On the number line, point $A$ is located at 0, point $B$ is located at 4, and point $C$ is located at 6. [asy]
defaultpen(1);
draw((0,0)--(6,0));
for (int i=0;i<7;++i){

draw((i,-.1)--(i,.1));

label(string(i),(i,-.1),(0,-1));
}
label("\(A\)",(0,0),(0,1));
label("\(B\)",(4,0),(0,1));
label("\(C\)",(6,0),(0,1));
[/asy] A dart randomly lands somewhere on the number line between $A$ and $C$.  What is the probability that it lands closer to $B$ than it does to $A$ or $C$?
First, suppose that the dart lands between $A$ and $B$.  It then has a probability of 1/2 of being closer to $A$ than to $B$, and it is always closer to $B$ than it is to $C$, so it is closer to $B$ than it is to $A$ and $C$ with probability 1/2.

On the other hand, if it lands between $B$ and $C$, then it is definitely closer to $B$ than to $A$, and it has a probability of 1/2 of being closer to $B$ than to $C$.  As before, it is closer to $B$ than it is to $A$ and $C$ with probability ${1/2}$.

[asy]
defaultpen(.7);

draw((0,0)--(6,0));
for(int i=0;i<=6;++i){

draw((i,-.1)--(i,.1));

label(string(i),(i,-.1),(0,-1));
}

label("\(A\)",(0,0),(0,1));
label("\(B\)",(4,0),(0,1));
label("\(C\)",(6,0),(0,1));

draw((2,0)--(5,0),linewidth(3.5));
[/asy] In either case, the probability that the dart lands closest to $B$ is 1/2, so the probability overall is $\boxed{\frac{1}{2}}$.