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?

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 $P(D) = 0.577$
