# Two probabilities of the same thing

Consider this minesweeper situation. The 3 I've circled tells us that there is a mine in one of A, B and C. So, this means the probability that the mine is in A is $\frac{1}{3}$. And the 4 I've circled (and the 2 below it) tells that there is a mine in A or D. This makes the probability of finding a mine in A = $\frac{1}{2}$.

So, how to combine these two probabilities to get the actual probability of finding a mine in A?

I thought of this: There are two possibilities: Either there is a single mine in A or there is one mine in D and one in B or C. These two possibilities are equally likely, I guess. So, the probability that there's a mine in A is $\frac{1}{2}$. If that was correct, then is there any other approach to this?

EDIT: Also, the circled 3 tells that that the probability that a mine is in B or C=$\frac{2}{3}$. But if we consider the 3 to the right of the circled 3, it tells us that there is a single mine in one of B,C,E and F, which makes the probability of finding a mine in B or C=$\frac{1}{2}$. Again, there are two probabilities of finding a mine in B or C.

• Is there only one mine? Can there be 4 mines, one each in ABCD? – Gaurang Tandon Oct 20 '17 at 6:18
• @GaurangTandon In this situation, there can be either one mine or two mines. The numbers tell us how many mines there are in the 8 blocks surrounding the number. For example, the number 4 tells us that there are four mines in the 8 blocks surrounding it. I've determined 3 of those 4 mines (that's where those 3 red flags are). This means the remaining mine is in A or D. – Dove Oct 20 '17 at 6:42
• The total number of possible cases in this situation is so limited you could count them by writing them by down. But, a general solution for a possibly broader situation is more suitable. – Gaurang Tandon Oct 20 '17 at 6:46
• @RobertIsrael The probability of the presence of a mine in ABCD would surely be affected after numbers in blocks surrounding them are uncovered. But, they don't affect the probability as long as they remain covered, at least? We've to base our answer on what we know, not on what we could know. Like, the probability of rain today is 0.75. But, if I come to know later that there is a hurricane approaching, I'd say rain=0.99. But, for now, I'm commenting based on what I know, not what I could know *more*. Willing to be clarified :) – Gaurang Tandon Oct 20 '17 at 7:09
• After doing some googling, the probability model of minesweeper seems to assume they are uniformly random - each mine configuration is equally likely, given that it is consistent with the current information. In your example it is lucky that the group of $6$ grids contains exactly $2$ mines, namely $\{D, B\}, \{D, C\}, \{A, E\}, \{A, F\}$ and thus the probabilities in this group will be independent of the global no of mines and the no of remaining grids. In general you need to count all the possible configurations, but now this case is simple, with $P(D)=P(A)=1/2$ and $P(B)=P(C)=P(E)=P(F)=1/4$ – BGM Oct 20 '17 at 8:58

The inconsistencies in your considerations stem from the fact that you're applying the principle of indifference in cases where it doesn't apply. The events that you're considering as equiprobable actually have different structures, and there is no reason to regard them as equiprobable.

As BGM has pointed out in a comment, you would usually need to take into account non-local information (how many mines are left in all, or with which density the mines were distributed), but in this case you can prove that there are exactly two mines in squares $A$ through $F$, so the local information is enough. Every arrangement of these two mines that is consistent with the displayed numbers is equiprobable, since the game is presumably equally likely to generate any given arrangement of the mines. (This does not imply that any individual properties of the configuration (such as "there is a mine on $A$" or "there is a mine on $B$") are equiprobable -- they may be more or less likely because they may be consistent with more or fewer arrangements; only the arrangements are equiprobable.)

Another interesting question is what is the best strategy to proceed in order to uncover all six labelled squares without dying (assuming for the sake of argument that these are the only relevant squares and the unlabeled squares don't contain mines and can't be clicked). It is not necessarily optimal to uncover the square with the lowest probability of containing a mine, since you also want to gain as much information as possible in order to minimize risks in the future.

If you click on $A$, you die with probability $\frac12$; otherwise you can uncover $B$, $C$ and $D$, but then you still don't know whether the second mine is in $E$ or $F$, so you have another $50\%$ chance of dying, for a total survival probability of $\frac14$.

If you click on $D$, you also die with probability $\frac12$, but in this case, when you survive you have complete information, so the overall survival probability is $\frac12$.

If you click on $B$, you have a $\frac14$ probability to die, a $\frac14$ probability of getting the complete information that the mines are at $C$ and $D$, and a $\frac12$ probability of being left with an unknown mine in $E$ or $F$, with another $50\%$ chance to die, for a total survival probability of $\frac14\cdot0+\frac14\cdot1+\frac12\cdot\frac12=\frac12$.

If you click on $C$, you're certain to get a $1$ if you survive, so that doesn't gain you any information and thus can't be an optimal move.

If you click on $E$ or $F$, you have a $\frac14$ probability to die, but if you survive, you can get complete information without further risks: either the number you get tells you that the remaining two mines must be in $D$ and $B$, or it tells you that $B$ is empty, so you can click it and the number will allow you to distinguish the two remaining possibilities.

Thus, the optimal strategy is to click on $E$ or $F$, and in this case you have a survival probability of $\frac34$ – a stark contrast to the survival probability of only $\frac14$ for the worst strategy, first clicking on $A$.