previously we thought that the six best cells each had a $40\%$ chance of safety. Now we see that some of them have a $69.0\%$ chance of safety, and some of them ...