A point $P$ is chosen at random in the interior of equilateral triangle $ABC$. What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP$?
Since the three triangles $ABP$, $ACP$, and $BCP$ have equal bases, their areas are proportional to the lengths of their altitudes.

Let $O$ be the centroid of $\triangle ABC$, and draw medians $\overline{AOE}$ and $\overline{BOD}$. Any point above $\overline{BOD}$ will be farther from $\overline{AB}$ than from $\overline{BC},$ and any point above $\overline{AOE}$ will be farther from $\overline{AB}$  than from $\overline{AC}.$ Therefore the condition of the problem is met if and only if $P$ is inside quadrilateral $CDOE.$

[asy]
pair A,B,C,D,I,F,O;
A=(0,0);
B=(10,0);
C=(5,8.7);
D=(2.5,4.3);
I=(7.5,4.3);
F=(5,0);
O=(5,2.3);
draw(A--B--C--cycle,linewidth(0.7));
draw(A--I,linewidth(0.7));
draw(B--D,linewidth(0.7));
draw(C--F,dashed);
label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,N);
label("$D$",D,NW);
label("$E$",I,NE);
label("$F$",F,S);
label("$O$",O,SW);
[/asy]


If $\overline{CO}$ is extended to $F$ on $\overline{AB}$, then $\triangle ABC$ is divided into six congruent triangles, of which two comprise quadrilateral $CDOE$. Thus $CDOE$ has one-third the area of $\triangle ABC,$ so the required probability is $\boxed{\frac{1}{3}}$.