..."approval style" range voting (where you give every candidate either the max or min possible score) is strategically optimum (or at least, the utility of the best approval-style vote is only smaller than that of the truly-best range vote by a negligibly small amount) if we can assume that the probability of an election-leading near-tie (which your vote can hope to alter) between two of the candidates is very unlikely and the probability of a 3-or-more-way near-tie is very very, i.e. neglectibly, unlikely.
Proof sketch: We assume your probability of breaking an AB tie is proportional to your vote's A-B score difference. Give maximum score to some subset of candidates A and minimum to the complementary subset B. Why would you want any other vote? You might want to lower some A's score by some amount Q if by so doing you could increase the vote difference for some other pair (or pairs) by Q (in total); but if that other pair (or pairs) has (have) less utility (where "utility" of a pair of candidates is the probability of that tie times candidate-quality difference) then that move would be bad for you for any Q>0. In other words, assuming situation is "generic" so that exact ties do not happen, the best maxxing-and-minning-only vote, is in fact your best vote, period. Hence your best vote (maximizing expected utility) then always maxes-out or mins-out every score you can. Q.E.D.
A different assumption which also suffices seems to be that the candidates can be pre-ordered in decreasing likelihood of election chances, and that each's chance is hugely greater than the next one's.
Proof: In this case you can assign your scores to the candidates one by one in order, only basing your decision on a candidate's score on information about him and the previous candidates – the later ones are irrelevant to that decision. Then, for each candidate X you score, his score cannot affect the winning chances of the previously-scored ones if X does not win. X's score then can only affect the chance X wins, not the chances for previous candidates if X does not win. And it does so in a monotonic manner. Hence the decision on X should be purely based on whether X is superior or inferior in utility to the expected utility among the previous winners given your previous votes: if it is larger, give X the maximum, otherwise the minimum score. Q.E.D.
Forest Simmons also points out the following argument that in "large" elections, best range voting strategy is approval-style, at least up to statistical "noise" (which, he argues, ought to be negligibly small in most circumstances in large elections):
For each candidate each voter uses dice to pick a random number between the min and max possible rating, and approves or disapproves that candidate depending on whether or not the random number is less than the voter's ideal rating for the candidate.
According to the "law of large numbers," if all approval voters use this technique, then the approval proportions for the candidates are asymptotically the same as the range proportions.
But the reader is warned to be careful; this Simmons argument actually does not work in the large election of puzzle 19 because in that scenario the "noise" would dominate the intentional effects.
However, those assumptions, while plausible in large real elections, usually are false in small ones.
Also if you have complete knowledge of all the other votes then an optimum range vote is always trivial to determine, and without loss of generality is an "honest approval-style" (or one could also demand it be a "plurality-style"), vote. E.g. give the maximum score to the best candidate who can win and to all better candidates, and the minimum score to all others. (This strategy's validity to a large degree forms the basis for the "first argument" in the pleasant surprise claim.)
In a 3-voter (or more voter) [0,1]-range-voting election, assume the totals of the votes of the other two (or more) voters are either
but you do not know which. Call the candidates A,B, and C. Your candidate-utilities:
In the following 3-voter election, call the candidates A,B, and C. We assume [0,1] range voting. Let Rj denote independent uniform random numbers in [0,1], so that voter#2 is regarded as completely unknown to you, while voter#1 is regarded as completely known.
| Voter | A | B | C |
|---|---|---|---|
| voter#1 | 0 | 0.798 | 0.618 |
| voter#2 | R1 | R2 | R3 |
| you(voter#3) | 1 | V | 0 |
(Or, what might be more realistic – but is entirely equivalent – is to replace voter #2 with three random voters each of whom votes 1 for one candidate, 0 for another, and a random value for the third as 10R, R10, 0R1.) Assume the election-utilities (for you) of the three candidates are
Then here are the expected utilities of various possible votes you could make (result of 25 million Monte Carlo experiments): It is in fact possible to evaluate the exact rational utility answers by evaluating the volumes of certain 3-dimensional polyhedra, but we did not try.
| Vote V | Utility |
|---|---|
| 1(exaggerated) | 35.4 |
| 0.83(honest) | 43.6 |
| 0.58(best?) | 51.2 |
| 0(exaggerated) | 35.9 |
As you can see, neither the honest range vote, nor either of the two "approval style" exaggerated ones, is strategically best.
In the following 4-voter election, call the candidates A,B, and C. We assume [0,1] range voting. Let Rj denote independent uniform random numbers in [0,1].
| Voter | A | B | C |
|---|---|---|---|
| voter#1 | 0 | 0.896 | 0.219 |
| voter#2 | R1 | R2 | R3 |
| voter#3 | R4 | R5 | R6 |
| you(voter#4) | 1 | V | 0 |
Assume the election-utilities (for you) of the three candidates are
| Vote V | Utility |
|---|---|
| 1(exaggerated) | 3.5 |
| 0.94(honest) | 4.1 |
| 0.46(best?) | 7.8 |
| 0(exaggerated) | 0.9 |
Again, neither the honest range vote, nor either of the two "approval style" exaggerated ones, is strategically best.
See puzzle 19.
I thank Boris Alexeev for stimulating correspondence.