**A myth about Condorcet voting:**
Although Condorcet is
vulnerable to strategic
voting, it supposedly is only vulnerable in cycle situations with no Condorcet
winner.

**The bust:**
In fact it is vulnerable in situations with an honest Condorcet winner
too:
Let there be 3 candidates A,B,C (or there could be more D,E,...; it'll
still work).
Suppose C is the honest-voter Condorcet winner, with slight pairwise victories over
A,B,...; and A has large pairwise victories over everybody but C.
The A-voters instead of honestly voting A>C>B>... strategically vote A>B>...>C.
This causes C no longer to be the Condorcet winner.
In fact C now loses pairwise (by moderate margins) to every candidate except A.
And A now wins pairwise over every candidate except C, who
beats A by slight margin.
Now A is the winner under all these popular Condorcet
methods (plus more too, if you arrange it right):
Schulze beatpaths, Tideman ranked pairs,
basic Condorcet, Simpson-Kramer minmax.

**Conclusion:**
Strategic Condorcet voters can fail to elect a Condorcet winner.
Strategic range or approval voters will, however, elect an
honest-voter Condorcet winner
under certain (in practice highly reasonable) assumptions, as is proven by
this theorem.

If the other voters "catch on" to the strategic action
and realize it is coming down to an A vs. C battle,
then they can also alter their votes to try to maximally impact that
battle by always ranking A and C top and bottom (or the reverse)
regardless of their honest rankings of A, C, and everybody else.
If all voters do that, *then* C (if it truly is an honest-voter Condorcet winner)
again will win. If C is *not* an honest-voter Condorcet winner,
then still either A or C will then always win.
However, observe that the voters had to pay a heavy price in dishonesty to
do that. With range voting, voters
who wish to rank A and C top and bottom are *not* thereby
prevented from ranking their truly favorite and truly most-hated candidates
top and bottom; range voting thus permits somebody other than A or C to win.

Some readers have insisted this was not obvious enough, they could not conceive of this
kind of phenomenon, etc, hence they require a fully-worked out
random-looking numerical example. (I personally thought it *was* obvious,
but "the reader is always right.")
So here is a set of 21 ballots for 4 candidates:

#voters | Their vote |
---|---|

6 | A>C>B>D |

2 | B>C>A>D |

3 | B>A>C>D |

2 | C>D>A>B |

2 | C>B>A>D |

5 | D>C>A>B |

1 | D>A>C>B |

The defeats-margins matrix is:

Option | A | B | C | D |

A | 0 | 14 | 10 | 13 |

B | 7 | 0 | 5 | 13 |

C | 11 | 16 | 0 | 15 |

D | 8 | 8 | 6 | 0 |

**C wins** (with every Condorcet voting system) because C has
a victory pairwise versus each rival, indicated by the all-green row.

Now **A**'s 6 supporters naturally are not happy about that.
What can they do to make their man win?
There is one, and basically only one,
change to their "A>C>B>D" vote they can make which causes A
to win.
That is to "bury" C with "A>B>D>C."
(Well, the alternate burials A>D>B>C and A>D>C>B
also work in this example, but they both are at-least-as-dishonest votes
and anyway still count as burying C.)
If they do this, we get

#voters | Their vote |
---|---|

6 | A>B>D>C |

2 | B>C>A>D |

3 | B>A>C>D |

2 | C>D>A>B |

2 | C>B>A>D |

5 | D>C>A>B |

1 | D>A>C>B |

whereupon **A wins** using every one of these Condorcet methods:
Tideman ranked pairs,
Basic Condorcet,
Simpson-Kramer min-max,
and
Schulze beatpaths. (Success!)
The new defeats matrix is

Option | A | B | C | D |

A | 0 | 14 | 10 | 13 |

B | 7 | 0 | 11 | 13 |

C | 11 | 10 | 0 | 9 |

D | 8 | 8 | 12 | 0 |

Some people think maybe it is very rare for Condorcet methods to suffer from this problem. But there are theorems to the contrary. Here we shall work out a crude estimate for the rarity.

We shall look at the (admittedly simplified versus reality, but at least it is clearly defined and not biased) random elections model. In this model, each voter casts a random vote chosen uniformly independently from among all possible votes. (E.g, with 4 candidates, there are 4!=24 possible votes. Each voter chooses one by flipping a dodecahedral die and a coin.)

Equivalent interpretation:
"All V-voter elections are considered equally likely" where
an "election" means a table saying, for each voter, how they voted.
(That is what I mean by "unbiased." Clearly I cannot be inserting any evil personal biases
favoring one particular kind of elections, because *all* elections are equally likely.)
We assume a *large* number V→∞ of voters.

In this model, it is known that a "Condorcet winner" candidate C exists (in 4-candidate elections) with probability 82.452%.

**ESTIMATE:**
With the "Tideman ranked pairs" Condorcet method, with
probability≈45% (or more) in the V→∞ limit,
some voter-type whose vote was "X>C>the others"
(for some X, where C is the honest-voting Condorcet winner),
can force victory for X (assuming all the other voter-types stay with their
honest votes) by strategically changing their vote to "bury" C, like this "X>others>C."

**DERIVATION:**
Restricted to the three non-C candidates, there is somebody who would
win (since ties have probability→0). Call that somebody X.
This strategic move by the X>C>others voters
will, with probability→100% (which follows from standard
Gaussian statistics), cause C to become pairwise-defeated by every candidate except X,
and always by the 2 largest pairwise margins.

Thanks to this strategic move, we now have a situation in which X would
win if C were omitted from the race, and in which C loses to every rival but X
by the hugest two margins. With Tideman's Condorcet
method, C's defeats are "locked in"
first. Now X will win if either of the one or two X>somebody defeats
is large enough to get locked in before the C>X defeat.
That happens with probability≈2/3 because there are 6 possible orderings
of these 3 defeats by severity, and only 2 of those orderings stop
X from winning – or in the case where X only defeats one of his
non-C rivals we get probability≈1/2. So on average say 58%.
(This is really cheating because the orderings are
not equally likely because X and C are not random candidates, etc.
So this argument is only approximate, which is why we call it an "estimate" not a "theorem."
One could do a computer Monte-Carlo measurement to overcome that objection and thus get
the exact probability to within 0.01%. One can also see this
for genuine theorems.)
Finally, 0.8245×0.58≈45%.
**Q.E.D.**

**Conclusion:**
This whole "burial strategy works" phenomenon seems *very* common.
In the random elections model using Tideman ranked pairs, it happens at least
about 45% of the time.
With many other Condorcet methods and other probabilistic models,
your mileage may differ, but it ought still to be quite common.

If you want a precise conjecture: I suspect *every* commonly-accepted Condorcet method
will exhibit "burial strategy works" over 30% of the time in random elections
with 4 candidates and #voters→∞; and with some number more than 4 candidates,
I believe burial will work even more often.
In the Dirichlet model,
I suspect again we will have >30% occurrence rates
provided the number of candidates is right.

Also, if we were to add in the chances that some voter-class could cause the election result to
*impove* in their view (despite not necessarily electing their favorite) by
burying the winner, then the estimate "45%" would rise, and if there were the right number of
candidates, it'd rise by a *lot*.