Actually, according to quantum mechanics, for every particle (matter), there is a complex valued wave function and it is the modulus of that wave function that is the probability that the particle is at a given location.
For example, a wavefunction for a uniform plane wave of matter could be:
$$\psi(\vec x, t) = A e^{i(\vec k \cdot \vec x + \omega t)} = A(cos(\vec k \cdot \vec x + \omega t) + i\cdot sin(\vec k \cdot \vec x + \omega t))$$
This wavefunction is a complex valued wave (notice the sine and cosine functions) that has a wavelength ($1/|\vec k|$ = the de Broglie wavelength) and a frequency ($\omega$). It is this complex wave that could result in quantum interference effects.
However the probability for a particle to be at a position $\vec x, t $ is:
$$P(\vec x,t) = |\psi(\vec x, t)|^2 = |A|^2 $$
So it is constant everywhere in space and time and is not a wave. Of course, an actual wavefunction for a particle would be something like this plane wave multiplied by a gaussian to result in a wave packet that is localized in space and time and that travels the way a classical particle would travel (see wave packet for more information).
But the key point is that there is a complex valued wavefunction with wave-like properties but that the probability does not have to have any wave like properties.
EDIT for edited question:
For a particle localized in space, there is still an average position and an average momentum so the average de Broglie wavelength will be the one that corresponds to that average momentum. According to Heisenberg there will be an uncertainty in position and an uncertainty in momentum and thus there will be an uncertainty in the de Broglie wavelength also.
What this really means is that whenever you have a localized particle, you have a wave packet which means there is a superposition of a number of plane waves with similar momentum, in other words a superposition of a number of different de Broglie wavelengths (see wave packet for more information).
If you are wondering if the concept of de Broglie wavelengths is useful in this situation, then the answer is that it is not very useful - just work with your mathematically defined complex valued quantum mechanical wavefunction instead.
Yes the product $\nu \lambda$ makes sense as a velocity. Defining $E = \hslash \omega$ and $p=\hslash k$ (the Planck constant $h=2\pi \hslash$, where the $2\pi$ is injected into the $\hslash$, since physicists usually prefer to discuss the angular frequency $\omega=2\pi\nu$ and the wave vector $k=2\pi p$ rather than the frequency $\nu$ and the momentum $p$ for Fourier-transform notational convenience), you end up with
$$\nu \lambda = \frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$$
which is the standard definition for the phase velocity. It corresponds to the velocity of the wave component at frequency $\nu$ propagating over distance $\lambda$ per unit time.
This is always true, but for more complicated situation, a system is represented by a superposition of different waves propagating at different phase velocity. It results a wave-packet propagating, as an ensemble, at the group velocity
$$v_{g}=\frac{\partial \omega \left(k\right)}{\partial k}$$
where the dispersion relation of the wave-packet is noted $\omega \left(k\right)$.
Only the group velocity has some physical clear interpretations. For instance, the phase velocity can be larger than the speed of light, but the group velocity can never been larger than the speed of light $v_{g}\leq c$, at least in vacuum.
For a photon in free space for instance, $\omega \left(k\right)=c k$ and thus it's group velocity is $c$. For a free non-relativistic particle of mass $m$, $\omega = \hslash k^2/2m$, and $v_{g}=\hslash k/m$. etc...
For a human body, you have to count all the atoms constituting the body. The individual frequency of one atom interferes with the frequencies of all the others, resulting in an almost flat dispersion relation. The group velocity is then ridiculously small. A human body does not move due to quantum effect ! You can get a basic idea of the group velocity of a human body supposing you are a free particle of mass $100 \textrm{kg}$ and wavelength $1 \textrm{m}^{-1}$ (i.e. the order of magnitude of your size is about $1 \textrm{m}$) the $\hslash$ kills $v_{g} \sim \left(10^{-36}-10^{-34}\right) \textrm{m.s}^{-1}$ !
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I think it would help you to study the theory of Fourier transforms! Then this momentum/position duality becomes much more apparent.
A wavepacket is, like you write, a sum of many momentum states (not probability densities). If you look in the momentum space, the wider the spread of the momentum states, the smaller the spread of the position states will be. This is quite obvious from the Fourier theory which is why I recommend studying this. The Heisenberg uncertainty relation of position/momentum is related to this duality - when you make the momentum spread thinner, the position spread will increase and vice versa so there is a non-zero minimum of uncertainty (width of the distribution) where you have made both position and momentum as localized as you can.
The de Broglie relations simply relate momentum p to wavelength lambda - actually not more interesting than the simple observation that a sinusodial wave has a frequency and a wavelength. Short wavelength means higher momentum. So if you have a state with an uncertainty in momentum you also have an uncertainty in the de Broglie wavelength. If you want to take an average of it, go ahead as long as you know it's an average of a distribution of a certain width. This will be OK for many applications, but for some the detailed spread will be crucial also of course.
Addendum: Also please note that the position and momentum descriptions are a duality. You cannot specify both, the full information of the state is in either one of them and then you can Fourier-transform between them to extract a better understanding of the problem or extract some numerical predictions etc. This point is lost in some introductions so I'll take the opportunity here to mention it :)