My friend asked me about this, and I couldn't give him a good answer. Say you have some favorable event $G$. You know that knowing either of events $A$ or $B$ will more likely than not result in $G$. That is, $P(G|A) = P(G|B) = .51$.
Under what circumstances is $P(G|A,B) > .51$? That is, under what circumstances is knowing both $A$ and $B$ together, better for you?
It seems intuitively reasonable, but then again, I can also draw venn diagrams where $P(G|A,B) = 0$.
Assume $\xi$ has uniform distribution on $[0, 1]$. Let event $A$ be an indicator for $\xi \in [0, 0.7]$, $B$ = $\xi \in [0.3, 1]$, and $G$ be $\xi \in [0.2, 0.8]$
From simple geometric arguments it follows that $P(G | A) = 5/7$ and $P(G | B) = 5/7$, while $P(G|A, B) = 1$ (obviously if $\xi$ is both greater than 0.3 and lesser than 0.7, then it's definitely falls into $[0.2, 0.8]$
Now how can one arrive at that? Note that $P(G|A)$ can be written in terms of $P(G|A, B)$: $$ P(G|A) = P(G, B|A) + P(G, \neg B|A) = P(G|A, B) P(B | A) + P(G|A, \neg B) (1 - P(B | A)) $$
So $P(G|A)$ is just an average of $P(G|A, B)$ and $P(G|A, \neg B)$ with weight $P(B | A)$. By manipulating this value you can control the position of $P(G|A)$ on a range $$[\min(P(G|A, B), P(G|A, \neg B)), \max(P(G|A, B), P(G|A, \neg B)) ]$$
This suggests that unless the dependence between $A$ and $B$ is deterministic (that is, $P(B|A)$ is 0 or 1), $P(G| A, B)$ would be greater than $P(G|A)$ alone (assuming both $A$ and $B$ "work" in favor of $G$).
If this is right the result is unsurprising and just kind of messy. Also @Bey had the right idea probably. The reduction of measure going from $A$ to $AB$ has to be less than the reduction going from $GA$ to $GAB$, which you can get just looking at the main inequality.
Method 1:
Call $c_a = P(GAB)/P(GA)$. We can see that $0 \le c_a \le 1$ because $GAB \subset GA$. As long as $GAB \neq \emptyset$, and $GAB$ is a strict subset of $GA$, then we can say $0 < c_a < 1$.
Then assume that given this $c_a$, $AB$ is sufficiently smaller than $A$. In other words, $$ P(AB) < \frac{P(GAB)P(A)}{P(GA)} = c_aP(A). $$ Rearranging the above gives you the result: $$ P(G|A,B) = \frac{P(GAB)}{P(AB)} > \frac{P(GA)}{P(A)} = P(G|A). $$
Method 2:
Let $d_a = P(AB)/(A)$. As long as $AB$ isn't impossible and $AB$ is a strict subset of $A$, then $0 < d_a < 1$.
Then assume $GAB$ is only a little bit smaller than $GA$ given this $d_a$, or $$ P(GAB) > d_a P(GA). $$ Then we get the desired result as well. And all this stuff can be repeated for finding the conditions to gaurantee $P(G|AB) > P(G|B)$.
This was my friend's original intuition..that $A$ and $B$ have to tell you different things about $G$. I guess these are the ways we can get that in math-speak.
While it seems simple, you cannot get to your answer solely with what you have (your post indicates as much). Going back to the basic definition of conditional probability for non-null events:
$$P(G|A,B)=\frac{P(A \cap B \cap G)}{P(A \cap B)}$$
Knowing that $P(G|A)=P(G|B)=0.51$ means that the set of outcomes that result in $G$ when either $A$ or $B$ has occurred "uses up" more than half the area of the resulting Venn diagrams (for the intersections) in each case.
In order for knowing $A$ and $B$ to make $G$ even more likely, we need the set of outcomes that produce $G$ when $A$ and $B$ have happened to take up even more "area". This comes down to finding the probability of $G$ when we restrict our attention (shrink out universe) to the space where $A \cap B$ has occurred.
This may sound like just a long re-hash of your question, but it is the general case for your statement to be true. @Barmaley.exe gave a nice example for a specific scenario.
As you can see, in that case, the "space" of outcomes where $G$ occurred given $A \cap B$ was the entire intersection.
In other words, the reduction in probability for events $A$ and $B$ when going to the event $A \cap B$ must be greater than the reduction in probability from $A \cap G$ or $B \cap G$ to $A\cap B\cap G$.
Its all saying the same thing, $A\cap B\cap G$ must take up proportionately more of the "area" $A\cap B$.