As this question explains in detail, the construction of a MOSFET is symmetrical w.r.t. to the drain and source, except for the decision to connect the substrate to one of them (the reasons for doing this are described in the question). As soon as you do this, it breaks the symmetry, and that pin becomes the source, and the other pin the drain. This is done internally on all discrete 3-pin MOSFETs.
This picture is from the above linked question, before the source is connected to the p+ substrate (body):
In the case of an enhancement-mode N-channel MOSFET, when the gate-source voltage is 0V, the MOSFET is OFF (no current from drain to source). Then the gate-source rises (e.g. 3V), the MOSFET begins to turn ON.
My question is, in the case of 4-pin MOSFETs, what gate-to-??? voltage is relevant for turning the MOSFET ON, given the source is no different from the drain?
4-pin MOSFETs are not common when buying discrete MOSFETs, but in CMOS design it's common to not connect the substrate to the source to build things like analogue switches or transmission gates. In the case of an N-channel CMOS analogue switch, the substrate is normally connected to ground, and both the drain and source may be at arbitrarily higher voltages. This has the neat behaviour of being able to block signals in both directions.