Lower bound on competitive ratio of $m$-machine scheduling
Consider first the sequence $m \times 1$. If the algorithm schedules all the 1's in different machines, then its competitive ratio is 1. Otherwise, its competitive ratio is at least 2 (since the maximum is at least 2 instead of the optimum 1). This case gives an average competitive ratio of at least $p \cdot 1 + (1-p) \cdot 2 = 2-p$.

Consider next the sequence $m \times 1, 2$. If the algorithm schedules all the 1's in different machines, then its competitive ratio is 3/2, since the maximum is 3 while the optimum is 2. If it doesn't schedule the 1's in different machines, we use the trivial bound on the competitive ratio. This case gives an average competitive ratio of at least $p \cdot (3/2) + (1-p) \cdot 1 = 1 + p/2$.

We conclude that the competitive ratio is at least $\max(2-p,1+p/2)$. If $p \leq 2/3$ then $2-p \geq 4/3$, whereas if $p \geq 2/3$ then $1+p/2 \geq 4/3$, and so whatever the value of $p$, we get that the competitive ratio must be at least $4/3$.