probablity of random pick up three points inside a regular triangle which form a triangle and contain the center what is  the probablity of random pick up three points inside a regular triangle 
which form a triangle and contain the center of the regualr triangle
the three points are randomly picked within the regular triangle and then form a new triangle
and the new triangle have to contain the center of the original regular triangle
what is the probability

Let $R$ be a convex region of the plane with unit area, and choose $n\ge 3$ points at random from the region.  We will derive a general expression for the probability $P_{n}$ that the convex hull of these points contains a particular point $\tau \in R$ (which we will take as the origin of our coordinate system). We can then evaluate that expression for the particular question posed (in which $n=3$, $R$ is an equilateral triangle, and $\tau$ is its center).
Suppose the hull does not contain $\tau$, and assume that $\tau$ is not collinear with any two of the points (this is true with probability 1).  Then we can draw a ray from the center point through one of the $n$ points (the "leftmost" point) such that the remaining $n-1$ points are to the right of the ray.  (Note that this condition is both necessary and sufficient.  Note also that if the convex hull does contain $\tau$, then there is no "leftmost" point.)  We can calculate the desired probability by integrating over all such configurations.  In particular, take $f(\varphi)$ to be the area of the subregion that lies to the right of the radial ray at angle $\varphi$, and take $r(\varphi)$ to be the distance from $\tau$ to the boundary of $R$ along that ray.  Then the probability that a particular point will lie in the angular interval $[\varphi, \varphi+d\varphi]$ is $da = \frac{1}{2}r(\varphi)^2 d\varphi$, and the probability that each of the other $n-1$ points will lie to its right is $f(\varphi)$.  Since there are $n$ points to choose as the leftmost, the overall probability of not capturing $\tau$ in the convex hull is
$$
1-P_{n} = n \int f(\varphi)^{n-1} da = n \int_{-\pi}^{\pi} \frac{1}{2} f(\varphi)^{n-1} r(\varphi)^2 d\varphi,
$$
We can make a few observations at this point.  First, if the region $R$ is symmetric under reflection through $\tau$, then $f(\varphi)$ is identically $1/2$: each radial line splits the region in half.  The result then is
$$
P_{n} = 1 - \frac{n}{2^{n-1}},
$$
which gives the known result $P_{3} = 1/4$ for polygons with an even number of sides.  Second, if the region $R$ instead has bilateral symmetry across the $x$-axis, then $r(\varphi)$ is an even function, and $f(\varphi) = 1/2 + f^{-}(\varphi)$, where $f^{-}$ is an odd function.  Then terms in the integral with odd powers of $f^{-}$ must vanish: in particular, we have
$$
P_{3} = \frac{1}{4} - 3 \int{{f^{-}(\varphi)}^2 da} = \frac{1}{4} - 3\int_{-\pi}^{\pi}\frac{1}{2}{f^{-}(\varphi)}^2 r(\varphi)^2 d\phi \le \frac{1}{4},
$$
and interestingly the relation $P_{4} = 2P_{3}$ continues to hold.

It remains to evaluate $f^{-}(\varphi)$, $r(\varphi)$, and the resulting integral in the case of an equilateral triangle.  The figure above shows an equilateral triangle with vertices at $(-x,0)$ and $(x/2, \pm x\sqrt{3}/2)$.  For $\varphi \in [0,\pi/3]$, the function $f^{-}(\varphi)$ is the area of the blue triangle minus the area of the red triangle,
$$
f^{-}(\varphi) = \frac{x^2}{8}\tan\varphi - \frac{x^2}{2}\frac{1}{\cot\varphi + \cot\frac{\pi}{6}} = \frac{x^2}{8}\tan\varphi \left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right);
$$
and the radius $r(\varphi) = \frac{1}{2}x\sec\varphi$ over the same domain. By symmetry, the full integral is just six times its value over $[0,\pi/3]$:
$$
\begin{eqnarray}
P_3 &=& \frac{1}{4} - 18\int_{0}^{\pi/3}\frac{x^6}{512}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&=& \frac{1}{4} - \frac{1}{36\sqrt{3}}\int_{0}^{\pi/3}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&=& \frac{1}{4} - \frac{1}{12}\int_{0}^{1} u^2\left(1 - \frac{4}{1+3u}\right)^2 du,
\end{eqnarray}
$$
where we have used $x = 2/\sqrt{3\sqrt{3}}$ (in order for the triangle to have unit area) and introduced the transformed variable $u = \tan\varphi / \sqrt{3}$ (so $du = (\sec^2\varphi / \sqrt{3}) d\varphi$).  The final integral is a straightforward exercise for the reader; the result is
$$
P_3 = \frac{1}{4} - \frac{1}{324}\left(57 - 80\ln{2}\right) = \frac{2}{81}\left(3 + 10\ln{2}\right) = 0.2452215...,
$$
in agreement with the brute-force and numerical results already given.  This result can be generalized easily to the regular $m$-gon for any odd $m$ (changing only the values of some constants), and to the case where $n>3$ (complicating the final integral).
