Alcoholic KOH mechanism with alkyl halides
With a strong base $E2$ will always predominate over $E1$.

Both elimination mechanisms have a preference for tertiary substrates over secondary substrates over primary substrates. This behavior is on contrast with the substitution reactions, where steric factors are important for $S_N2$. Steric factors are less important for the proton transfer necessary for $E2$.

Since $E2$ and $E1$ have the same substrate preference order, you may feel that you cannot decide which is more likely. However, $E2$ is usually faster because $E2$ is bimolecular. Consider the rate laws:


$\mathrm{rate}_{E2}=k\mathrm{[substrate][base]}$
$\mathrm{rate}_{E1}=k\mathrm{[substrate]}$


If we make the assumption that the rate constants are of similar magnitude, then increasing the concentration of the base increases the rate of $E2$.