Conditional Probability of Conditional Variance Question
Let a random variable $X$ be defined on a probability space $\left(
\Omega ,\mathcal{F},P\right)$. For example, if $\Omega =\{a,b,c\}$, then the
a $\sigma$-field  $\mathcal{F}=\left\{  \varnothing ,\left\{
a\right\} ,\left\{ b\right\} ,\left\{ c\right\} ,\left\{ a,b\right\}, \left\{ a,c\right\} ,\left\{ b,c\right\} ,\Omega \right\}$.
Consider a $\sigma$-field $\mathcal{H}$, such that  $\mathcal{H\subset F}$, then  I wonder how to compute $\Pr \left[ \operatorname{Var}\left[ X\mid\mathcal{F}\right] >0 \mid \mathcal{H}\right]$?
Since $\operatorname{Var}\left[ X\mid\mathcal{F}\right]$ is a random variable the question should make sense.
For example, how to evaluate: $\Pr \left[ \operatorname{Var}\left[ X\mid \mathcal{F}\right] >0\mid \{a\} \right] =?$
Current thoughts
My line of thought is as follows:
$$\Pr \left[ \operatorname{Var}\left[ X\mid\mathcal{F}\right] >0 \mid \{a\} \right] = \frac{1}{\Pr [\left\{ a\right\} ]}\Pr \left[ \left\{ \operatorname{Var}\left[ X\mid\mathcal{F} \right] >0\right\} \cap \{ a\} \right] $$
I conjecture that, the latter equals $\frac{1}{\Pr [\{ a\} ]}\Pr %
\left[ \operatorname{Var}\left[ X\mid\mathcal{F}\cap \{ a\} \right] >0\right] ,$
however, I am not sure.
Any suggestions would be welcomed.
 A: The set $\{a\}$ is not a $\sigma$-field.  It is an event.  Hence the conditional probability of a random variable given $\{a\}$ is simply a number rather than another random variable.
One error in a detail is where you wrote $\mathcal{F} = \{ \{ \varnothing\},\{a\}, \{ b\} ,\{ c\}, \{ a,b\}, \{ a,c\}, \{ b,c\}, \Omega \}$.
That should instead say $\mathcal{F} = \{ \varnothing,\{a\}, \{ b\} ,\{ c\}, \{a,b\}, \{ a,c\}, \{ b,c\}, \Omega \}$ (with $\varnothing$ where you had $\{\varnothing\}$).
This is the finest of all $\sigma$-fields on the set $\Omega$, i.e. it includes all subsets.  A consequence it that the conditional variance $\operatorname{var}(X\mid\mathcal{F})$ will be $0$ with probability $1$.
So let's look at a coarser $\sigma$-field on the same set: $\mathcal{G} = \{ \varnothing, \{a\}, \{b,c\},\Omega \}$. What is $\operatorname{E}(X\mid\mathcal{G})$ and what is $\operatorname{var}(X\mid\mathcal{G})$?  Let's get concrete and suppose, for example that
\begin{align}
P\{a\} & = 0.23, \\
P\{b\} & = 0.31, \\
P\{c\} & = 0.46.
\end{align}
Then $P\{b,c\} = 0.77$. Further suppose that
\begin{align}
X\{a\} & = 10, \\
X\{b\} & = 18, \\
X\{c\} & = 20.
\end{align}
We have
\begin{align}
\operatorname{E}(X\mid \{a\}) & = X(a) = 10, \\[8pt]
\operatorname{E}(X\mid \{b,c\}) & = \frac{0.31 X\{b\} + 0.46X\{a\}}{0.31+0.46} = \frac{0.31(18)+0.46(20)}{0.31+0.46} \approx 19.1948\ldots.
\end{align}
Therefore
$$
\operatorname{E}(X\mid\mathcal{G}) = \begin{cases} 10 & \text{with probability } 0.23, \\
19.1948\ldots & \text{with probability }0.77. \end{cases}
$$
This is a random variable whose variance and expected value are readily found. (Its expected value is the same as that of $X$ itself, and its variance is smaller than that of $X$.)
Now let's look at conditional variances.
\begin{align}
\operatorname{var}(X\mid\{a\}) & = 0, \\[8pt]
\operatorname{var}(X\mid\{b,c\}) & = \frac{0.31}{0.31+0.46} \cdot \frac{0.46}{0.31+0.46} \cdot (18-20)^2 \approx 0.962\ldots
\end{align}
I.e. the variance of a random variable that is equal to $18$ with probability $0.31/(0.31+0.46)$ and equal to $20$ with probability $0.46/(0.31+0.46)$ is that number.
Hence we have
$$
\operatorname{var}(X\mid\mathcal{G}) = \begin{cases} 0 & \text{with probability } 0.23, \\
0.962\ldots & \text{with probability } 0.77. \end{cases}
$$
So $P(\operatorname{var}(X\mid\mathcal{G})>0) = 0.77$.
