Random Voting and Rawls June 5, 2025

Caveat lector: It’s quite possible that my musings below are either entirely wrong or simply represent trivial, well-known results in social choice theory. Still, I thought they were funny, so maybe you’ll enjoy them, too.

If the lottery is an intensification of chance, a periodical infusion of chaos in the cosmos, would it not be right for chance to intervene in all stages of the drawing and not in one alone? Is it not ridiculous for chance to dictate someone’s death and have the circumstances of that death—secrecy, publicity, the fixed time of an hour or a century—not subject to chance?

Borges, The Lottery in Babylon, translation by John M. Fein

Consider an election with two candidates, $A$ and $B$, and say some fraction $0 \le a \le 1$ of the population votes for candidate $A$. How should we decide the result of the election?

This seems like an almost silly question: surely, if $a > 1/2$, then we should pick $A$, and if $a < 1/2$, we should pick $B$. (In the event of $a = 1/2$, we would presumably run some sort of tie-breaker.) This is what pretty much any sane democracy would do; what other reasonable choices might one make?

Chwazi for president

To start off with, there are a couple of downsides to the simple majority-wins-every-time strategy. One obvious one is the so-called tyranny of the majority, in which the majority’s preferences override the minority’s preferences every single time. In the real world, we typically try to get around this by setting up our institutions to protect minority interests; for instance, the United States Senate gives every state two senators, so that less-populous states still have adequate representation in Congress.

But could we instead solve this problem with our voting system? Here’s a funny proposal: say we randomly select candidate $A$ with probability $a$ and candidate $B$ with probability $1-a$! If you like, this is equivalent to selecting one ballot uniformly at random and using that ballot to dictate the winner.¹

Now I appreciate that it takes a special kind of person—someone who like random algorithms, say—to enjoy this kind of thinking, but indulge me here. In some vague sense, this system is quite fair: the more people who vote for $A$, the more likely she is to win, but the minority still has a chance of getting their candidate in; in fact, this chance is precisely proportional to how large the minority is. Equivalently, you could imagine that, in $a$ percent of all universes, $A$ wins, and in $1-a$ percent of all universes, $B$ wins.

An objection from Rawls

Okay, so the math is elegant, but something still feels a bit off about this. Surely, if the majority selects someone, that person should just win the election? Can we provide a more compelling argument against random ballot voting than “it just feels wrong?”

Here’s one, based on the famous veil of ignorance thought experiment² due to John Rawls. The argument goes like this: to determine what constitutes a fair society, pretend for a moment that you are behind a “veil of ignorance” that prevents you from knowing who you are going to be in society; for all you know, it could be anyone—man or woman, rich or poor, sick or healthy. The guiding principle here is that you should construct a society that you’d be happy entering as a randomly selected person.

A straightforward application of this principle to our voting problem is that we should set up our system for determining the election outcome such that it maximizes our utility, assuming that we will end up as a voter selected uniformly at random. Note that this means that you end up as a supporter of $A$ with probability $a$, and as a supporter of $B$ with probability $1-a$.

Perhaps the simplest utility function that one could imagine is: $$X = \begin{cases} 1 & \textrm{if your candidate wins} \\ 0 & \textrm{otherwise.} \end{cases}$$ Now we’re equipped to analyze a whole range of probabilistic strategies. Let $p$ be the probability that $A$ wins. By setting $p$, we can determine how random we want the outcome of the election to be. We can straightforwardly compute the expectation: $$\begin{align*} E[X] &= E[X \mid \textrm{I support $A$}] \cdot a + E[X \mid \textrm{I support $B$}] \cdot (1-a) \\ &= a p + (1-a)(1-p) \\ &= (2a-1)p + (1-a). \end{align*}$$ What does this suggest? Notice that if $a > 1/2$, then the slope is positive, so we should set $p = 1$ to maximize $E[X]$. On the other hand, if $a < 1/2$, the slope is negative, so $p = 0$ is optimal. In other words, the veil of ignorance suggests that we should actually apply the simple majority-wins-every-time strategy!

Everyone loves an underdog

Okay, that’s a little disappointing, but it suggests an interesting avenue for exploration: what do more interesting utility functions give us? Here’s a somewhat bizarre-looking one that I’ll propose calling the underdog utility: $$X = \begin{cases} 2 - \Pr\{\textrm{your candidate wins}\} & \textrm{if your candidate wins} \\ 0 & \textrm{otherwise.} \end{cases}$$ An intuitive explanation of this function is that:

1. If your candidate wins, you always get some utility. But, you like excitement and backing the underdog. The less favored your candidate was, the more excited you are if she wins.
2. If your candidate loses, you get no utility.

Okay, now let’s compute the expectation. We have: $$\begin{align*} E[X] &= E[X \mid \textrm{I support $A$}] \cdot a + E[X \mid \textrm{I support $B$}] \cdot (1-a) \\ &= (2-p) a p + (2-(1-p)) (1-a) (1-p) \\ &= 2 a p - a - p^2 + 1. \end{align*}$$ This is a downward-facing parabola whose vertex—and therefore maximum—occurs at $p = a$. In other words, with this utility function, the veil of ignorance precisely supports the random ballot voting system, in which the probability of $A$’s winning is equal to the share of the vote that she received!

Now, I don’t know if I would call this an excellent argument for random ballots, but it’s at least kind of funny, in my opinion.