"Burial" strategy in Condorcet methods

By Warren D. Smith, Feb 2008      (skip to conclusion)

We report on some computer Monte-Carlo simulations and theorems about how often the "burial" strategy works in Condorcet methods.

My simulations were for 4-candidate elections, but the theorems hold for any number N≥3 of candidates. The program supports either the "random election model" or the "Dirichlet model" (see glossary for definitions of those probability models).

An interesting computer finding, which I then backed up with a human proof, is the following. "Black's Condorcet method" is to use Borda if there is no Condorcet winner. It has generally been looked-down-on by internet Condorcet advocates. But it turns out to have a nice property.

THEOREM (BLACK BURIAL-RESISTANCE): In N-candidate elections (N≥3 fixed) in the random elections model, using pure-rank-orderings as votes, consider the "burial" strategy where everybody who voted X=top C=second, now artificially votes C=bottom (but aside from that, all votes unchanged). Here C denotes the Black-winner with honest votes. Then: with probability→100% in the #voters→∞ limit, the new winner will be neither C nor X. And when it is not X, that means burial "does not work."

PROOF:

1. If C originally was a Condorcet winner: The new election will with probability 100% have no Condorcet winner since X will still be beaten pairwise by C, but C will with probability→100% be beat by the others thanks to the burial. (All the "→100%" statements we make arise easily from standard Gaussian statistics in the limit of large #voters. The point is that the amount of typical statistical variation away from "all vote-counts equal" is of order √V, and variations larger than that are arbitrarily unlikely; meanwhile the strategy pushes stuff by a far larger amount of order V, in the limit V→∞.) So Black goes to Borda. But the burial raises the Borda scores of all rivals Y,Z,... of C and X, while leaving X's Borda-score unaltered and lowering C's. Hence with probability→100% one of those rivals will win.
2. If there originally was no Condorcet winner (so C was the Borda winner): The new election will with probability→100% have Borda winner∈{Y,Z,...}. This will be the Black-winner too because the only way to avoid that is if the burial creates a Condorcet winner. The burial causes the Y:C and Z:C comparisions to improve for Y,Z and worsen for C, but does not affect any other pairwise comparisons. Hence it cannot create a Condorcet winner except if that winner is among {Y,Z,...}.

In the Dirichlet model, however, burial still works with Black voting. My simulation finds, in fact, that it works (29.50±0.03)% of the time for X (where X is the candidate who would have won if C were disallowed) with 4 candidates; as opposed to tending to 0% in the Random Elections Model.

Note that the BURIAL-RESISTANCE THEOREM is not going to have a nice analogue for most other commonly proposed Condorcet variants. It is pretty special to Black's method.

In particular(!), we have the exact opposite property for Nanson-Baldwin's Condorcet method (which consists in repeatedly eliminating the Borda-loser):

THEOREM (BURIAL-VULNERABILITY FOR NANSON-BALDWIN): In the same scenario but now using "Nanson-Baldwin's Condorcet method," burying C works for the X=top C=second voters (where X is the candidate who would have won with honest votes if C had not been in the race) with probability→100% to make X become the winner.

PROOF: Because C will with probability→100% have the lowest Borda score hence be eliminated first in the modified election. Then all the rankings among X, Y, Z,... (after C's elimination) are unaltered by the strategy and hence X still will beat the others with certainty. Q.E.D.

But (again) if we switch to the alternative Dirichlet model, then burial no longer reliably works, albeit the computer finds it works at least about 17.1% of the time in 4-candidate situations (this figure is only a lower bound).

Most common Condorcet-methods proposals are somewhere in between the two extremes represented by Black and Nanson-Baldwin. For example, N.Tideman in his book and D.Woodall both have recommended (two different) variants of a Condorcet method one might call "Smith-IRV." It involves eliminating both candidates not in the "Smith set" as well as eliminating those top-ranked the least. (Woodall's version makes more sense to me, but those details will not matter for our purposes here.) In this method, burial works, in the 4-candidate random elections model, with some probability which tends to a nonzero limit as #voters→∞. (Proof sketch: In pretty much any situation where C was the Condorcet winner but would be eliminated early in IRV voting, we have vulnerability to C-burial.)

As another example, the "Simpson-Kramer min-max" Condorcet method consists in electing the candidate whose strongest-pairwise-defeat, is the weakest. My Monte-Carlo simulations indicate that burying C works for the X-top voters (where X is the second-place candidate, i.e. who has the second-weakest strongest-pairwise-defeat) with probability (27.40±0.11)% in the Dirichlet and (42.89±0.07)% in the random-election model with 4 candidates.

The "Copeland" method is to elect the candidate who defeats the most rivals pairwise. (We shall break ties by using a random fixed candidate-ordering.) My Monte-Carlo simulations indicate that burying C works for the X-top voters (where X is the second-place candidate) with probability (32.28±0.1)% in the Dirichlet and (82.18±0.1)% in the random-election model with 4 candidates.

If we instead break Copeland ties using Borda, then Monte Carlo shows the Dirichlet burial-works probability shrinks to (29.9±0.1)%, and (we prove in puzzle 86) the random-election model probability shrinks to →0 in the 4-candidate case, but stays a positive constant if there are at least 5 candidates.

Chris Benham points out things may not be so cut-and-dried

Despite the fact Black is asymptotically 100% burial-resistant in the random-elections model, while some other Condorcet methods like Smith-IRV and Schulze are not (and Nanson-Baldwin is asymptotically 100% burial-vulnerable), Chris J. Benham points out that in the following election (the "B>C>A" votes are insincere and arise from a strategic decision to "bury" A; they honestly would be "B>A>C")...

... Black awards the victory to B, the one who orchestrated the burying. (A>B>C>A cycle; Borda scores: A=97, B=139, C=64. (Also saying B wins: Tideman ranked-pairs, Schulze-beatpaths, Simpson-Kramer min-max, and basic Condorcet.) Meanwhile Nanson-Baldwin and Smith-IRV both elect the honest-votes winner, A!

The main reason for this discrepancy is that the "random elections model" is not a terribly realistic probability model for elections – at least not for the elections Benham mainly cares about involving only two strong candidates. (The Dirichlet model comes closer.) On the other hand in a hypothetical future world in which elections are run using Condorcet voting, perhaps the random elections model would become a better statistical model than it is now.

Benham favors Woodall's Smith-IRV Condorcet method above the others. As an example of its superiority, Benham gives

where the (*) votes insincerely bury A. Almost every Condorcet method known to man elects C due to the burial-strategy employed by C-voters. That includes Black's method (since the Borda scores are A=85, B=86, C=129 and we have an A>C>B>A cycle). But not Woodall's Smith-IRV method, which successfully elects A (the honest-vote Condorcet winner).

Conclusion

Our results greatly-contradict the usual previous thinking that the differences between Condorcet methods in vulnerability to strategy, are small. That's interesting and new. However, as Benham points out, just because that result holds in the "random elections model" does not necessarily mean it should be taken too seriously in real life.

So maybe the internet Condorcet community should now re-evaluate which Condorcet variants they prefer. (As well as re-evaluate how well they claim to have "understood" Condorcet strategy. It appears they were overconfident in their assertions of understanding.) Or maybe not.

In the random elections model, some Condorcet variants are highly burial-resistant (Black); others are highly vulnerable (Nanson-Baldwin); and most are in between (Simpson-Kramer min-max, Smith-IRV, and I presume also Tideman ranked pairs, Schulze, etc.; I estimate crudely that these too have burial-works probabilities on the order of 50% in the 4-candidate random elections model).

These results are also relevant to the following important debate. I had contended that real voters, if made to use Condorcet voting, would usually artificially rank the two major-party candidates top and bottom. They would do this for psychological/strategic reasons to maximize the impact of their vote (reasons include: the effectiveness of "burial"; the usefulness of "favorite betrayal" in every Condorcet method; also the theorem that the "rank the major threat bottom and the most-likely-to-win better rival top" strategy is usually near-maximally effective in a very wide class of single-winner election methods). If enough of that behavior occurred, a third-party candidate could never win. Thus, Condorcet voting might lead to self-reinforcing 2-party domination in which every voter robotically followed the recommendations of the "how to vote" cards from the two major-parties, since history kept proving no other course was worth bothering with.

But given that voters only ever had 2 real choices, the "improvement" caused by having Condorcet voting versus plain Plurality, wouldn't be much of an improvement, since every voting method is essentially the same as plain plurality in the 2-candidate case.

So if and when this contention is correct, Condorcet voting is practically dead on arrival. But is it correct? Nobody really knows, but here are arguments pro and con:

Pro:	Con:
Yes, based on data that a huge percentage of Australian voters behave this way even though burial actually makes less strategic sense in Australia's IRV system than with Condorcet methods, even Black's. (That is: with IRV, the "X>C>others" voters can never make X win by voting "X>others>C." With Condorcet, they sometimes can; indeed even with Black's Condorcet method, they sometimes can – the probability→0% result only happens in a limit which is never actually achieved in a probabilistic model which does not correspond perfectly to reality.)	No, the contention is severely undercut provided we use Black's method, by the fact that Burial is highly strategically ineffective with Black under the Random Elections Model. (Of course, you'd need to believe Joe Voter would appreciate this theorem and method-flavor-distinction. Which seems doubtful in view of the Australian facts at left.)
Yes, rationally, based on theoretical analysis indicating Burial works in a lot of Condorcet situations for a lot of Condorcet methods (about 20-80% of the time depending on the probability model and the Condorcet variant); if this percentage greatly exceeds the percentage of the time third-parties win, then the exaggeration strategy is worth it overall.	Despite the huge percent of Australians who just follow the major party "how to vote" recommendations, these majors-top-and-bottom votes can still lead to third-party Condorcet winners if the two majors are precisely-balanced-enough and there is a small percentage of votes that rank the third-party candidate top. That just might happen often enough that two-party domination will not ensue with Condorcet. (Certainly, Condorcet is more likely to avoid duopoly, than IRV; but unfortunately IRV experimentally does lead to self-reinforcing 2-party domination.)

Probability that "burying" the honest-voter winner C, by those who'd honestly voted "X>C>..." causes X to win (4-candidate elections). Here X denotes the candidate who "finished second" with one particular definition of "finished second"; with the alternative definition which maximized the probability burial would work, we would have gotten at least as large numbers.

Condorcet variant→ Prob Model↓	Black	Nanson-Baldwin	Simpson-Kramer min-max	Copeland (fixed pref-order tiebreak)	other (Tideman, Schulze, Smith-IRV etc)
Random Elections (#voters→∞)	→0%	→100%	42.9%	82.2%	≈50%?
Dirichlet	29.5%	≥17.1%	27.4%	32.3%	≈30%?

Return to main page