Flaw in a conditional probability argument Imagine an experiment where you roll two fair, six-sided dice. Someone peeks at the dice, and (truthfully) tells you that "at least one of the dice is a 4". What is the probability that the total of the dice is 7?
It seems straightforward to calculate that the probability the total is 7 is 2/11.
However, the person who peeked at the dice could equally well have said "at least one of the dice is a 1" and you would come to the same conclusion - 2/11. Or they could have said "at least one of the dice is a 2" or "at least one of the dice is a 3", or indeed any number from 1 to 6, and you would still conclude that the probability that the total is 7 is 2/11.
Since you will always conclude that the probability that the total is 7 is 2/11, you could block your ears as they speak, and you'd still come up with 2/11. From there it's a short hop to conclude that even if they don't say anything the probability that the total is 7 is 2/11.
However, clearly if they don't say anything, the probability that the total is 7 is not 2/11, but rather 1/6.
Where is the flaw in the argument?
 A: The flaw in the argument is that the conditioning random variable is not well-defined.
The ambiguity lies in  how our friend peeking at the dice decides to report the information back to us. If we let $X_1$ and $X_2$ denote the random variables associated with the values of each of the dice, then it is certainly true that for each $k \in \{1,2,\ldots,6\}$,
$$
\mathbb P(X_1 + X_2 = 7 \mid X_1 = k \cup X_2 = k) = \frac{2}{11} \>,
$$ 
independently of $k$.
However, the events $\{X_1 = k \cup X_2 = k\}$ are clearly not mutually exclusive, and so clearly we cannot claim
$$
\begin{align}
\mathbb P(X_1 + X_2 = 7) &\stackrel{?}{=} \sum_{k=1}^6 \mathbb P(X_1 + X_2 = 7 \mid X_1 = k \cup X_2 = k) \mathbb P( X_1 = k \cup X_2 = k )  \cr
&\stackrel{?}{=} \frac{2}{11} \sum_{k=1}^6  \mathbb P( X_1 = k \cup X_2 = k )  \cr
&\stackrel{?}{=} \frac{2}{11}
\end{align}
$$
Formally, we need to properly define a random variable, say $K$, that encodes the knowledge imparted by our peeking friend.
Our peeking friend could always report the value of the left-most die, or the right-most, or the larger of the two. She could flip a coin and then report based on the coin flip, or employ any number of more complicated machinations.
But, once this process is specified, the apparent paradox vanishes.
Indeed, suppose that $K = X_1$. Then, we have
$$
\begin{align}
\mathbb P(X_1 + X_2 = 7) &= \sum_{k=1}^6 \mathbb P(X_1+X_2 = 7, K=k) \cr
 &= \sum_{k=1}^6 \mathbb P(X_1+X_2 = 7 \mid K=k) \mathbb P(K=k) \cr
 &= \sum_{k=1}^6 \frac{1}{36} = \frac{1}{6} \>.
\end{align}
$$
Similar arguments hold if we choose $K = X_2$ or $K = \max(X_1,X_2)$, etc.
A: If $B$ is an event with the property that $P(B\mid D_i) = p$ for all events 
$\{D_1, D_2, \ldots\}$ in a countable partition of the sample space $\Omega$,
(that is, $D_i \cap D_j = \emptyset$ for all $i \neq j$ and 
$\bigcup_i D_i = \Omega$), then the
law of total probability tells us that
$$P(B) = \sum_i P(B\mid D_i)P(D_i) = p\sum_i P(D_i) = p.$$  However,
the law of total probability does not apply if the events $D_i$ are
not mutually exclusive (even though their union is still $\Omega$), and 
we cannot conclude that $P(B)$ equals the common value of $P(B\mid D_i)$.
Let $A_i$ denote the event that at least one of the dice shows the number $i$ and $B$ the event that the sum of the two numbers on the die is $7$.  We know that
$P(B) = \frac{1}{6}$ and that $P(A_i) = \frac{11}{36}$.  Also, 
$P(B\mid A_i) = \frac{2}{11}$.  Now, 
$A_1\cup A_2\cup A_3 \cup A_4\cup A_5\cup A_6$
is the entire sample space $\Omega$
but we cannot use the fact that $P(B\mid A_i)$
is the same for all choices of $i$ to conclude that $P(B) = \frac{2}{11}$ 
because the $A_i$ are not mutually exclusive events.
However, notice that regarded as a multiset,
$A_1\cup A_2\cup A_3 \cup A_4\cup A_5\cup A_6$ contains each outcome
$(i,j)$ exactly twice, once as a member of $A_i$ and again as a member of 
$A_j$.  Therefore,
$$\sum_{i=1}^6 P(B \mid A_i)P(A_i) 
= \sum_{i=1}^6 \frac{2}{11}\times\frac{11}{36} = \frac{1}{3} $$
which is exactly twice the value of $P(B)$. 
