Overlapping Probability in Minesweeper I play a lot of minesweeper in my spare time. Sometimes I'll be in the situation where probabilities of hitting a mine overlap:
Situation on a board


*

*There's one mine in the squares A, B, C, and D (25% chance per square).


AND


*There's one mine in the squares D and E (50% chance).


Questions
a) How does this information change the probability in the square D? 
b) What are the odds of there being a mine in A, B, C, and E? 
c) And if I were to click a square, which would give me the lowest odds of hitting a mine?
 A: Sorry for this being a long answer. Actual probabilities are at the end if you want to jump straight there.
To work this out we need the underlying probability of any random square being a mine, knowing nothing about the square. Let's call this probability $q$.
$$q = \dfrac{\text{#mines in the grid}}{\text{#squares in the grid}}.$$
So,
\begin{eqnarray*}
\text{Level} && \qquad \qquad q \\
\hline && \\
\text{Beginner} && \qquad \frac{10}{9 \times 9} &=& \frac{10}{81} \approx 0.12346 \\ && \\
\text{Intermediate} && \qquad \frac{40}{16 \times 16} &=& \frac{5}{32} = 0.15625 \\ && \\
\text{Advanced} && \qquad \frac{99}{16 \times 30} &=& \frac{33}{160} = 0.20625.
\end{eqnarray*}
Now, using the information you provide, we define the following events:
\begin{eqnarray*}
U &=& \text{Exactly $1$ of $A,B,C,D$ is a mine} \\
V &=& \text{Exactly $1$ of $D,E$ is a mine} \\
M_X &=& \text{Square $X$ is a mine - where $X$ can be $A,B,C,D$ or $E$.} \\
\end{eqnarray*}
Your three questions are asking for conditional probabilities.
(a) We want the probability of $M_D$ given both $U$ and $V$.
\begin{eqnarray*}
P(M_D \mid U \cap V) &=& \dfrac{P(M_D \cap U \cap V)}{P(U \cap V)}.
\end{eqnarray*}
\begin{eqnarray*}
P(M_D \cap U \cap V) &=& P(M_D \cap \neg M_E \cap \neg M_A \cap \neg M_B \cap \neg M_C) \\
&=& (1-q)^4q.\qquad\qquad\text{(See Note 1)}
\end{eqnarray*}
\begin{eqnarray*}
P(U \cap V) &=& P(M_D \cap \neg M_E \cap \neg M_A \cap \neg M_B \cap \neg M_C) \\
&& + P(\neg M_D \cap M_E \cap M_A \cap \neg M_B \cap \neg M_C) \\
&& + P(\neg M_D \cap M_E \cap \neg M_A \cap M_B \cap \neg M_C) \\
&& + P(\neg M_D \cap M_E \cap \neg M_A \cap \neg M_B \cap M_C) \\ && \\
&=& (1-q)^4q + 3(1-q)^3q^2.
\end{eqnarray*}
\begin{eqnarray*}
\therefore P(M_D \mid U \cap V) &=& \dfrac{(1-q)^4q}{(1-q)^4q + 3(1-q)^3q^2} \\ && \\
&=& \dfrac{1-q}{(1-q) + 3q} \qquad\text{dividing through by $(1-q)^3q$} \\ && \\
&=& \dfrac{1-q}{1 + 2q}.
\end{eqnarray*}
(b) I think you're asking the probability of $D$ not being a mine, so that $E$ will be a mine and so will exactly one of $A,B,C$. This is just $P(\neg M_D \mid U \cap V)$:
\begin{eqnarray*}
P(\neg M_D \mid U \cap V) &=& 1 - P(M_D \mid U \cap V) \\ && \\
&=& 1 - \dfrac{1-q}{1+2q} \\ && \\
&=& \dfrac{3q}{1+2q}.
\end{eqnarray*}
(c)
\begin{eqnarray*}
P(M_E \mid U \cap V) &=& P(\neg M_D \mid U \cap V) \\ && \\
&=& \dfrac{3q}{1+2q}.\end{eqnarray*}
Similarly, if $M_D$ doesn't occur then exactly one of $M_A, M_B, M_C$ does occur and by symmetry, each of them is equally likely. Thus,
\begin{eqnarray*}
P(M_A \mid U \cap V) &=& \dfrac{1}{3} P(\neg M_D \mid U \cap V) \\ && \\
&=& \dfrac{q}{1+2q}.\end{eqnarray*}
Now, to find the least likely square to be a mine, we want the smallest of the above conditional probabilities.
Obviously,
$$P(M_A \mid U \cap V) = \dfrac{q}{1+2q} \lt \dfrac{3q}{1+2q} = P(M_E \mid U \cap V).$$
\begin{eqnarray*}
P(M_D \mid U \cap V) - P(M_E \mid U \cap V) &=& \dfrac{1-q}{1+2q} - \dfrac{3q}{1+2q} \\ && \\
&=& \dfrac{1-4q}{1+2q} \gt 0 \text{ if and only if } q \lt 0.25 \\
&& \qquad \text{which is true for all $3$ levels.}
\end{eqnarray*}
Therefore $A,B,C$ are equally the least likely to be a mine and $D$ the most likely (as intuition probably tells you).
Finally, some actual (conditional) probabilities for the three levels:
\begin{eqnarray*}
\text{} \quad && D \text{ is a mine} \quad && E \text{ is a mine} \quad && A \text{ is a mine} \\
\hline && \\
\text{Formula}\quad  && \frac{1-q}{1 + 2q} && \frac{1-q}{3q} && \frac{q}{1 + 2q} \\ && \\\text{Beginner} \quad && 0.703 && 0.297 && 0.099 \\ && \\
\text{Intermediate} \quad && 0.643 && 0.357 && 0.119 \\ && \\
\text{Advanced} \quad && 0.562 && 0.438 && 0.146.
\end{eqnarray*}
Note 1: This is not the exact calculation but is an approximation by assuming that $P(M_X) = q$ independently of other squares. This is not true because they are in fact dependent on each other. But it's a reasonable approximation given the large number of squares, especially in the Advanced level. And it simplifies the calculations quite a bit. An example of the true value (assume the Beginner level with $10$ mines and $81$ squares):
$$P(M_D \cap \neg M_E \cap \neg M_A \cap \neg M_B \cap \neg M_C) = \dfrac{10 \cdot 71 \cdot 70 \cdot 69 \cdot 68}{81 \cdot 80 \cdot 79 \cdot 78 \cdot 77}.$$
So exact results are certainly calculable; it's just more tedious to get them.
A: My attempt at calculating the probability that there is a mine in $D$:
If there is one mine in $\lbrace A, B, C, D \rbrace$ and one mine in $\lbrace D, E \rbrace$, then there are two cases:


*

*There are two mines, placed in either E and A, or E and B, or E and C

*There is one mine, placed in D


Let $S$ denote the number of mines in $\lbrace A, B, C, D, E \rbrace$.
Define $p_1 = P(S=1 | S \in \lbrace 1,2 \rbrace)$. Let $P(D)=P(\text{mine in D})$.
Since $P(D|\text{one mine})=P(\text{one mine}|D)=1$, we have $P(D)=p_1$. Meaning, the probability of the mine being in $D$ is the same as the probability of the no. of mines being $1$. So we calculate the latter:
$$
p_1 = P(S=1 | S \in \lbrace 1,2 \rbrace) = \frac{P(S=1 \cap S \in \lbrace 1,2 \rbrace)}{P(S \in \lbrace 1,2 \rbrace)}  = \frac{P(S=1)}{P(S \in \lbrace 1,2 \rbrace)} =  \frac{P(S=1)}{P(S=1)+P(S=2)}
$$
The probabilities in the last fraction are 'a-priori' probabilities, which are calculated thus: Let $N$ denote the total number of mines, and $T$ the total number of squares.
$$
P(S=1)=\frac{ {5 \choose 1} {N\choose 1} {T-N \choose 4}}{T\choose 5} \hspace{1cm}\text{and} \hspace{1cm} P(S=2)=\frac{ {5 \choose 2} {N\choose 2} {T-N \choose 3}}{T\choose 5}
$$
Now caluclating $p_1$, after substituting the formulas and do some algebra, we get
$$
p_1 = \frac{T-N-3}{3N+T-7}
$$
So for example for a 'medium' sized minesweeper with $N=40$ mines and $T = 16 \times 16$ squares, one gets $P(D) = 0.577$
