'Fermi-Dirac'-like occupation probability at high temperature Consider an ensemble of $N\to\infty$ free particles, each of which can assume energy states $E_i\in\{0,E\}$. Using the canonical ensemble one can compute the occupation probability for a single of those particles to be in the excited state $E_i=E$ (or equivalently the expectation value for what fraction of all particles is in the excited state). The result is:
$$n_T(E)=\frac{1}{e^{\frac{1}{k_B T}E}+1}$$
Now, if we check this expression in the limit $T\to 0$, we properly obtain $n_0(E)=0$, telling us that at low temperatures almost no particles will be in the excited energy state. But then, in the opposite limit $T\to\infty$ we get $n_\infty(E)=1/2$, so apparently at infinite temperature there will be equally many particles in the ground and the excited state! I kind of feel like all the particles should go into the excited state for $T\to\infty$, so that this goes against intuition. But maybe I am wrong? What should I expect to happen for $T\to\infty$?
 A: Perhaps the following (which basically involves investigating what happens for a general system with discrete energy spectrum) will help.
The canonical partition function for a quantum system with discrete spectrum $\{E_n\}$ is
\begin{align}
  Z = \sum_ne^{- E_n/(kT)}
\end{align}
and the population fraction of the systems in the ensemble with energy $E_n$ is given by
\begin{align}
  p_n = \frac{e^{-E_n/(kT)}}{Z}.
\end{align}
Now, consider two energy levels $E_n$ and $E_m$, then the ratio of their population fractions is
\begin{align}
  \frac{p_n}{p_m} = e^{-(E_n - E_m)/(kT)}
\end{align}
Now here's the key point.  As long as the difference $E_n-E_m$ is finite (which of course it will be for any two energies in the spectrum), the $T\to\infty$ limit of this expression always gives $1$!  The difference in the energies gets "washed out" by the largeness of $T$.  This tells us that at high temperature, any two levels in the spectrum will have an equal likelihood of being populated!
In particular, if the system has a finite-dimensional Hilbert space, say of dimension $N$, then the probabilities must add to $1$ and must all be equal in the high-temperature limit;
\begin{align}
  p_1+p_2+\cdots+p_N = 1, \qquad \text{$p_n = p_m$ for all $m,n\in\{1,\dots,N\}$}
\end{align}
which gives $p_n = 1/N$ for all $n\in\{1,\dots, N\}$.
Now, to complete your intuition, you simply need to understand "why" the partition function looks the way it does, but that's another story all together for which, admittedly, my own intuition isn't the greatest.
