Question: How do I calculate how long a battery operated product will run?

Here's what I've got:

1. 2 AA, 1.5V, 2700mAH batteries
2. Voltage Regulator with a Iq of 25 uA
3. Voltage Regulator Eff = 80%
4. Active Current = 50mA
5. Sleep Current = 1uA
6. Duty Cycle = 99.9% (only active 0.1% of the time)
7. Active Voltage is 3.3V

I've gone the current route and got an answer. I went the power route and got a TOTALLY different answer (days vs years different).

How do you do this?

Answer: 
My calculation, probably missing something, but here's what I did:

$$
1 \mathrm{\ \mu A} + (50 \mathrm{\ mA} \times 0.1\%) + 25 \mathrm{\ \mu A} =
76 \mathrm{\ \mu A}
$$

$$
\frac{76 \mathrm{\ \mu A}}{ 80 \, \% \mbox{ efficiency}} = 88 \mathrm{\ \mu A}
$$

Round up to \$100 \mathrm{\ \mu A} = 0.1 \mathrm{\ mA}\$

$$
\frac{2700 \mathrm{\ mAh}}{ 0.1 \mathrm{\ mA}} \approx 3 \mbox{ years}
$$

If you're using rechargeable batteries, they'll discharge on their own long before that. Or if any of your other calculations are off (like maybe it's a 98% instead of 99.9% sleep), that will affect it a lot too.
The output current required from the 3.3V 
regulator is 

\$1 \mu A \times 0.999 + 50mA \times 0.001 + 25uA = 75.999uA \$

The output power is

\$ 3.3V \times 75.999 \mu A = 250.8\mu W\$

The input power to the regulator is

\$ \frac{250.8\mu W}{0.8} = 313.5\mu W \$

When the batteries are fully charged 
the input current to the regulator is

\$ \frac{313.5 \mu W}{3V} = 104 \mu A \$

If the batteries have a flat discharge
curve then you will get a life of

\$ \frac{2700mAh}{104 \mu A} \$ = 25837 hours = 2.95 years

Since your batteries are 2700mAh 1.5V AA
I am guessing that the discharge curve
is not flat. You will need to draw
higher currents as the voltage drops.

Also your regulator efficiency probably
drops at lower voltages. Again I am
guessing since I have not seen the
design. 

Be careful when calculating using currents.
You may be inadvertently assuming that 
the input and output voltage of the regulator
is the same. With an input of 3V and an output
of 3.3V it is not a big error. If you do
a more accurate estimate of the battery
discharge curve it will matter.