hi, for an independent event, like flipping a fair coin does Pr(A\mid B) always equal to Pr(B\mid A)? for an independent event, like flipping a fair coin does  $P(A\mid B) = P(B\mid A)$?
Example You flip a fair coin, independently, three times,
 Event A. The first flip results in heads
 Event B. The coin comes up heads exactly once.
will  $P(A\mid B) = P(B\mid A) = \frac78$
 A: If $A,B$ are independent (and have positive probability), then $\Pr(A\mid B)=\Pr(A)$ and $\Pr(B\mid A)=\Pr(B)$. Hence your equality will hold iff $\Pr(A)=\Pr(B)$.
A: Actually, it is not true that $P(A\mid B) = P(B\mid A)$ iff $P(A) = P(B)$. If I have $\frac16$ chance of rolling a $1$ on a die, and I have a $\frac12$ chance of rolling an even number, then $P(A\mid B) = P(B\mid A) = 0$ EVEN THOUGH $P(A) \ne P(B)$.
Here is my complete solution:
We can say that if $P(A\mid B) = P(B\mid A)$ iff $P(A)=P(B)$ or $P(A\mid B) = P(B\mid A) = 0$.
By definition, $P(A\mid B)=\frac{P(A\cap B)}{P(B)}\implies P(A\cap B)=P(A\mid B)\cdot P(B)$ and $P(B\mid A)=\frac{P(A\cap B)}{P(A)}\implies P(A\cap B)=P(B\mid A)\cdot P(A).$
So we know that $P(A\mid B)\cdot P(B)=P(B\mid A)\cdot P(A).$ (This is actually called Bayes Theorm). We can easily see that if $P(A)=P(B)\ne0$, $P(A\mid B) = P(B\mid A)$, and if $A$ and $B$ cannot both happen, then $P(A\mid B) = P(B\mid A) = 0$.
A: Suppose $A$ is the event that you get heads on the first three tosses and $B$ is the event of heads on the fourth toss.  Then
$$
\frac 1 8 = \Pr(A) = \Pr(A\mid B)
$$
and
$$
\frac 1 2 = \Pr(B) = \Pr(B\mid A)
$$
so in this case $\Pr(A\mid B)\ne\Pr(B\mid A)$.
