Formalize the notion that "$A$ has no relevant information about $X$ that $B$ doesn't have" in probability theory Assume a probability space $(\Omega, \mathcal B, \mathbb P)$. Assume two $\sigma$-algebras on $\Omega$: $A,B$, that represent information about $\Omega$.
Consider some random variable $X:\Omega\to O$. How do we formally state the notion that "$A$ has no relevant information about $X$ that $B$ doesn't have"?
I at first thought that we could represent this by $$\forall x\in O, \quad\mathbb P (X=x|BA)= \mathbb P (X=x|B)$$
But then I realized that this is a necessary but not sufficient condition for the desired intuitive notion: It can be that $A$ DOES contain relevant information in the sense that $A$ consists of $2$ "observations" that exactly balance each other out so as to leave the conditional distribution unchanged. 
For example, if $X$ denotes whether the $9$th coin flip of a possibly unfair coin is heads or not, and $B$ contains no information, and $A$ contains the information that half of the first $8$ were heads, (and the prior was 50%), then $\forall x\in \{H,T\}, \quad\mathbb P (X=x|BA)= \mathbb P (X=x|B)$, but it is not the case that the observations were "irrelevant", in an intuitive sense.
EDIT: Probably this has to be done by some kind of subset condition on sigma algebras?
 A: Perhaps you should consider conditional expectation as way to state this. If $A\oplus B$ is the $\sigma$ algebra generated by the union of  $A$ and $B$, then:
$\mathbb{E}[X\vert A\oplus B]= \mathbb{E}[X\vert A]$
Recall that for a probability space $(\Omega, \mathcal{F},P)$ and a random variable $X:\Omega\rightarrow \mathbb{R}$, given a sub sigma algebra $\mathcal{G}\subseteq \mathcal{F}$ a $\mathcal{G}$ measurable function $Y:\Omega\rightarrow \mathbb{R}$ is said to be a conditional expectation of $X$ with respect to $\mathcal{G}$ if:
$\mathbb{E}\Big[ X\cdot1_F \Big]= \mathbb{E}\Big[ Y\cdot 1_F \Big]$ for all $F\in \mathcal{G}$.
It is a known fact that the conditional expectation exists and is unique almost surely. 
A: Please forgive if this is too glib in not invoking any structure! Although a tad perplexed by the word 'relevant' as in one interpretation this could imply an intersection with a third $\sigma-$algebra $C$ which defines 'relevance', how about simply that $\{\sigma (X) \cap A\} \subset B$? Is that too much of a shortcut by invoking $\sigma (X)$ (intersection of all $\sigma-$algebras generated by Borel set pre-images under $X$ and therefore the smallest such $\sigma-$algebra, i.e all the information content afforded by $X$)? Again, not sure if such a structureless answer is what you are looking for, happy to delete of course if it isn't / I've missed the point! 
A: I actually think "$\forall x \in O, P(X=x | BA) = P(X=x | B)$" is sufficient. In your example about the possibly bias coin, you secretly slipped in "and the prior was $50\%$". This means you have some probability distribution over the probability $p$ that a given flip lands on heads! For example, if instead you knew that $p = .5$, then knowing whether the first 8 coin flips contained 4 heads is completely irrelevant to knowing whether the 9th coin flip is heads. Similarly if you knew that $p=.7$. The point is that there is some uncertainty as to the value of $p$. And this uncertainty makes it so that $P(X=x | A) = P(X=x)$ is actually false! 
Consider, for example, the case of the probability distribution over $p$ being $\frac{1}{3}$ that $p=.1$ and $\frac{2}{3}$ that $p=.5$. Then, for $x = \text{heads}$, $P(X=x) = \frac{1}{3}.1+\frac{2}{3}.5 = .3666$, while $P(X=x | A) = \frac{P(X=x \& A)}{P(A)} = \frac{\frac{1}{3}{8 \choose 4}.1^4(1-.1)^4.1+\frac{2}{3}{8 \choose 4}.5^4(1-.5)^4.5}{\frac{1}{3}{8 \choose 4}.1^4(1-.1)^4+\frac{2}{3}{8 \choose 4}.5^4(1-.5)^4} = .4966687$, which is intuitive since it's much more likely that $p=.5$ once we know that $4$ out of the first $8$ flips were heads. 
The moral is that if $A$ actually did "balance out" observations, then $A$ wouldn't have relevant information about $X$, since, as my/your example showed, relevant information is not balanced out!
A: The condition you have proposed is correct and it is the counterexample that is wrong. I'm not sure which of the following cases you had on your mind so I'll provide argumentation for each of them:


*

*$p=const,\ \xi_j\overset{iid}{\sim} Be(p),\ X=\xi_9, B=\sigma(\emptyset), A=\sigma(\sum^8_{j=1}\xi_j=4)$. Then $X\bot A, B$ and indeed observation contained in $A$ is irrelevant.

*$P\sim\theta\delta_{1/2} + (1-\theta)\nu,\ \xi_j|P\overset{ciid}\sim Be(P),\ X=\xi_9, B=\sigma(\emptyset), A=\sigma(\sum^8_{j=1}\xi_j=4, P=1/2)$ and $\nu$ being some probability distribution on $[0,1]\backslash\{1/2\}$ with expected value $\mu$. Then 


$$P|A \sim \delta_{1/2}1_{\{P=1/2\}} + \nu1_{\{P\neq1/2\}}$$
$$X|A \sim Be(\frac121_{\{P=1/2\}} + \mu1_{\{P\neq1/2\}})$$
which is different from
$$X\sim Be(\Bbb{E}P) \sim Be(\theta/2+(1-\theta)\mu)$$
unless expected value of distribution $\mu$ is $1/2$. So observation contained in $A$ is relevant, unless $\mu=1/2$.


*Same as above but now $A=\sigma(\sum^8_{j=1}\xi_j=4)$ and denote distribution of $P$ as $\mu$.
$$\Bbb{P}(P\in dp | \sum^8_{j=1}\xi_j=4) = \frac{p^4(1-p)^4\mu(dp)}{\int p^4(1-p)^4\mu(dp)}$$ and conversely
$$\Bbb{P}(P\in dp | \sum^8_{j=1}\xi_j\neq4) = \frac{[1-p^4(1-p)^4]\mu(dp)}{\int [1-p^4(1-p)^4]\mu(dp)}$$ which gives $P|A$. Heuristically after observing $\sum^8_{j=1}\xi_j=4$ the distribution of $P$ is tightened around $1/2.$ This gives you $X|A$ which is in general different from bare $X$ (for example it is the same when $\mu$ is symmetrical around $1/2$), hance again observation contained in $A$ is relevant (in general).

