Good point. You can solve this issue by electing more than one representative from ordered (per-party) lists: the first picked vote for Labour elects Labour#1, the second Labour#2, etc. Labour#4 could still be lucky or unlucky, but Labour#1 will almost certainly be picked (assuming e.g. 50% of the vote for Labour, with probability 1 - 2^-10 = 1023/1024.)

However, I'm not sure how much of a (perceived) advantage over proportional representation this gives you.