Why is voltage properly siphoned off in this circuit?
Well, there's a problem with this instructable, in that voltage does not leach off to anywhere. Voltage does not posses the ability to leach. Only current does.

In fact, nothing is being leached, by its official definition.

The sensor acts like a variable resistor on its sensor pins. So by adding another resistor to ground you just allow a current to flow, from the supply through the sensor and through the resistor to ground. The point between the sensor and the resistor then become the centre point of a voltage divider, the exact voltage of which is determined by the resistance of the sensor. This resistance is in turn determined by the amount of contamination in the air/gas inside the sensor. 

Without the external resistor you would just measure 3.3V, as the current into the input pin is microscopic and would not cause a measurable voltage drop.



EDIT1: Add to that the fact that to make a full-range adjustability, according to the later section on the resistive ladder R-2R network on the other pins (homework: verify), the resistor would be much better chosen to be about double the value, let's say about 91kOhm. (Or stay at 47kOhm and the whole ladder divided by two, depending on whether you want higher sensitivity (91kOhm and same ladder) or deeper measurement (47kOhm and all ladder values halved)).