Sorry for not explaining myself Chris. I assumed you hadn't looked closely at the circuit and thought it was the usual arrangement using pin 7.
The pin 2 and pin 6 threshold voltages are symmetrical around 1/2 VCC and pin 3 swings (nominally - ideally) between VCC and GND, so the duty cycle is roughly 50%. Any error is due to pin 3 not pulling up as well as it pulls down.
How much of a problem this is depends on the type of output stage, and the loading on the output. In the original bipolar design, the output is pulled low by an NPN with its emitter grounded, but it's pulled high by an NPN emitter follower arrangement. So it will pull down very close to 0V (assuming minimal load on the output) but even with only 1 mA load, it will only pull up to about 1.4V less than VCC.
So the output doesn't swing symmetrically around 1/2 VCC. That's why the duty cycle isn't exactly 50%. The lower the supply voltage, the more the asymmetry (as a proportion of the supply voltage) and the worse the duty cycle error.
Other types of 555 probably have better output stages that will swing closer to the positive rail. Also, for the bipolar version, if the output is loaded heavily, the positive and negative dropout voltages will become closer to each other, because the emitter follower pulls high more strongly than the common emitter (apart from the inherent voltage drop in the emitter follower).
So in summary, if the output stage was perfect the duty cycle would be exactly 50%. With an old bipolar 555 the output will not be exactly 50% and the error will be worse at lower supply voltages.