Most likely velocity vector of ideal gas? 
$$\left( \text{probability of a molecule having velocity }\vec{v}\right) \propto e^{-mv^{2}/2kT}$$
So the most likely velocity vector for a molecule in an ideal
gas is zero. Given what we know about Boltzmann factors, this result should hardly be surprising.

Schroeder's Thermal Physics, page-232.
First of all I don't understand how is that true. I can see that average velocity should be zero. But why should most likely velocity vector should be zero? 
And if that is true then, after that we derive the speed distribution. And it goes to zero for $v=0$. So that means that we won't find any molecule with zero speed (at rest in container's frame).
But that also means that there is no molecule with zero velocity! 
SO then how could zero velocity be the most likely velocity if there is no molecule with zero velocity??
 A: The distribution is
$$
w(v_x,v_y, v_z)=Ae^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2k_BT}},
$$
where A is the normalization constant that can be easily found by integration (see Gaussian integral):
$$
\int dv_xdv_ydv_z w(v_x,v_y, v_z) = 1.
$$
The average velocity along any direction is zero, e.g.,
$$
\langle v_x\rangle = \int dv_xdv_ydv_z v_x w(v_x,v_y, v_z)=0
$$
This can be shown either by direct integration by parts or from the symmetry reasons: the odd function is integrated in symmetric limits (from $-\infty$ to $\infty$).
On the other hand, the speed, i.e., the magnitude of this vector, is not zero:
$$
v=\sqrt{v_x^2+v_y^2+v_z^2},
\langle v\rangle =\int dv_xdv_ydv_z v w(v_x,v_y, v_z)>0,
$$
since the integrand is positive everywhere. However, usually one prefers to calculate $\langle v^2\rangle$, since in this case integration is easier.

But why should most likely velocity vector should be zero?

The most likely velocity corresponds (by definition) to the peak of the distribution, which here is at $v_x=v_y=v_z=0$, so it is zero as well.
