Why does parLapplyLB not actually balance load?

parLapplyLB is not balancing the load because it has a semantic bug. We found the bug and provided a fix, see here. Now, its up to the R devs to include the fix.

NOTE: Since R-3.5.0, the behavior/bug noted by the OP and explained below has been fixed. As noted in R's NEWS file at the time:

* parLapplyLB and parSapplyLB have been fixed to do load balancing
  (dynamic scheduling).  This also means that results of
  computations depending on random number generators will now
  really be non-reproducible, as documented.

---

ORIGINAL ANSWER (Now only relevant for R versions < 3.5.0 )

For a task like yours (and, for that matter, for any task for which I've ever needed parallel) parLapplyLB isn't really the right tool for the job. To see why not, have a look at the way that it's implemented:

parLapplyLB
# function (cl = NULL, X, fun, ...) 
# {
#     cl <- defaultCluster(cl)
#     do.call(c, clusterApplyLB(cl, x = splitList(X, length(cl)), 
#         fun = lapply, fun, ...), quote = TRUE)
# }
# <bytecode: 0x000000000f20a7e8>
# <environment: namespace:parallel>

## Have a look at what `splitList()` does:
parallel:::splitList(1:4, 2)
# [[1]]
# [1] 1 2
# 
# [[2]]
# [1] 3 4

The problem is that it first splits its list of jobs up into equal-sized sublists that it then distributes among the nodes, each of which runs lapply() on its given sublist. So here, your first node runs jobs on the first and second inputs, while the second node runs jobs using the third and fourth inputs.

Instead, use the more versatile clusterApplyLB(), which works just as you'd hope:

system.time(
  parallel::clusterApplyLB(cl, 1:4, function(y) {
    if (y == 1) {
      Sys.sleep(3)
    } else {
      Sys.sleep(0.5)
    }}))
# user  system elapsed 
# 0.00    0.00    3.09