With Schulze or Ranked Pairs the pathology is DH3: If a third party can win, you can often help them win by voting a “dark horse” candidate ahead of their two real competitors… but if enough people try that then the “dark horse” actually wins. That could actually end up much worse than plurality.
That’s very, very hard to pull off in practice, and it’s obvious to anyone who descends from the omniscient view to an actual campaign.
Suppose there’s the A party(45%), the B party(40%), and the generally-disliked C party (15%). In order for B not to be the Condorcet winner already but make it as easy as possible for DH3, let’s suppose that C is evenly split on the A vs B issue. This is nearly the ideal case for DH3. Picking a lower- or higher-scoring party would present a somewhat different set of challenges.
The DH3 strategy is for B to lie and say that C is better than A, so as to create a cycle, and then win it. A has a lead over B of 45-40=5%, and a lead over C of 85-15=60%. B also has a lead over C of 85-15=60%.
B needs to get A to lose to C by an amount greater than A’s lead over B, but less than B’s lead over C. Each percentage point of B’s strategic votes of C over A shifts A’s lead over C down by 2%. So, if they can get that +60 down to −7 or something, then A’s lead over B will be the smallest one.
But look at what actually has to happen for B to win this cycle:
1)First, they need to successfully create the cycle, which means getting a huge number of their people on board (>13/16 of them). This cannot be done secretly.
2) A voters need to reliably prefer B over C. If B was campaigning in favor of C, then some A voters could be lured away legitimately. If B was issuing ballot instructions to put C above A, some A voters could think that’s so bad they will put B at the bottom, again legitimately. The entire strategy relies on A voters cooperating with the B party in stealing the election, and B’s pursuing this strategy could well make them worse in A’s eyes than C would have been.
Each percentage point of A voters who votes for C over B pushes B’s lead over C down from 60% by 2%. Supposing that B got 90% of their voters on board with the strategy (good luck with that), then C’s lead over A would be 12%. A voters would have a smaller swing to achieve (+ 60% down to +11% instead of +60% down to −6%) and more voters to do it with (45% instead of 40%). So, instead of having to strategically vote with 65⁄80 of their party, they need to partially-strategically, partially-honestly vote with 44⁄80 of their party.
3) ALSO, the C voters could stay neutral or break either way in this, and become decisive. B needs them not to break against B, at least. If B gets C voters to rank B above A (as they might considering how B has been acting), then sure, B wins, but they have done it by forming a coalition with C, which is, you know, a legitimate political move, and not a failure of an election system.
Only if all of these go just right does B succeed. Failing in 1 does nothing good, and helps the disliked C party, making it look much more popular, which is a bad idea even in the medium term. Failing in 2 gives the win to C, which is of course terrible. Failing in 3 loses to A again like 1; and doing better than marginally succeeding in 3 makes B’s victory actually legitimate.
Anyway, in the thousands of actual elections held by the CIVS, nothing like this has happened, and for good reason. It’s lunacy.
That’s very, very hard to pull off in practice, and it’s obvious to anyone who descends from the omniscient view to an actual campaign.
Suppose there’s the A party(45%), the B party(40%), and the generally-disliked C party (15%). In order for B not to be the Condorcet winner already but make it as easy as possible for DH3, let’s suppose that C is evenly split on the A vs B issue. This is nearly the ideal case for DH3. Picking a lower- or higher-scoring party would present a somewhat different set of challenges.
The DH3 strategy is for B to lie and say that C is better than A, so as to create a cycle, and then win it. A has a lead over B of 45-40=5%, and a lead over C of 85-15=60%. B also has a lead over C of 85-15=60%.
B needs to get A to lose to C by an amount greater than A’s lead over B, but less than B’s lead over C. Each percentage point of B’s strategic votes of C over A shifts A’s lead over C down by 2%. So, if they can get that +60 down to −7 or something, then A’s lead over B will be the smallest one.
But look at what actually has to happen for B to win this cycle:
1)First, they need to successfully create the cycle, which means getting a huge number of their people on board (>13/16 of them). This cannot be done secretly.
2) A voters need to reliably prefer B over C. If B was campaigning in favor of C, then some A voters could be lured away legitimately. If B was issuing ballot instructions to put C above A, some A voters could think that’s so bad they will put B at the bottom, again legitimately. The entire strategy relies on A voters cooperating with the B party in stealing the election, and B’s pursuing this strategy could well make them worse in A’s eyes than C would have been.
Each percentage point of A voters who votes for C over B pushes B’s lead over C down from 60% by 2%. Supposing that B got 90% of their voters on board with the strategy (good luck with that), then C’s lead over A would be 12%. A voters would have a smaller swing to achieve (+ 60% down to +11% instead of +60% down to −6%) and more voters to do it with (45% instead of 40%). So, instead of having to strategically vote with 65⁄80 of their party, they need to partially-strategically, partially-honestly vote with 44⁄80 of their party.
3) ALSO, the C voters could stay neutral or break either way in this, and become decisive. B needs them not to break against B, at least. If B gets C voters to rank B above A (as they might considering how B has been acting), then sure, B wins, but they have done it by forming a coalition with C, which is, you know, a legitimate political move, and not a failure of an election system.
Only if all of these go just right does B succeed. Failing in 1 does nothing good, and helps the disliked C party, making it look much more popular, which is a bad idea even in the medium term. Failing in 2 gives the win to C, which is of course terrible. Failing in 3 loses to A again like 1; and doing better than marginally succeeding in 3 makes B’s victory actually legitimate.
Anyway, in the thousands of actual elections held by the CIVS, nothing like this has happened, and for good reason. It’s lunacy.