First I’d like to add that what Joel propose is not simply Condorcet (for that we would need to ask participants to rank all possible distributions which is impractical), but that the proposed method is:
- Ask participants to supply their preferred distribution.
- Assume that participants prefer distributions closer to their owns.
- Simulate a Condorcet vote on all proposed distributions.
Here is a vibe example:
Domains
A, B and C.
Options (candidate distributions)
- X = (A=60, B=30, C=10) — from Voter 1
- Y = (A=20, B=50, C=30) — from Voter 2
- Z = (A=10, B=40, C=50) — from Voter 3
Distances (L1) and rankings
Voter 1 (prefers X)
- d(X,X)=0
- d(X,Y)=|60−20|+|30−50|+|10−30|=40+20+20=80
- d(X,Z)=|60−10|+|30−40|+|10−50|=50+10+40=100
→ Ranking: X > Y > Z
Voter 2 (prefers Y)
- d(Y,X)=|20−60|+|50−30|+|30−10|=40+20+20=80
- d(Y,Y)=0
- d(Y,Z)=|20−10|+|50−40|+|30−50|=10+10+20=40
→ Ranking: Y > Z > X
Voter 3 (prefers Z)
- d(Z,X)=|10−60|+|40−30|+|50−10|=50+10+40=100
- d(Z,Y)=|10−20|+|40−50|+|50−30|=10+10+20=40
- d(Z,Z)=0
→ Ranking: Z > Y > X
Pairwise Condorcet comparisons
- X vs Y: V1→X, V2→Y, V3→Y → Y wins 2–1
- X vs Z: V1→X, V2→Z, V3→Z → Z wins 2–1
- Y vs Z: V1→Y, V2→Y, V3→Z → Y wins 2–1
Result
Condorcet winner = Y (A=20%, B=50%, C=30%).
Note that the method extends quite naturally when voters do not provide an allocation to all domains (which is different from preferring a 0 allocation). For example you could put:
- W = (A=50) as your preferred distribution
- Which would then have for distances:
- d(W,X)=|50-60|=10
- d(W,Y)=|50-20|=30
- d(W,Z)=|50-10|=40
→ Ranking: X > Y > Z
We can do that, but keep in mind that we may not see the full picture here. The main goal of Condorcet is to reduce opportunities for strategic voting, so if voters do not know the method which will be used, it’s unclear how this benefit different methods (you may end up having people not voting strategically because they fear condorcet will be used, making averaging look better than what it actually is).
Indeed, with mean voting, if you expect other people to value A less than you do, you can just vote (A=100,B=0,C=0) even if your true preference would have been (A=50,B=50,C=0). The proposed voting system doesn’t have those kind of issues (in most case, per arrow theoreme there may still be some edgecases of strategic voting, but they are orders of magnitude less problematic than mean voting).
The proposed method on the other hand would lead to a lot of domains allocated exactly 0. So we can get both the selected domains and the allocation in one shot.
Median - like method
If we had only 2 domains to allocate, we would only need to specify one domain (if A=x, we have B=100-x). And if we were to apply this method, we should end up with the median.
Here with more than 2 domains, we can see this method as a generalization of the concept of median in higher dimensional spaces. With those two nice properties of:
- If the majority of voters want to allocate 0 to a domain, it should get 0.
- Overstating/Understating one’s preference strategically will be unlikely to affect the result.