**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.")

**Answer:**

**(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."