Question: Given a simple circuit of a draw around 28mA (when supplied 3.3v). I'm trying to maximize run time on a battery, but working with what I've got on hand... I have the following batteries:

* 5000mAh 3S LiPO (12.6V fully charged)
* 1700mAh 2S LiFE (7.2V)
* 600mAh 1S LiPO (4.2V)

I have the following regulators:

* MCP1700 3.3V LDO
* LM2931 5.0V LDO

The MCP1700 Can't handle the larger batteries. So I see my options are:

* Use the tiny 1S and the MCP1700
* Use the 3S and the LM2931 output into the MCP1700
* As above but with the 2S

Which should last longer?

Obviously the 5000mAh battery is by far the largest, but that's a pretty inefficient setup. The 2S is somewhat more efficient and still about 3x the size of the 1S. The 1S should be by far the most efficient, but the battery is tiny.

I can, of course, just try each and measure. Thought I'd ask here for insights.

Answer: 
LDO regulators don't "transform" energy, they are "linear" regulators. Whichever goes into the load, nearly the same current will be drawn from the battery.

So, within conditions of your task, the largest battery has 5 A-h capacity, the medium one has 1.7 A-h, and the smallest one has 0.6 A-h.

Since the same current (0.028 A) will be consumed from a battery in all cases, the run time will be 178 hours (5/0.028), 60.7 hours, and 21.4 hours correspondingly. I assume that regulators have negligible control losses (quiescent current) in their feedback/control circuitry. If not, this "control current" needs to be added to the 0.028 A load. 

ADDITION: MCP1700 uses MOSFET as regulating element, with control circuitry consuming a negligible 1.6 uA (max 4 uA) additional current from input. For LM2931 the "ON" current is 2.5 mA typical, up to 30 mA, so LM2931 might last half of MCP1700 time. True, the MCP1700 can't handle input voltages above 6V, but there must be other low-quiescent-current regulators. And, as Tony Stewart hints, DC-DC switching converters would do a much better job.
The rather quick estimation is this: the power output needed for your intended purpose is 92.4 mW.

Let's then assume that each LDO is 70% efficient. That means that for one LDO the input power drained from the battery is 132 mW. For two, the overall efficiency is 50% and will drain 188.57 mW.

Your three batteries can respectively output 63 000 mW, 12 240 mW and 2 520 mW for one hour.

Autonomy for each battery is therefore 334h, 65h and 19h. 

I'm not familiar enough with the inner workings of LDO, but I know that standard linear regulator are more wasteful when the output and input voltage are far appart (I.E that 7 in 5 out is more efficient than 15 in 5 out). Therefore, the two daisy chained losses of the bigger batteries are probably underestimated. 

In the end, I would either take the largest battery for longevity or the smallest battery for efficiency. Why ignore the 2s? You get roughly the lack of efficiency of the 3s combined with almost 6 times less autonomy.