Given the exponential nature of the relationship between current and voltage, even the smallest variation in forward voltage results in significant current change.
Even among "matched" LEDs, there is some variability due to the intrinsic non-ideal characteristics.
A simple way to look at this is by examining the well known exponential curve for a diode:
$$I_D = I_S e^{V_D/V_T}$$
Where \$I_S\$ is the scale current, \$V_D\$ is the forward voltage across the diode, and \$V_T\$ is the thermal voltage at room temperature (\$V_T=26\text{mV}\$).
Say now that you have two diodes, "matched" (\$I_S\$ the same for both) diodes with currents \$I_{D1}\$ and \$I_{D2}\$, then the ratio between those two comes back to be:
$$ \dfrac{I_{D1}}{I_{D2}}=e^{\frac{V_{D1}-V_{D2}}{26\text{mV}}}$$
Which means that if \$V_{D1}-V_{D2}=26\text{mV}\$, that the ratio of \$I_{D1}\$ to \$I_{D2}\$ is 2.7. A difference of 60\$\text{mV}\$, represents a ratio of 10, you get the idea. It shows that one of the diodes will be taking up a lot more current than the other pretty quickly. Now imagine that \$I_S\$ isn't exactly the same for the diodes, which is the real case, then you could see that the ratio of the \$I_S\$ will scale the current ratio further.
You may ask why should there be a difference in the voltage if the diodes are place in parallel and that goes back to the non-ideal characteristics (like \$I_S\$ being different). Think of each of the diodes having some internal resistance in series, in that case the perfect parallel connection situation falls apart.
All this analysis could have been done without using the exponential model, and instead using a linear approximation of the exponential model around the nominal \$V_F\$ of the diode--that would more clearly show the "internal" resistance since the model would have the form \$V_D = V_{F0} + I_D R_{Di}\$, where \$R_{Di}\$ is the slope of the linear approximation of the exponential I-V plot around the nominal voltage \$V_{F0}\$.
Hope it helps.
