Dartboard paradox and understanding independence By definition, events $A$ and $B$ are independent if $$P(A \cap
    B) = P(A)\:P(B).$$ Thus if an event $A$ happens almost never
$\left(P(A)=0\right),$ then
$A$ is independent of all events, including itself. So, hitting the
exact centre of a dartboard (happens almost never) and hitting within its inner ring are independent events.

On the other hand, it is standard to characterise
independence as follows:

Two events are independent if the occurrence of one does not
affect the probability of occurrence of the other.

Since hitting the exact centre of a dartboard guarantees hitting
within its inner ring (which is otherwise not guaranteed), the
two events are dependent.


Isn't this a contradiction?? If yes, then do we accept that the definition of independence is not meant to fully correspond to its verbal/intuitive characterisation?

Is there any semantic difference between the following two versions?

$(i)$ Two events are independent if the occurrence of one does not
affect the probability of occurrence of the other.


$(ii)$ Two events are independent if the occurrence of one does not
affect the occurrence of the other.




ADDENDUM
On further pondering, I have resolved question 2:

Consider this experiment: flip two fair coins, letting $H_1$ be the
event that the first coin lands on Heads, and $X$ be the event that
the coins land on different sides. Then $$ P\left(H_1 \cap
X\right)=\frac14=P(H_1)\:P(X);$$ i.e., $H_1$ and $X$ are independent
events.
Knowledge that $H_1$ has occurred reduces the known number of ways that $X$ can eventuate—from $2$ (HT, TH) to $1$ (HT)—but does not alter the probability $(\frac12)$ assigned to $X.$


The following revision characterises pairwise independence more clearly and accurately:

Events $A$ and $B$ are independent if $A$ having occurred does not
change the probability of $B.$

It still doesn't apply when $A$ is an impossible event though; and remains problematic when $A$ is an almost-never event. (What does it even mean to say that an almost-never event has occurred: in what sense have I hit the exact centre of a dartboard?)
Events $A$ and $B$ are indeed defined to be independent if $P[AB] = P[A]P[B]$.  This definition holds for all cases, including cases when $P[A]=0$ and/or $P[B]=0$. The definition leads to your desired interpretation &quot;knowing if $B$ occurred does not affect the likelihood of $A$&quot;  in the special case $P[B]&gt;0$, since it means $P[A|B] = P[A]$ in that case. It does not allow your interpretation in the case when $P[B]=0$, since we cannot condition on a zero-probability event.  So indeed, there are peculiarities when $P[B]=0$ since intuition fails in that case. Nevertheless, your intuition is still useful since it applies when $P[B]&gt;0$.
It may also be helpful to observe that if $P[B]=0$ then $P[A|B^c] = P[A]$, which is intuitively what we would want when $A$ and $B^c$ are independent.

A situation about darts similar to the one you posed can be addressed by the definitions of conditioning with respect to a random variable. For a dartboard, we can imagine $(X,Y)$ uniformly distributed over a 2-d ball of radius 1.  Then define $R = \sqrt{X^2+Y^2}$ as the distance from the center.  Consider the events:
$$ A = \{R \leq 1/10\}, B = \{R = 1/10\}$$
Now $P[B]=0$ so $A$ and $B$ are independent.  Nevertheless this is similar to the situation you describe because $B \subseteq A$, so &quot;intuitively&quot; we would want $P[A|B]=1$. This holds if we use:
$$ P[A|R=r] = \frac{f_{R|A}(r)P[A]}{f_R(r)} $$
which shall be applied when $f_R$ and $f_{R|A}$ are continuous at point $r$ and $f_R(r)\neq 0$,
and we observe:
\begin{align}
f_R(r) &amp;= \left\{\begin{array}{cc} 
2r  &amp;  \mbox{ if $r \in [0,1]$}\\
0 &amp; \mbox{else} 
\end{array}\right.\\
f_{R|A}(r) &amp;= \left\{\begin{array}{cc}
\frac{2r}{P[A]} &amp; \mbox{ if $r \in [0,1/10]$} \\
0 &amp; \mbox{ else} 
\end{array}\right.
\end{align}
so
$$ P[A|R=r] = \left\{\begin{array}{cc}
0 &amp; \mbox{if $1/10 &lt; r&lt;1$} \\
1 &amp; \mbox{ if $0&lt;r\leq 1/10$} 
\end{array}\right.$$
Our expression $P[A|R=r]$ cannot be viewed as an instance of the definition of conditional probability $P[A|B]$ for events $A$ and $B$, since the event $\{R=r\}$ has probability 0.  Rather, it is an expression related to the definition of conditioning on a random variable $R$. Formally, you can say the random variable $Z$ defined by
$$ Z = \left\{\begin{array}{cc}
0 &amp; \mbox{if $R&gt;1/10$} \\
1 &amp; \mbox{ if $0\leq R \leq 1/10$} 
\end{array}\right.$$
is a &quot;version&quot; of the conditional expectation $E[1_A|R]$.  As such, it is almost surely the same as any other version. Thus, while this version takes the liberty to define $Z=1$ when $R=0$, and $Z=0$ when $R\geq 1$, this is fine because it only affects a set of measure 0.
When two events A and B are disjoint it means that if one of them occurs, the other one cannot occur, i.e., $A\cap B= \phi$. Suppose $A=${hitting the exact centre of a dartboard} is an impossible event, then based on the above argument non-happening of A implies the happening of other events, and hence there is no independence between A and the other events. Further, the dartboard is an example of continuous sample space and an event like A may be a rare event, but one claim that it is an almost impossible event.