## DH3 utility calculation

By Clay Shentrup & Warren D. Smith

Imagine that my sincere election utilities are

UA=10,    UB=2,    UC=1,    UD=0.

Suppose I believe that, if I alter my vote to "bury the rivals B and C" (as opposed to voting A>B>C>D honestly), that can either

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

UA=10,    UB=9,    UC=8,    UD=0.
Suppose I believe my burying-vote can either

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.)