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.
The sign $\lambda$ in de Broglie's equation,
$$\lambda=\frac h p,$$
is indeed the de Broglie wavelength of the object involved. It is the only wavelength one can meaningfully give a material particle, and "normal wavelength" is meaningless in that context. de Broglie's relation is also true for a photon (though it amounts to a calculation of a photon's momentum), where the wavelength is the usual wavelength of light.
Though this is a flawed picture, you can see de Broglie's relation as describing a "wave of probability" associated with the material particle. This wave must have total "weight" equal to $1$, which means that the particle must be somewhere. (The relative "weights" of the wave in different volumes give the relative probabilities of us finding the particle there.) Particles with higher mass have smaller de Broglie wavelengths, so they can be localized better. If they are located in smaller volumes their amplitudes (specifically: their probability densities) must be bigger, to keep the total "weight" equal to $1$.
The amplitude of this "wave of probability" is of course not related to the energy - that has more to do with its momentum and therefore with its wavelength.
Best Answer
@Coopercape is almost right, but it all works so it that de Broglie's relation with $E$ dependent on $f$ and $p$ on $\lambda$ is still right. The $m$ in the equation is in fact the rest mass, so that $f\lambda$ is not $c$ for matter waves, or any waves other than those from massless particles. I'll explain a bit more below.
The relationship $E^2 = (pc)^2 + (mc^2)^2$, which is fully valid in special relativity gives the energy as the sum of a kinetic energy ($=pc$) and the rest mass energy ($mc^2$ with $m$ being the particle's rest mass). Notice that in relativity (actually both special and general) the rest mass term includes the particle's potential energy due to internal forces.
You can write the equation in terms of $f$ and wavelength $\lambda$ as $$(hf)^2 = (hc/\lambda)^2 + (mc^2)^2$$ with $E = hf$ and $p= h/\lambda$.
This actually holds for all particles and systems, in special relativity. In general relativity you have to insert the other off-diagonal metric terms and it's a little more complex but still straightforward.
Notice that f is no longer equal to $c/\lambda$, unless the rest mass $m$ is zero. The relationship of $f$ and $\lambda$ depends on $m$, i.e., the mass of the particle. Only for $m=0$ is $f\lambda=c$. The relationship between f and $\lambda$ in general is called the dispersion relation. With $\omega = 2\pi f$, and k = $2\pi$/$\lambda$ one gets a simple relation if one sets natural units with $2\pi h$ = c = 1, $$\omega^2 = k^2 + m^2$$ with $m$ still the rest mass. This is called the wave's dispersion relations.
It is well understood in physics.
See the Wikipedia article in the section Matter Waves. https://en.m.wikipedia.org/wiki/Energy_momentum_relation
See also about dispersion relations in general at https://en.m.wikipedia.org/wiki/Dispersion_relation