DH3 utility calculation

By Clay Shentrup & Warren D. Smith

Imagine that my sincere election utilities are

X.

Have no winner-altering effect. (The most likely possibility, by far.)

Y.

If I choose to "bury the rivals" that unfortunately might cause D to win, whereas someone else (whose expected utility is [10+2+1]/3 = 13/3 = 4.3 assuming equal chances for each of {A,B,C}) would have otherwise won. My utility loss in this case is –4.3.

Z.

If I choose to "bury the rivals" that might work and cause A to win, whereas someone else (whose expected utility is [2+1+0]/3 = 1 if all three among {B,C,D} are equally likely; but no matter what the likelihoods the expected utility is at most 2) would have otherwise won. My utility gain in this case is somewhere between +8 and +10.

The expected alteration in value for me if I choose to bury is   ≥8×P(Z) - 4.3×P(Y).   If this is positive, then it is strategically wise for me to bury. If P(Z) and P(Y) are approximately equal – or if I consider Z to be more likely than Y (or even if I consider Y more likely by a factor of 186%) – then the correct strategy for me is to bury the rivals.

Most people who object to DH3 as a criticism of Condorcet voting methods, claim event Z is very unlikely because it requires a lot of voters to strategize. (For example, if 1000 out of 2000 voters needed to strategize to make Z happen and each one did so with independent probability≤33.3%, then the likelihood of Z would be below 10^-51.) Because Z is unlikely (the objectors continue) strategizing is not worth it, so we do not have to worry about it.

However, Y tends to require even more voters to strategize, hence by their reasoning is even less likely. (For example, if 1100 out of 2000 voters need to strategize to make Y happen, and each one does so with independent probability≤33.3%, then the likelihood of Y would be below 10^-30 times the likelihood of Z!) If so, then 8×P(Z) - 4.3×P(Y) is positive. Then burying will rationally happen. Then the DH3 pathology will happen. Note that this happens even if the DH3-objectors were exactly correct. Indeed, the more-correct they are that strategic Condorcet voting is unlikely, the more justified it becomes for any given voter to strategize.

Let's do that again but now with A,B,C nearly equal and good, D bad:

The most severe form of the DH3 pathology supposes A,B,C are nearly equal in utility to all voters, while D is a lot worse. (What we just analysed was a less-severe form of the DH3 pathology, with numbers altered to make it less severe but more-clearly likely to happen.) Let's now examine this most-severe form.

Imagine that my sincere election utilities are

X.

Have no winner-altering effect. (The most likely possibility, by far.)

Y.

If I choose to "bury the rivals" that unfortunately might cause D to win, whereas someone else (whose expected utility is [10+9+8]/3 = 27/3 = 9 assuming equal chances for each of {A,B,C}) would have otherwise won. My utility loss in this case is –9.

Z.

If I choose to "bury the rivals" that might work and cause A to win, whereas someone else (whose expected utility is [9+8+0]/3 = 17/3 = 5.7 if all three among {B,C,D} are equally likely; but no matter what the likelihoods the expected utility is at most 9) would have otherwise won. My utility gain in this case is somewhere between +1 and +10.

The expected alteration in value for me got by choosing to bury is   ≥1×P(Z) - 9×P(Y).   If this is positive, then it is strategically wise for me to bury. If Z is viewed as lots more likely than Y then burial is a good idea. (Burying is always a good idea if Z at least 9 times more likely. Burying is never a good idea if P(Y)≥1.25×P(Z). If 0.8×P(Y)≤P(Z)≤9×P(Y) then burying might be a good idea.)

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