One has to be careful with the various "velocities" related to waves. One must be aware, what they mean and under which circumstances one can use these notions. Take e.g., "group velocity". A general solution for the time evolution of a wave equation often can be written as
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{k} \frac{1}{(2 \pi)^{3/2}} \tilde{\psi}_0(\vec{k}) \exp[-\mathrm{i} \omega(\vec{k}) t + \mathrm{i} \vec{k} \cdot \vec{x}].$$
The reason is that often the Laplace operator occurs in the wave equation, and the plane waves are a complete set of generalized orthonormal functions.
E.g., for the free Schröinger equation (with ##\hbar=1##),
$$\partial_t \psi(t,\vec{x})=-\frac{1}{2m} \Delta \psi(t,\vec{x}),$$
you get the dispersion relation
$$\omega(\vec{k})=\frac{1}{2m} \vec{k}^2.$$
To make sense of the group velocity you need the assumption that ##\tilde{\psi}_0(\vec{k})## is narrowly peaked around ##\vec{k}_0##. Then one can expand the exponent in powers of ##\vec{k}-\vec{k}_0##:
$$\mathrm{i} [\vec{k} \cdot \vec{x}-\omega(\vec{k})t] \simeq \mathrm{i} [\vec{k} \cdot \vec{x}-\omega_0 t - t (\vec{k}-\vec{k}_0) \cdot \vec{\nabla}_k \omega(\vec{k}_0)].$$
Then you get with ##\vec{v}_g=\vec{\nabla}_k \omega(\vec{k}_0)##
$$\psi(t,\vec{x}) \simeq \exp[-\mathrm{i} (\omega_0 -\vec{k}_0 \cdot \vec{v}_g) t+\mathrm{i} \vec{k}_0 \cdot \vec{x}] \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{k} \tilde{\psi}_0(\vec{k}) \exp[\mathrm{i} \vec{k} \cdot (\vec{x}-t \vec{v}_{\text{g}})],$$
i.e., finally
$$\psi(t,\vec{x}) \simeq \exp[-\mathrm{i} (\omega_0 -\vec{k}_0 \cdot \vec{v}_g) t+\mathrm{i} \vec{k}_0 \cdot \vec{x}] \psi_0(\vec{x}-\vec{v}_{\text{g}} t).$$
This means, in this approximation up to a phase the initial wave packet moves undeformed with group velocity ##\vec{v}_{\text{g}}##. The position-probability density is
$$|\psi(t,\vec{x})| \simeq |\psi_0(\vec{x}-\vec{v}_{\text{g}} t)|.$$
It is important to note that this only holds for sufficiently narrow-peaked ##\tilde{\psi}(\vec{k})##, i.e., for a particle with a quite well prepared momentum!
With the dispersion relation for the free-particle Schrödinger wave you get the expected result
$$\vec{v}_{\text{g}}=\frac{1}{m} \vec{k}_0,$$
i.e., the wave packet moves with this velocity, which is pretty intuitive from the classical particle picture.
One must, however be aware that this interpreation relies on the assumptions made, i.e., that you have a wave packet that is narrowly peaked in momentum space, such that dispersion can be neglected.
It's a good exercise to solve the initial-value problem for a Gaussian wave packet as the initial state exactly. Most conveniently one gives it in terms of the momentum representation,
$$\tilde{\psi}(\vec{k})=N \exp \left (-\frac{(\vec{k}-\vec{k}_0)^2}{4 \sigma_k^2} \right ),$$
where ##N## is an unimportant normalization factor and ##\sigma_k## the standard deviation of the momentum components. Here the time-evolution integral can be easily solved exactly since it's a Gaussian integral, and you can also easily compare it to the approximation involving the group velocity ##\vec{v}_{\text{g}}##.