In Instant Runoff Voting (IRV), it can occur that the winner is the same candidate as the winner would have been if all voters reversed all rank orderings ("trying to elect the worst"). Here's a simple example.
| #voters | their vote |
|---|---|
| 2 | B>C>A |
| 2 | A>B>C |
| 1 | C>A>B |
At right is the simplest
example (in the sense of having the fewest candidates and voters):
In this 5-voter 3-candidate
IRV election, A wins (after C is eliminated). But if we reverse all votes
(trying to determine the worst candidate rather than the best!) then A still "wins"
(after B is eliminated).
That is one reason it is rather hard to maintain (as IRV propagandists Rob Richie and S. Hill unfortunately often nevertheless do) the bogus claim that "IRV always produces 'true-majority winners'."
[ In some Condorcet methods, winner=loser can also occur; but it is never possible for a beats-all Condorcet winner to be the same as the loses-to-all Condorcet loser if both exist; and in some Condorcet methods such as Schulze beatpaths, "reversal failure" i.e. winner=loser is never possible, at least if ties are disregarded. Ditto in range and approval voting: winner=loser is never possible if ties disregarded. ]
How often does IRV reversal failure actually happen? In the usual simplistic probabilistic "random elections" model (all elections equally likely – that is, all voters independently choose a rank-order at random, all orders equally likely), the answer (each datapoint based on ≥20K Monte Carlo elections, but bold datapoints based on ≥100K; probably I give too many significant figures in percentages) is:
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This particular probablistic model of course is usually politically unrealistic but in my experience (perhaps surprisingly) it nevertheless usually delivers estimates of frequencies of goofy phenomena, which happen to be pretty close to the frequency estimates you get by trying to look at real elections throughout history. So until further evidence comes along, use this as your frequency estimate.
I have also done two Monte Carlo simulations each using an infinite number of voters. They find:
| probabilistic model | percentage |
|---|---|
| Dirichlet Model | (1.69±0.09)% |
| Random Elections Model | (2.47±0.14)% |
This is the question of the frequency of the "Peru Scenario" (we call it that because this happened in the Peru 2006 presidential election) in which a candidate who would beat every opponent in a head-to-head race, is eliminated by IRV. (Same thing also happened in the France 2007 and Chile 1970 Presidential elections.) This is a different phenomenon which again makes it rather hard to maintain that "IRV always produces 'true-majority winners'." (!)
In the same probabilistic model, we showed in puzzle 17, this has frequency 3.71% in 3-candidate races (and 3.23% in the Dirichlet model). However in "1-dimensional" politics it is more like 25-33% common.
Although the winner=loser and eliminated-Condorcet-winner phenomena both are very embarrassing for IRV, they are fairly uncommon. Nevertheless in the 659-member UK house of commons, if elected via IRV we would expect a dozen or so of the races each election to be winner=loser races, and even more to be eliminated-Condorcet-winner races. Then there would be consequent screaming (unless of course they made the votes a secret so nobody knew to scream, which in the past has been the usual method of preventing screaming in IRV elections – but in the last few years there has started to be a trend toward actually posting IRV election data on the internet)...
So this is not a neglectible problem. However, it is fairly rare.
What is much more common is the IRV "favorite-betrayal pathology" where ranking your true favorite top is strategically unwise, so therefore you (not being a strategic idiot) vote dishonestly. That problem happens in 19.6% of 3-candidate elections in this model. But if all a voter knows is that third-party candidates win a lot more rarely than 19.6%, then that voter is always strategically better off dishonestly betraying her favorite in her vote. We repeat. Always. I.e. in 100% of elections in this model. This problem is not rare.
(Note: it has sometimes been wrongly claimed by IRV advocates that you cannot vote strategically without knowing how all the other IRV voters are voting, and without doing a tremendous amount of difficult thinking. False. The truth is, you don't need to know a damn thing besides "third party candidates rarely win" and you don't need to think at all.)
So is there any sense in which it is true that "IRV always produces 'true-majority winners'."?
Well, you could just define "true-majority winner" to mean "IRV winner" and then it would be true. That'd be laughable. We could similarly define "true-majority winner" to mean "plurality winner" and then the plurality voting system, instead, would always deliver a true-majority winner. Or the Coombs voting system. Or whatever (a random-winner dice-roll system...). This is a joke.
Another joke is to note that IRV's final round (A versus B) always makes A win over B exactly when a majority of the voters prefer A over B. (That is false in the top-3 variant of IRV used in San Francisco, but never mind that.) Therefore "A is a true-majority winner." By that logic, every candidate not a lose-to-all loser, would be a "true majority winner"! I.e, in the limit of a large number of candidates, every candidate in almost every election would be a "true majority winner"! That'd be happy for them all – but totally ridiculous.
The truth is, for this to be a meaningful assertion we need to agree with what political scientists normally mean when they say "majority winner" – one whom, a majority would agree, should win in a race versus any opponent. And then, sorry, our counterexamples show the statement is just false. (Here's another interesting counterexample.)
PS: If you think I use this "random elections" model for my simulations generally, you are confused; the Bayesian Regret studies that make range voting look good compared to other systems, are based on trying a very large number of different probability models that are mostly not at all as simplistic as this one, and Range's superiority is highly robust across models.