You're comparing the wrong things - you must take the appropriate limits when using statistical physics (or statistical mechanics as you called it) to recover quantum and classical results. Also, recall that when we use statistical physics, we must consider the system in an ensemble formalism, i.e. the canonical ensemble. That is, we consider a (statistically) large number of identically prepared copies of our system: a single harmonic oscillator. Additionally, we must use the quantum version of statistical physics, where the ensemble is characterized by the density operator. For a nice intro to this, see Sakurai.
This article should answer your questions regarding the case of a single 1D harmonic oscillator potential viewed statistically. For a given temperature, we use the quantized energy of the oscillator to find the partition function in the canonical ensemble.
Consider the two limits:
1) Classcial: the thermal energy is much greater than the spacing of oscillator energies. Here, one recovers the classical result that the energy is $kT$.
2) Quantum: the thermal energy is much less than the spacing of oscillator energies. Here, one recovers the quantum result that for a given frequency the energy is $\frac{1}{2}\hbar \omega$
Further, using the appropriate density operator, you can recover the probabilities that you sought:
the density operator represented as a matrix in the basis $\{\lvert \phi_1 \rangle, \lvert \phi_2 \rangle \}$, for the system given by pure state $\Psi$ is
$$ \rho = \lvert \Psi \rangle \langle \Psi \lvert = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} = \frac{1}{2} \begin{pmatrix}1 & 1 \\\ 1 & 1\end{pmatrix}$$
To find the probability of being in state $i$, take,
$$\mathcal{P}_i = Tr(\rho P_i)$$
where $P_i$ is the projection operator of the $i^{th}$ state, and Tr means trace.