What is the probability that 3 points randomly plotted on the boundary of an equilateral triangle, contain the center of said triangle? I'm aware of the same problem, but with a circle instead. In that case $\text{probability}=\frac{1}{4}$,
but I didn’t know where to start from with this question.
With circles, we didn't account for the initial point, because it was a circle. Is it the same here?
Also, in both cases, we can visualize where the third point should be if we have the first two. That is, by drawing two lines connecting each of the first $2$ points with the center. To contain the center, the third point should be on the part of the boundary that cut by the 2 lines.
 With circles, however, we could easily calculate the expected perimeter bounded between the 2 initial points (smaller one), and divided by the perimeter of the circle. However, I don't know how to approach this problem with triangles. 
Any help would be greatly appreciated.
By the way, I ran this on python.
For $10^7$ runs, the answer was $0.182$.
 A: We split the problem into three cases.
Case 1: All three points lie on the same side. This occurs with probability $\frac19$. In this case the probability of the centre lying in the triangle is $0$.
Case 2: All three points lie on different sides. This occurs with probability $\frac29$.
Let our equilateral triangle be $ABC$, with side length $1$ and centre $O$, and let $X,Y,Z$ be uniformly chosen at random on $AB,BC,CA$ respectively. We may assume that $X$ lies closer to $A$ than to $B$.
Lemma: Let $XO$ meet $BC$ at $P$. Then $BP=\frac{BX}{3BX-1}$.
Proof: Let $M,N$ be points on $AB,BC$ respectively such that $BM=BN=\frac13$. Then $BMON$ is a rhombus of side length $\frac13$, so
\begin{align*}
\triangle XMO\sim\triangle ONP&\implies XM\cdot NP=OM\cdot ON=\frac19\\
&\implies (3BX-1)(3BP-1)=1\\
&\implies BP=\frac{BX}{3BX-1}.\qquad\square
\end{align*}
Note that $O$ lies in $\triangle BXY$ if and only if $Y$ lies on $CP$, which happens with probability
\begin{align*}
&\quad\int_{1/2}^1\!\left(1-\frac x{3x-1}\right)\,dx\\
&=\frac13\left[2x-\frac13\ln(3x-1)\right]_{1/2}^1\\
&=\frac13-\frac29\ln2.
\end{align*}
Now $O$ must lie in one of $\triangle BXY$, $\triangle CYZ$, $\triangle AZX$ or $\triangle XYZ$ (with probability $1$). But the first three cases have the same probability given above, so $O$ lies in $\triangle XYZ$ with probability
$$1-3\left(\frac13-\frac29\ln2\right)=\frac23\ln2.$$
Case 3: Two points lie on one side, and the third point lies on another side. This happens with probability $\frac23$.
Let $A,B,C,O$ be as before, and let $X\in AB$ and $Y,Z\in BC$ be chosen uniformly.
If $X$ is closer to $B$ than to $A$, then the segment $BC$ lies entirely on one side of the line $XO$, so $\triangle XYZ$ cannot contain $O$.
If $X$ is closer to $A$ than to $B$, then let $XO$ meet $BC$ at $P$ as before. Then $\triangle XYZ$ contains $O$ if and only if $Y,Z$ lie on different sides of $P$. By the Lemma above, this happens with probability
\begin{align*}
&\quad2\int_{1/2}^1\!\frac x{3x-1}\left(1-\frac x{3x-1}\right)\,dx\\
&=\frac29\int_{1/2}^1\!\left(2+\frac1{3x-1}-\frac1{(3x-1)^2}\right)\,dx\\
&=\frac29\left[2x+\frac13\ln(3x-1)+\frac13\frac1{3x-1}\right]_{1/2}^1\\
&=\frac19+\frac4{27}\ln2.
\end{align*}
Combining the results above, the total probability that $\triangle XYZ$ contains $O$ is equal to
\begin{align*}
&\quad\frac29\left(\frac23\ln2\right)+\frac23\left(\frac19+\frac4{27}\ln2\right)\\
&=\boxed{\frac2{27}+\frac{20}{81}\ln2}\approx0.245222.
\end{align*}
A: Thinking more about Case 2 of my previous answer, I now have a more general approach that works for any starting shape (in fact, any measure on the plane).
Let $\triangle ABC$ be equilateral with side length $1$ and centre $O$. Suppose $X,Y,Z$ are our randomly chosen points. Note that:


*

*If $O$ lies in $\triangle XYZ$ then $Y,Z$ lie on different sides of line $OX$, and similarly for $OY$ and $OZ$. In this case we call $OX,OY,OZ$ separating lines.

*If $O$ lies outside of $\triangle XYZ$, then exactly one of $OX,OY,OZ$ is a separating line.


Assume now that $X$ lies on side $AB$, closer to $A$ than to $B$. We need the same lemma as before:

Lemma: Let $OX$ meet $BC$ at $P$. Then $BP=\frac{BX}{3BX-1}$.
Proof: Let $M,N$ be points on $AB,BC$ respectively such that $BM=BN=\frac13$. Then $BMON$ is a rhombus of side length $\frac13$, so
\begin{align*} \triangle XMO\sim\triangle ONP&\implies XM\cdot NP=OM\cdot ON=\frac19\\&\implies (3BX-1)(3BP-1)=1\\&\implies BP=\frac{BX}{3BX-1}.\qquad\square\end{align*}

Write $BX=x$. Then the line $OX$ divides the perimeter of $\triangle ABC$ into two parts, with lengths $x+\frac x{3x-1}$ and $3-(x+\frac x{3x-1})$ respectively. Hence the probability that $OX$ is a non-separating line is
\begin{align*}
&\quad2\int_{1/2}^1\!\frac19\left(x+\frac x{3x-1}\right)^2+\frac19\left(3-x-\frac x{3x-1}\right)^2\,dx\\
&=2\int_{1/2}^1\!\frac{x^4+(x^2-3x+1)^2}{(3x-1)^2}\,dx\\
&=\cdots\\
&=\frac{50}{81}-\frac{40}{243}\ln2.
\end{align*}
(The $\frac19$ in the first line comes from the fact that the perimeter of $\triangle ABC$ is $3$.) Hence the expected number of non-separating lines is $\frac{50}{27}-\frac{40}{81}\ln2$. But this is equal to
$$2\cdot\mathbb P(O\not\in\triangle XYZ)+0\cdot\mathbb P(O\in\triangle XYZ),$$
so
\begin{align*}
\mathbb P(O\in\triangle XYZ)&=1-\frac12\left(\frac{50}{27}-\frac{40}{81}\ln2\right)\\
&=\boxed{\frac2{27}+\frac{20}{81}\ln2}\approx0.245222.
\end{align*}
