A die versus a quantum experiment Let suppose you roll a die, and it falls into a hidden place, for example under furniture.
Then although the experiment has already been made (the die already has a number to show), that value can not be known, so the experiment was not fully realized.
Then till you see the die's top side, the probability remain p = 1/6.
I see no difference between this and the wave function collapse, at least as an analogy.
Can someone explain a deeper difference?  
Thanks
 A: You're absolutely right: the probabilities predicted by quantum mechanics are conceptually fully analogous to probabilities predicted by classical statistical mechanics, or statistical mechanics with a somewhat undetermined initial state - just like your metaphor with dice indicates. In particular, the predicted probability is a "state of our knowledge" about the system and no object has to "collapse" in any physical way in order to explain the measurements.
There are two main differences between the classical and quantum probabilities which are related to one another:


*

*In classical physics - i.e. in the case of dice assuming that it follows classical mechanics - one may imagine that the dice already has a particular value before we look. This assumption isn't useful to predict anything but we may assume that the "sharp reality" exists prior to and independently of any observations. In quantum mechanics, one is not allowed to assume this "realism". Assuming it inevitably leads to wrong predictions.

*The quantum probabilities are calculated as $|c|^2$ where $c$ are complex numbers, the so-called probability amplitudes, which may interfere with other contributions to these amplitudes. So the probabilities of outcomes, whenever some histories may overlap, are not given as the sum over probabilities but the squared absolute value of the sum of the complex probability amplitudes: in quantum mechanics, we first sum the complex numbers, and then we square the result to get the total probability. On the other hand, there is no interference in classical physics; in classical physics, we would surely calculate the probabilities of individual histories, by any tools, and then we would sum the probabilities.
Of course, there is a whole tower of differences related to the fact that the observable (quantities) in quantum mechanics are given by operators that don't commute with each other: this leads to new logical relationships between statements and their probabilities that would be impossible in classical physics.
A closely related question to yours:

Why is quantum entanglement considered to be an active link between particles?

The reason why people often misunderstand the analogy between the odds for dice and the quantum wave function is that they imagine that the wave function is a classical wave that may be measured in a single repetition of the situation. In reality, the quantum wave function is not a classical wave and it cannot be measured in a single case of the situation, not even in principle: we may only measure the values of the quantities that the wave function describes, and the result is inevitably random and dictated by a probability distribution extracted from the wave function.
A: In QM one can also deal with probabilities instead of probability amplitudes when one describes the prepared beam in terms of density matrix. For example (I assume there is six states $|n>$):
$\rho = (1/6)\sum_n {|n><n|}$.     (1)
Mixture is generally different from a pure (superposition) state with the same probabilities:
$\psi = (1/6)^{1/2}\sum_n {|n>}$.   (2)
The dice experiment corresponds to (1). It would be nice if you could read something about difference between mixtures and superpositions in QM (see sources of coherent and non coherent light, for example).
A: The other answers provide an account of quantum theory as a source of the answer. But I would like to consider the question from a more general perspective: that of the meaning of "probability" especially in the classical context.
Specifically what is meant by "probability" of the dice value in this question and where has the value 1/6 come from? This may seem like a trivial point, but there are (at least) three different perspectives on the answer:
(1) Frequentist. This is the view that probability is a number determined empirically by a series of tests. So this view says that such tests have been conducted on the dice and in 1/6 of cases (subject to some statistical factor) a given number was returned.
(2) Absolutist. This is the view (although I am not sure anyone will promote it) that the probability (with its value here of 1/6) is a derived physical property of the dice (like temperature, say). As a physical property it has been determined by some detailed analysis. One possible such analysis might be on the phase space of all possible motions, which has been appropriately partitioned and it is found that 1/6 of the volume is associated with each resultant value. This analysis might be done for that dice, or for some idealised dice. So the absolutist would claim that physics determines (classical) probabilities, and would view the frequentist data as simply experimental (dis-)confirmation of a specific model.
(3) Relativist. This is really the Bayesian viewpoint as espoused by Jaynes. This is using conditional probability $P(A|C)$ to define a probability in a relative way, with Bayes Theorem - which also allows us to construct $P(C|A)$ - this allows discussion of Inference and Prior Probabilities. The Bayesian view (especially as promoted by Jaynes) would be that there is no absolute probability, but only levels of knowledge, "observer inferencing" and (changing) prior assumptions. So the 1/6 value would be determined subjectively, as a measure of your "expectation" for example. It would have nothing to do with any intrinsic physics of the dice.
So to return to Quantum Mechanics. It too deals with probabilities, which are calculated from $|\Psi(x)*\Psi(x)|$. So what do these probabilities mean now that we have done some analysis?
Jaynes liked to promote his Bayesian view (ie 3 above) that he claimed reconciled Bohr with Einstein, thus interpreting $\Psi$ as somewhat epistemological. Phrases I have seen in other Stack answers like "the $\Psi$ collapse is all in your head" when discussing $\Psi$ might suggest that some physicists share an aspect of this view.
However an alternative view might be that $\Psi$ really solves the Absolutists problem: namely it provides a well validated physical basis for calculating probabilities in an absolute sense.
If this were true then the only physical way to get to 1/6 in the dice would be to do some giant quantum calculation from its atoms!
A: I was going to make this a comment but here it comes, since maybe it is a simpler formulation of a fundamental difference than all the rest of the answers:
The difference is that the value of the classical die under the couch can be seen/measured by many people.They will not tell you, and you still will not know,  but  the value that was  thrown will not change. In quantum mechanics each observation/measurement  changes the system, from one measurement to the other . The die's throw will not change  from one look of a hidden to you observer,  to the next.
A: The deeper difference between the classical and quantum dice is this - quantum dice obey the superposition principle. Consequently you can create entangled states of the form:
$$|\Psi \rangle = \frac{1}{\sqrt{2}}(|x \rangle_{D1} |y \rangle_{D2} \pm |y \rangle_{D1} |x \rangle_{D2} ) $$
where the subscripts denote the particular die (1 or 2), $|x\rangle_{D}$ and $|y\rangle_{D}$ are two different possible states of the die. The simplest case is when the role of the "die" is played by qubits. Then the above expression becomes:
$$|\Psi \rangle = \frac{1}{\sqrt{2}}(|\uparrow \rangle_{D1} |\downarrow \rangle_{D2} \pm |\downarrow \rangle_{D1} |\uparrow \rangle_{D2} ) $$
Entanglement appears to have no classical analog and it actually plays the role of a usable information resource in quantum computation.
A: A somewhat similar (yet inverted) question was posed, infamously, by Einstein, Podolsky, and Rosen as an argument against the same phenomenon you are challenging.
The basic argument was that such quantum wave collapses are indistinguishable from dice falling under couches, but under opposite grounds as yours -- both must be "determined" by what were later named "hidden variables".  That a quantum wave collapse, like a falling dice, is completely deterministic by the nature of its initial conditions, and its apparent randomness is due to our current inability to measure these initial conditions.
It was John S. Bell who thought up of an experiment to test this.
The basic idea was statistical, and involved quantum entanglement -- if hidden variables existed (and the particles already having a determined, yet unknown state at their inception), statistics should yield a certain correlation between entangled particles.  If there are no hidden variables, then this correlation would not show up.
Surprisingly, this correlation did not manifest by even a large margin of experimental error.  This proves an inherent difference in the randomness of quantum wave collapse as opposed to simply not knowing the landing of a dice.
The result culminated in Bell's theorem.
