Condorcet paradox

Note that in the graphical example, the voters and candidates are not symmetrical, but the ranked voting system "flattens" their preferences into a symmetrical cycle.[6] Cardinal voting systems provide more information than rankings, allowing a winner to be found.[7][8] For instance, under score voting, the ballots might be:[9]

Candidate A gets the largest score, and is the winner, as A is the nearest to all voters. However, a majority of voters have an incentive to give A a 0 and C a 10, allowing C to beat A, which they prefer, at which point, a majority will then have an incentive to give C a 0 and B a 10, to make B win, etc. (In this particular example, though, the incentive is weak, as those who prefer C to A only score C 1 point above A; in a ranked Condorcet method, it's quite possible they would simply equally rank A and C because of how weak their preference is, in which case a Condorcet cycle wouldn't have formed in the first place, and A would've been the Condorcet winner). So though the cycle doesn't occur in any given set of votes, it can appear through iterated elections with strategic voters with cardinal ratings.

We can calculate the probability of seeing the paradox for the special case where voters preferences are uniformly distributed between the candidates. (This is the "impartial culture" model, which is known to be unrealistic,[11][12][13]^:40 so, in practice, a Condorcet paradox may be more or less likely than this calculation.[14]^:320[15])

For ${\displaystyle n}$ voters providing a preference list of three candidates A, B, C, we write ${\displaystyle X_{n}}$ (resp. ${\displaystyle Y_{n}}$, ${\displaystyle Z_{n}}$) the random variable equal to the number of voters who placed A in front of B (respectively B in front of C, C in front of A). The sought probability is ${\displaystyle p_{n}=2P(X_{n}>n/2,Y_{n}>n/2,Z_{n}>n/2)}$ (we double because there is also the symmetric case A> C> B> A). We show that, for odd ${\displaystyle n}$, ${\displaystyle p_{n}=3q_{n}-1/2}$ where ${\displaystyle q_{n}=P(X_{n}>n/2,Y_{n}>n/2)}$ which makes one need to know only the joint distribution of ${\displaystyle X_{n}}$ and ${\displaystyle Y_{n}}$.

If we put ${\displaystyle p_{n,i,j}=P(X_{n}=i,Y_{n}=j)}$, we show the relation which makes it possible to compute this distribution by recurrence: ${\displaystyle p_{n+1,i,j}={1 \over 6}p_{n,i,j}+{1 \over 3}p_{n,i-1,j}+{1 \over 3}p_{n,i,j-1}+{1 \over 6}p_{n,i-1,j-1}}$.

The following results are then obtained:

${\displaystyle n}$	3	101	201	301	401	501	601
${\displaystyle p_{n}}$	5.556%	8.690%	8.732%	8.746%	8.753%	8.757%	8.760%

The sequence seems to be tending towards a finite limit.

Using the Central-Limit Theorem, we show that ${\displaystyle q_{n}}$ tends to ${\displaystyle q={1 \over 4}P(|T|>{{\sqrt {2}} \over 4})}$ where ${\displaystyle T}$ is a variable following a Cauchy distribution, which gives ${\displaystyle q={\dfrac {1}{2\pi }}\int _{{\sqrt {2}}/4}^{+\infty }{\frac {dt}{1+t^{2}}}={\dfrac {\arctan 2{\sqrt {2}}}{2\pi }}={\dfrac {\arccos {\frac {1}{3}}}{2\pi }}}$ (constant quoted in the OEIS).

The asymptotic probability of encountering the Condorcet paradox is therefore ${\displaystyle {{3\arccos {1 \over 3}} \over {2\pi }}-{1 \over 2}={\arcsin {{\sqrt {6}} \over 9} \over \pi }}$ which gives the value 8.77%.

Some results for the case of more than three objects have been calculated.[16]

When modeled with more realistic voter preferences, Condorcet paradoxes in elections with a small number of candidates and a large number of voters become very rare.[13]^:78

Many attempts have been made at finding empirical examples of the paradox.[17]

A summary of 37 individual studies, covering a total of 265 real-world elections, large and small, found 25 instances of a Condorcet paradox, for a total likelihood of 9.4%[14]^:325 (and this may be a high estimate, since cases of the paradox are more likely to be reported on than cases without).[13]^:47. While examples seem to occur occasionally in small settings (e.g., parliaments) very few examples have been found in larger groups, although some have been identified.[18]

One important implication of the possible existence of the voting paradox in a practical situation is that in a two-stage voting process, the eventual winner may depend on the way the two stages are structured. For example, suppose the winner of A versus B in the open primary contest for one party's leadership will then face the second party's leader, C, in the general election. In the earlier example, A would defeat B for the first party's nomination, and then would lose to C in the general election. But if B were in the second party instead of the first, B would defeat C for that party's nomination, and then would lose to A in the general election. Thus the structure of the two stages makes a difference for whether A or C is the ultimate winner.

Likewise, the structure of a sequence of votes in a legislature can be manipulated by the person arranging the votes, to ensure a preferred outcome.

The structure of the Condorcet paradox can be reproduced in mechanical devices demonstrating intransitivity of relations such as "to rotate faster than", "to lift and be not be lifted", "to be stronger than" in some geometrical constructions.[20]