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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 $P(A\mid B) = P(B\mid A) = \frac78$

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  • $\begingroup$ What is an independent event? For independence, you need a least two events. $\endgroup$ – Hagen von Eitzen Nov 16 '14 at 11:38
  • $\begingroup$ Pr(B|A)=7/8? Does this mean that, if the first flip comes up heads, you'd bet 7 to 1 that there will be no more heads? Hmm. $\endgroup$ – bof Nov 16 '14 at 12:16
  • $\begingroup$ Isn't Pr(A|B) = Pr(A) + Pr(B) = 1/2 + 3/8 then shouldn't Pr(B|A) = 3/8 + 1/2 ? $\endgroup$ – user2551612 Nov 16 '14 at 12:29
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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 $\Pr(A)=\Pr(B)$.

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  • $\begingroup$ I updated my post, does it still hold? $\endgroup$ – user2551612 Nov 16 '14 at 11:43
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    $\begingroup$ I changed every instance of \operatorname{Pr} to \Pr and also added proper use of \mid. $\endgroup$ – Michael Hardy Oct 1 '15 at 0:41
  • $\begingroup$ It is actually false that $\Pr(B\mid A)=\Pr(B)$ will hold iff $\Pr(A)=\Pr(B)$. (Look at my solution). $\endgroup$ – thepiercingarrow Oct 1 '15 at 2:33
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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 $P(A\mid B) = P(B\mid A) = 0$.

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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 $\Pr(A\mid B)\ne\Pr(B\mid A)$.

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