The wave-function for a single particle in a potential well of width $L$ is given by the relation $$\Psi_n(x)=\sqrt{\frac 2L} \sin(K_n x)$$ where $K_n$ is $(n+1)\pi/L$ and $n$ is a positive integer.
Quantum Mechanical Description of the Probability Distribution
The ground state wave function has $n=0$, so $K_n = \pi / L$ and so $\text{Re}(\Psi_0(x))$ looks like
The probability distribution for this wave function is $|\Psi_0|^2$, which is a $\sin^2$ function and on the same domain the graph is
Classical Mechanical Description of the Probability Distribution
Consider the same problem classically: a particle is in the box with some energy $E$. Classically it has some momentum and it moves around the box, bouncing back and forth off the walls. This particle spends the same amount of time at all points in the box as it is moving with constant velocity. So intuitively, the probability distribution should be constant, i.e. I am just as likely to find it near the wall as the center of the box. Similarly, the probability distribution looks like this:
Why are the two probability distributions so different? In the quantum case, the particle is more likely to be near the center and have a low chance of being near the walls. Classically, the probability should be uniform, so the quantum case is unintuitive. I know that the quantum mechanical description is correct at low energies, but is there an intuitive explanation for why the quantum probability distribution differs so strikingly from the classical one?