We all know that Condorcet and Approval would produce the same results in a situation where every voter's preferences are know to every other voter (provided there are no cycles). What about the case where partial information is known? The following example shows a statistical dead heat between A and C. Voter utilities are between 0 and 1. I avoid worrying about cycles by assuming the examples are symmetrical. Predicted Votes Candidate(utility) ------------------------------- 49 A(1.0) B(0.1) C(0) 02 B(1.0) A,C(0) 49 C(1.0) B(0.1) A(0) If A and C both truncate, they each have a 0.5 probability of winning, so the expected utility to each of these voters is 0.5. If they rank sincerely, expected utility is only 0.1. It looks as though voters should bullet vote just as they would under approval voting. I had been using this example as my "bad Condorcet" example, but in this situation Condorcet seems to do the same thing as Approval whether under full or partial info. A group should truncate whenever the von Morgen-whatsit utility of the sincere CW is lower than the probability of that group's favorite defeating its least favorite. Of course, under Condorcet unsophisticated voters might be tricked into thinking that sincere voting is their best strategy, rather than truncating when called for. Two key assumptions are that each group is homogenous, and that each side can count on the opposite side truncating. The latter shouldn't be a problem in this case, since the AB lottery is in both sides' interests. Given those conditions, I wonder if this is true for all three-candidate cases? If the groups are not homogenous, there is the possibility that sub-groups with low utilities for the sincere CW will use order reversal to overrule groups who really do like the CW. Example based on the first: Predicted Votes Candidate(utility) ------------------------------- 04 A(1.0) B(0.1) C(0) 45 A(1.0) B(0.9) C(0) 02 B(1.0) A=C(0) 45 C(1.0) B(0.9) A(0) 04 C(1.0) B(0.1) A(0) Condorcet strategy: 04 AC 45 AB 02 B 45 CB 04 CA 53 AB 47 53 CB 47 49 AC 49 ... either A or C wins This assumes the two 4-voter groups trust each other. The problem is something like the prisoner's dilemma -- it may not be feasible with a single isolated election, but with repeated elections cooperation might be possible. Even if each group thinks there is only a 50-50 chance that its counterpart will cooperate, it would be the correct strategy so long as A and C are in a dead heat and the utility assigned to B is less than 0.25. This is because p(A/C) given cooperation between both groups = 0.5 (dead heat), and probability of cooperation is estimated to be 0.5, so overall p(A/C) = 0.25. Thus the value of this strategy to the 4 AC voters is 0.25 (assuming failure means that C wins). In the example above, sincere strategy is only valued at 0.1, so order reversal seems the correct strategy in this case. Using Approval voting, the CW would have won with ~92% of the vote.