Am I measuring it wrong, or am I misunderstanding the calculations?
The answer by PhysicsDave gets close to what I believe is at the root of your measurements.
According to theory, an ideal voltmeter (with infinite input impedance) would measure the following voltages across the capacitors:
$$V_{.05} = V\frac{C_{eq}}{0.05\,\mathrm{\mu F}}$$
$$V_{10} = V\frac{C_{eq}}{10\,\mathrm{\mu F}}$$
$$V_{100} = V\frac{C_{eq}}{100\,\mathrm{\mu F}}$$
where
$$C_{eq} \equiv \frac{1}{\frac{1}{100} + \frac{1}{10} + \frac{1}{0.05}}\,\mathrm{\mu F} $$
However, a physical voltmeter has finite input resistance and so, there is a time constant associated with each capacitor. If the time constant is relatively short, you may significantly discharge the capacitor before the voltmeter reading stabilizes enough to record a value.
You can estimate the time constant for measuring each capacitor voltage as follows:
Assume, for example, an input impedance of $1\,\mathrm{M\Omega}$ and find that the time constant associated with the smallest capacitance is just
$$\tau_{.05} = 1\,\mathrm{M\Omega}\cdot 0.05\,\mathrm{\mu F} = 50 \,\mathrm{ms}$$
while the time constant for the largest capacitance is
$$\tau_{100} = 1\,\mathrm{M\Omega}\cdot 100\,\mathrm{\mu F} = 100 \,\mathrm{s}$$
So, it makes sense that you're measuring the largest voltage across the largest capacitance because you can take your time with the measurement and find that the voltage across the capacitor is falling very slowly as you watch the voltmeter reading.
But, for the smallest capacitance, you've essentially discharged the capacitor in about a 1/4 of a second so it's unlikely that you can get an accurate measurement of the voltage unless your meter has a very quick "peak hold" function.
