Here's a novel thought for the scenario where the infamous haybales task is replicated:

It all depends on the ratio of clues to hay bales. The reason that most teams finished quickly was that the field started with 20 clues in about 270 bales. That gave each team a 1/13.5 or 7.4% chance at getting one. That means they would have to unroll just 14 bales on average to find a clue. The odds on unrolling over 100 bales and not finding one are astronomical; it would not have happened if Lena was efficient in seeing clues in some of her bales, as there were 12 clues left after all other teams had finished and there should have been approximately 270 - 8 X 13.5 = 162 bales left. Up until that point, her odds of finding a clue in a bale (assuming efficiency on her part) were higher, but even then it had to be something like 12/162 on each try or still 7.4% each try. It should again take her on average at that point only another 14 bales to finish.

So, if you were Bertrand van Munster (who is really a very clever guy), what would you do if you needed to assure that a repeat of the Lena debacle did not happen again? You would increase the ratio of clues to bales, that's what you would do. Doing so would decrease the differentiation between teams coming out of this task, but that's OK since it will be a total BUNCHING for the ferry anyway. Even if they used the original ratio, it should still not take very long for all teams to finish.