You're getting the expected result. What you see is the normal behavior of diodes in series and it's completely normal to have one resistor and a string of LEDs connected after it.
What's basically happening is this: When they told you that the forward voltage is 2 V, they lied. It actually depends on the current going through the LED and you can consider the 2 V some sort of nominal value, but the exact drop should be read in the datasheet (if it's available).
In general case when you want to connect diodes in series, you use this formula for resistor:
$$ R= \frac {V_{supply}-NV_{f}}{I_{f}}$$
where the N is number of diodes you have.
This way it turns into simple Ohm's law. But in your case, you're approaching the border at which the above formula will not hold. You basically have a circuit with one branch only and the current going through that branch isn't going to much change with the number of LEDs if the voltage of the supply is high enough to be higher than LED forward voltage.
Take a look at this diagram from Wikipedia:
Notice the point marked \$ V_d\$. For this diode, once the voltage at the diode terminals reaches that point, the current will start quickly increasing with only a small change in voltage. That is why adding more LEDs doesn't immediately affect current. The voltage is high enough that all LEDs will conduct. Should you for example put 10 LEDs in series, the voltage will be too low and they will either show barely noticeable light levels or stay off.
Next, let's take a look at the different voltages you got at the LEDs. Again take a look at the curve for the diode from the Wikipedia. The \$V_d\$ point for each diode made is different and there are some tolerances here. So some diodes of same model number will at same current have a bit larger voltage drop and others will have a bit smaller voltage drop.
Next about LEDs in series. There is nothing wrong with that, but you're still not doing it right. Using the formula I provided, you should set the resistor so that the LEDs will be within their rated current. If you fulfill that condition, there's absolutely nothing wrong with having multiple LEDs connected in series, should you have voltage to spare.
You have to start with a closed circuit, so that there can flow current. Do you have the resistor connected between the +
and the -
of the power supply? Then the voltage difference is 3.3 V. And you use Ohm to calculate the current.
The LED. Did you place that in series with the resistor? Which is how a LED circuit is built: the resistor makes sure that there's not too much current through the LED. Always use one.
If the LED's voltage would be 3 V then the difference between your supply voltage and the LED's voltage would be across the resistor. Kirchhoff is to blame for that. Kirchhoff's Voltage Law (KVL) says that the total of the voltages in a closed loop is zero. So we'll have 1.1 mA through LED and resistor. The 3 V was specified at 20 mA, so we're an end below that. It's normal for a LED to have a lower voltage at low currents.
But note that the 1.1 mA was true for 3 V LED voltage. We're apparently at 2.4 V, so the difference is now 0.9 V, and the current 3.3 mA. If you decrease the resistor value so that the current increases, you'll notice that the LED's voltage will increase as well.
How do you calculate the value? (Here we go again)
\$ R = \dfrac{\Delta V}{I} = \dfrac{3.3 V - 3 V}{20 mA} = 15 \Omega \$
edit re your update of the question
Welcome to the real, imperfect world. What you have at hand is measurement error. This is an important issue in engineering, and handling it properly can be a painstaking process.
You're giving your numbers in three significant digits, that's probably what the multimeter gives you. A multimeter's precision is most of the time expressed as a percentage (relative error) + a "count" (absolute error). A hobby quality meter may for instance have 2 % precision +/- 1 count. The 2 % should be clear: a 100 V reading may actually represent anything between 98 V and 102 V. The 1 count is an error in the last digit. A 5 may actually be a 4 or 6. That's an absolute error and doesn't depend on the value the meter gives you. If you measure 100 V then 1 count represents 1 %, if you read 900 V (same number of digits!) then 1 count is 0.11 %.
Let's presume you have a decent multimeter with 1 % +/- 1 digit precision. Then worst case your values may become
3.28 V - 1 count = 3.27 V, - 1 % = 3.237 V
267 Ω + 1 count = 268 Ω, + 1 % = 270.7 Ω
11.7 mA + 1 count = 11.8 mA, +1 % = 11.92 mA.
3.237 V / 270.7 Ω = 11.96 mA, which agrees well with the 11.92 mA we calculated for worst case. If your multimeter has a 1.5 % precision the calculated current will fall perfectly within the measured value's error range.
Best Answer
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) Just a resistor. (b) Series LEDs. (c) Parallel LEDs.
First consider the resistor on its own - Figure 1a. From Ohm's Law we can calculate the current through it as \$ I = \frac {V}{R} = \frac {9}{10k} = 0.9~mA \$. This is the maximum we can get and adding LEDs will reduce it.
Looking at Figure 1b you can see that the same current goes through all the elements in the circuit.
Figure 2. A typical LED graph of current vs forward voltage. Note that even at tiny currents (< 1 mA) the forward voltage drop doesn't change much.
We can see from Figure 2 that the 1.65 V reading you are getting across each of the LEDs is correct.
Now what happens when you connect the LEDs in parallel? We will still be limited to a maximum of 0.9 mA through the resistor but now this has to split two-ways through D3 and D4. They will get about 0.45 mA each. Checking back to Figure 2 we can see that they will still drop about 1.6 V or so.
Note that adding the LEDs reduces the voltage across the resistor and therefore reduces the current available. For Figure 2b the actual current will be \$ \frac {9-3.3}{10k} = 0.57~mA \$. For Figure 2c it will be \$ \frac {9-1.65}{10k} = 0.735~mA \$ which helps a bit when the current is shared.
The water analogy is not great, but let's try. To resemble an electrical circuit the water will have to flow in a closed loop such as a hot water central heating system. The hot water pump raises the pressure at the outfeed of the pump (the battery raises the voltage). We feed through the pipes (the resistor) to two check valves which piped in series or in parallel.
Figure 2. The check-valve is the plumber's diode. Water can only flow one way. A certain pressure is required to overcome the spring. This causes a pressure drop across the valve but once the valve is open the pressure drop doesn't change much even at high currents. Source: Modified from Gentec image.
Now let's put some numbers on our system. The pump raises the pressure to 9. (Think psi, bar, pascals, inches of mercury or whatever you like.) The valves will open at 1.65. If we pipe them in series as in Figure 1b it should be clear that if the valves open the pressure reading at the top of D2 will be 1.65 and above D1 it will be 3.3. If we connect them as shown in Figure 1c they will both open when the pressure reaches 1.65.
LEDs and resistors don't "consume" voltage in the same way your hose doesn't consume pressure. Voltage is dropped or "voltage decreases" across the resistor or LED in the same way that pressure losses occur in the water system.
While they have resistance we do not call them resistors because they are non-linear just as the check-valve is. Also the current doesn't "get lowered" because what leaves the battery on one terminal must come back on the other.
I hope that helps.