For example, in the following election of 3 candidates:

60% vote A1>A2

40% vote B1>…>B10 (where there are 10 candidates from party B.)

The Hare Quota is 33.333%. The first set of candidates considered is {A1,B1}, since it is the smallest possible set containing everybody’s first choice. Both meet quota, but A1 has more support then B1, so A1 is elected, and 33.333 percentage points worth of A votes are used up. Going into the second round, the count is:

26.667% vote A2

40% vote B1>…>B10

The first set considered is {A2, B1}. Only B1 makes quota, so B1 is elected, and 33.333 percentage points worth of B votes are used up. So for the third round, the count is:

26.667% vote A2

6.667% vote B2>…>B10

The sets considered would be {A2, B2}, {A2}, and {B2} (there are other sets, but they are are of the form {A2, some B candidates}, in which no one makes quota, or {some B candidates}, which is dominated by {B2}.)

{A2, B2} is considered first, but since no one achieves quota, the next set is chosen.

With the set {B2}, it takes 1 choice demotion for the A ballots to run out of candidates, so there are a total of 0.26667 * Votes choice demotions

With the set {A2}, it takes 9 choice demotions for the B ballots to run out of candidates, so there are a total of 0.06667 * 9 * Votes= 0.6 * Votes choice demotions.

So {B2} is considered first and wins.

If you treat ballots that run out of candidates as requiring number_of_candidates-1 choice demotions (in this case, 9 for the third round), the problem goes away.