## Answer to Puzzle #40 – Feel alike ⇒ vote same?

Puzzle: A single-winner voting system satisfies FAVS if the best strategy for any set of "feel alike" voters is for all members of the set to vote in the same manner.
(a) Show that instant runoff voting (IRV) fails FAVS.
(b) Show that Borda voting fails FAVS.
(c) Show that the Schulze beatpaths Condorcet method (with Simpson-Kramer min max as a tiebreaker, i.e. of two tied Schulze winners, the one with a milder worst-defeat is preferred) fails FAVS.
(d) Show that Approval voting satisfies FAVS – but not if we allow incomplete-information scenarios (i.e. in which the strategic-voter-set does not have complete information about all the other votes).
(e) Show that range voting always satisfies FAVS in both complete and incomplete information scenarios.
(It is not clear whether satisfying FAVS is desirable, but this result is of interest for refuting the vague claim that strategy in range voting is comparatively "complicated" or "messy.")

(a) In this IRV election, if 4 additional voters (all of whom want to elect A) vote: If all 4 vote A top: C wins. If all 4 vote C top: C wins. If all 4 vote B top: B wins. Best group strategy is 2 vote A and 2 vote B – then A wins.

#voters their vote
8 A>B>C
6 B>C>A
4 C>B>A
3 C>A>B

(b) If the current Borda totals are as shown, then 7 more voters can elect "Good" by voting (2,1,0) thrice and (2,0,1) four times, but any 7 copies of the same vote will fail to elect Good.

Candidate their current total
Good 0
Lousy 9
Cruddy 9

(c) Here is an example to show that Schulze beatpaths (or Tideman ranked pairs) Condorcet voting, with "Simpson Kramer minmax" used as tiebreaker in both cases, fails FAVS.

There are 7 candidates, labeled 0,1,...,6. There are 3 strategic voters who wish to elect 0. Suppose after these 3 voters cast 3 votes of the form 0>1=2=4>3=5=6 the situation is as follows: every candidate X beats X+1, X+2, and X+4 (mod 7) by a 5-vote margin.

In this situation, everything is completely symmetrical and every candidate would have an equal chance to win. Perfect tie. That is good (since without these 3 votes, candidate-0 would have lost for sure) but not good enough since only 1/7 chance that 0 will win.

The 3 voters can do better by reasoning as follows: "1,2, and 4 form a Condorcet cycle. So do 3,5, and 6. By changing our "=" to appropriately devised ">" and "<" we can strengthen or weaken these two Condorect cycles (if we so choose)."

Suppose they strengthen them both. (This cannot be done by all voting the same, but can be done by a mixture of 3 kinds of votes, e.g. 1>2>4, 2>4>1, 4>1>2.) In that case, no member of either of these cycles can win, because each has a worse defeat than usual and hence by the worst-defeat-severity tiebreaking, must lose. Hence, 0 must be the winner.

(d) With approval voting, the feel-alike voters can (collusively) figure out the best candidate they are capable of electing, then approve him (and those candidates they regard as superior) only.

But in an incomplete information scenario, the first example here scaled up by a factor of 10 can be used to see that 10 identical approval-style votes will be less good than a mixture of 5 and 5 approval-type votes of two different types.

(e) Range voting satisfies FAVS because suppose it didn't, then just have all the N voters use the average of whatever N vote-vectors they used in their supposedly better-than-all-voting-same voting strategy – to prove it actually wasn't "better."