Probably the simplest answer to your main question is this equation (I'm calling the supply voltage \$V_\text{CC}\$):

$$\%\,I_\text{LED}=-\frac{\%\,V_\text{LED}}{\frac{V_\text{CC}}{V_\text{LED}}-1}$$

(See ***Appendix*** below.)

This equation approximates how much the LED current will change for some tiny percent change in the LED voltage. It's an interesting equation to examine.

 1. What if \$V_\text{CC}=V_\text{LED}\$? This is the question you are asking, by the way, when you ask about supplying the exact voltage specified for the LED on a piece of paper (which isn't real, but only "typical.") In this case, the denominator becomes zero and the percent change in the LED's current tends towards very large numbers even for very tiny LED voltage changes. If nothing else tells you why, this alone should scare you.

 Why is this so? Well, LED voltages are never particularly exact. They will range over several tenths of a volt for any two of them you grab out of a bag. And unfortunately, LEDs "go exponential" when the voltage climbs just a little bit over their required voltage. For example, a \$100\:\text{mV}\$ increase might multiply the current in the LED by a factor of 5 or even 10!! So just small errors in how you guess at the LED voltage vs the supply voltage you use can lead to either destruction of the LED or almost no light coming out, at all.

 So the conclusion here is that you ***must*** use a supply voltage that is not only larger than the typical LED voltage value. But you must use a supply voltage that is more than ***all*** of them might even remotely be. And since you must use a supply rail that is larger than any of the LEDs can ever themselves require, and since the LEDs "go exponential" when given too much voltage, you ***must*** include a resistor (or a different method, perhaps an active one) to limit the current.

 The reasons a resistor is suggested (beyond the fact that it is cheap and easy) is that the voltage drop across a resistor is proportional to the current through it. Since LEDs try to "go exponential" when the voltage across them rises even by a tiny amount, while a resistor stays nice an linear about it, the LED can attempt to exponentially increase its current ... but this would then imply that the resistor would counter that attempt by increasing its own voltage drop *equally exponentially* in response. So the LED may try, but the resistor very quickly counters by dropping more voltage and so the LED finds that it cannot increase its current by much. So it kind of works.

 2. You can also see from the equation above that if \$V_\text{CC}\gg V_\text{LED}\$, then the regulation is pretty good. In fact, the greater the difference the better, as the denominator becomes big enough to really help limit things.

 For example, if \$V_\text{CC}\approx 2\, V_\text{LED}\$ then the percent variation in the LED current will be about the same as the percent voltage variation in the LEDs. If an LED requires \$3.2\:\text{V}\pm 200\:\text{mV}\$, the voltage variation of these LEDs would be \$\pm 6.25\,\%\$. So if we designed the circuit and resistor value in order to yield some specified LED current and used \$V_\text{CC}=6.4\:\text{V}\$, then we'd expect around \$\pm 6.25\,\%\$ variation in the LED currents as we plugged in different ones from a bag.

The resistor value is actually pretty easy to compute. But you do need to find a datasheet on the LED or else you need to do some testing or else make some educated guesses about the typical LED voltage and typical LED current. (If other things are important to you, then you may need to use a somewhat different process.) Once you have these estimated typical values and know the power supply rail you have available, you can just compute:

$$R=\frac{V_\text{CC}-V_\text{LED}}{I_\text{LED}}$$

Now, there are lots of other ways to go. But that's easy and should work for many useful cases. Just remember to be wary of cases where \$V_\text{CC}\$ is close to \$V_\text{LED}\$.

To emphasize that last comment I made, suppose \$V_\text{MARGIN}=V_\text{CC}-V_\text{LED}\$? Then:

$$\%\,I_\text{LED}=-\%\,V_\text{LED}\cdot\frac{V_\text{LED}}{V_\text{MARGIN}}$$

If you only reserve, say, \$V_\text{MARGIN}=1\:\text{V}\$ for an LED with \$V_\text{LED}=3.2\:\text{V}\$, then the percent change in LED current will be \$3.2\times\$ larger than the percent voltage variation for your LEDs. So in that case, a \$\pm 6.25\,\%\$ voltage variation for your LEDs might imply \$\pm 20\,\%\$ current regulation. That may be okay. But it may not be, too.

So you now have not one, but two (and maybe three) useful equations. One to compute the resistor value and the other to provide reasons why you need a resistor as well as how to estimate how closely you can control the currents by using that computed resistor value.


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##Appendix

Because of G36's comments/questions below this answer, of late, I'm editing this answer to include the development of the equation I provided at the outset. It's not complicated.

We start out with the simple KVL equation:

$$V_\text{CC}-I_\text{LED}\cdot R_\text{LIMIT}-V_\text{LED}=0\:\text{V}$$

And solve it for \$I_\text{LED}\$:

$$I_\text{LED}=\frac{V_\text{CC}-V_\text{LED}}{R_\text{LIMIT}}$$

Now, our goal is to compute sensitivity equations from the above. A sensitivity equation just quantifies the output uncertainties based upon the effects of input uncertainties. I didn't find a really easy paper on the topic, but I did find a reasonably readable one here: [Sensitivity Analysis for Uncertainty](https://biomath.usu.edu/files/Sensitivity.pdf). So, feel free to read that if you have any doubts about the rest of what I write, below.

We want to find the % variation of something with respect to the % variation of something else. In calculus form, % variation looks like \$\%\,x = \frac{\text{d}\,x}{x}\$. This is the exact % variation, which is much better than the finite approximation variation that is \$\%\,x \approx  \frac{\Delta\,x}{x}\$. It turns out that the calculus mindset is actually not so hard to do.

First, we apply the implicit product rule (or multivariate chain rule):

$$\text{d}\,I_\text{LED}=\frac{\text{d}\, V_\text{CC}}{R_\text{LIMIT}}-\frac{\text{d}\, V_\text{LED}}{R_\text{LIMIT}}$$

The we divide both sides by \$I_\text{LED}\$:

$$\begin{align*}\%\, I_\text{LED}=\frac{\text{d}\,I_\text{LED}}{I_\text{LED}}&=\frac{\text{d}\, V_\text{CC}}{R_\text{LIMIT}\,I_\text{LED}}-\frac{\text{d}\, V_\text{LED}}{R_\text{LIMIT}\,I_\text{LED}}\\\\&=\frac{\text{d}\, V_\text{CC}}{V_\text{CC}-V_\text{LED}}-\frac{\text{d}\, V_\text{LED}}{V_\text{CC}-V_\text{LED}}\end{align*}$$

Now, we need to convert the infinitesimals on the right side into % variations. This is simple to do:

$$\begin{align*}\%\, I_\text{LED}&=\frac{\frac1{V_\text{CC}}}{\frac1{V_\text{CC}}}\cdot\frac{\text{d}\, V_\text{CC}}{V_\text{CC}-V_\text{LED}}-\frac{\frac1{V_\text{LED}}}{\frac1{V_\text{LED}}}\cdot\frac{\text{d}\, V_\text{LED}}{V_\text{CC}-V_\text{LED}}\\\\&=\frac{\frac{\text{d}\, V_\text{CC}}{V_\text{CC}}}{1-\frac{V_\text{LED}}{V_\text{CC}}}-\frac{\frac{\text{d}\, V_\text{LED}}{V_\text{LED}}}{\frac{V_\text{CC}}{V_\text{LED}}-1}\\\\&=\frac{\%\, V_\text{CC}}{1-\frac{V_\text{LED}}{V_\text{CC}}}-\frac{\%\, V_\text{LED}}{\frac{V_\text{CC}}{V_\text{LED}}-1}\end{align*}$$

This allows us to focus on \$\%\,V_\text{LED}\$, by taking the last term and its sign on the right side, or to focus on \$\%\,V_\text{CC}\$, by taking the first term and its sign on the left side. (Or, of course, to take both into account at the same time.)