{"id":2732,"date":"2026-05-01T07:00:09","date_gmt":"2026-05-01T11:00:09","guid":{"rendered":"https:\/\/mathvoices.ams.org\/featurecolumn\/?p=2732"},"modified":"2026-04-30T21:15:06","modified_gmt":"2026-05-01T01:15:06","slug":"the-story-your-ballot-doesnt-tell","status":"publish","type":"post","link":"https:\/\/mathvoices.ams.org\/featurecolumn\/2026\/05\/01\/the-story-your-ballot-doesnt-tell\/","title":{"rendered":"The Story Your Ballot Doesn\u2019t Tell"},"content":{"rendered":"

Every ballot quietly asks a question, and depending on the voting method we use we choose to listen to or discard little bits of the answer…<\/em><\/span><\/p>\n

The Story Your Ballot Doesn\u2019t Tell: The Mathematics of Single-Winner Elections<\/h1>\n

Anna Haensch
\nUniversity of Wisconsin – Madison<\/b><\/p>\n

Introduction: Who\u2019s the
\nSpoiler?<\/h1>\n

The great majority of national and state-level elections
\nin the United States are contests between two candidates representing
\nour two major political parties. Democrats and Republicans exist on the
\nleft and right end of the political spectrum, respectively, and while it
\nis a well-studied
\nfact<\/a> that candidates tend to creep as close as they can towards the
\ncenter to ensure the broadest base possible, they typically stay on
\ntheir side of the spectrum. Because of this, the results of elections
\noften come down to what\u2019s happening in the margins.<\/p>\n

Occasionally, however, it will happen that a third candidate gains
\ntraction in the race and this can have election-upending consequences.
\nIn these cases, the third candidate who takes their own bite out of the
\nvoter base is called the \u201cspoiler.” The weird thing though, is that
\nwhether or not this spoiler actually spoils, and for whom they spoil
\nthings is all a matter of how votes are cast and counted.<\/p>\n

In August 2022, Alaska held a special election for its single U.S.
\nHouse seat, and in the process, it became a case study<\/a> in how a
\nvoting method can shape an outcome. The race featured three candidates:
\nDemocrat Mary Peltola, and Republicans Nick Begich and Sarah Palin.
\nAlaska was using a voting system called \u201cInstant Runoff Voting,” in
\nwhich each voter cast a single ballot with candidates ranked from
\nfavorite to least favorite, with possible elimination rounds. Begich, who
\nwas closer to the center than Palin, was considered the compromise
\ncandidate, and according to most predictions, should
\nhave won<\/a> the election. But that\u2019s not what happened. Even though
\nBegich had the necessary broadness of support, he lacked that intensity
\nof first-choice preference, and Peltola won.<\/p>\n

The choice of voting method (a choice made by a vote of its own)
\noften determines what kind of support is the most valuable. Common
\nmethods like plurality and instant runoff both prioritize first-choice
\ndominance, but other methods, like Borda count give more value to broad
\nif only mildly-enthusiastic support. Given this range of possibilities
\nwhen it comes to setting up a voting regime, it\u2019s reasonable to wonder:
\nis the candidate really the spoiler, or is the spoiler democracy itself?
\nI know, provocative. To understand this better, let\u2019s strip away some of
\nthe specifics and get down to the geometry of the problem.<\/p>\n

What
\nDo Voters Look Like? Modeling Preferences with Gaussian Mixtures<\/h1>\n

Although most people\u2019s beliefs are multidimensional, to keep things
\nsimple let\u2019s assume that voters live in a one-dimensional space of
\npolitical ideology, with Democrats on the left and Republicans on the
\nright. Many polls exist that try to get at the precise distribution of
\nvoters along this left-right spectrum, but we\u2019ll use the distribution
\nfrom the Pew Research Foundation\u2019s National
\nPublic Opinion Reference Survey<\/a>. The 2025 results from that survey,
\nshown below, place respondents along a 5-point
\nspectrum from Democrat ($\u22122$), Lean
\nDemocrat ($\u22121$), No Lean \/ Refused ($0$), Lean Republican ($1$), and Republican ($2$). These results are plotted and fit to a
\n5-component Gaussian Mixture Model with standard deviation $\\sigma = 0.5$, and probability density
\nfunction, $f(x)$ (also
\nshown below).<\/p>\n

\n\"A
The 2025 Pew Research National
\nPublic Opinion Reference Survey<\/a> results are shown here as colored
\nbars with shades of blue indicating Democrat and shades of red
\nindicating Republican. These results are fit to a 5-component Gaussian
\nmixture model, with $\\sigma = 0.5$,
\nshown as a black line.<\/figcaption><\/figure>\n

We can imagine political candidates as being situated along the
\nleft-right spectrum. For example, imagine a 3-candidate election where a
\nleft candidate is placed at $\\ell = \u22121.8$, a center candidate is
\nplaced at $c = 0.4$, and right
\ncandidate is placed at $r = 1.6$. These candidates are shown
\nas colored triangles in the figure below.<\/p>\n

Assuming that voters prefer the candidate closest to them on the
\npolitical spectrum and in descending order according to proximity (also
\nsituation known as a single-peaked
\npreference<\/a>) we can begin to carve up the electorate into regions of
\ncandidate support. Any voter to the left of the midpoint between $\\ell$ and $c$ (shown as the left-hand dashed
\nblack line in Figure 2<\/a>) will prefer the left candidate, and
\nwe\u2019ll call this region $r_1$. On the other end of
\nthe spectrum, any voter to the right of the midpoint between $r$ and $c$ (shown as the right-hand dashed
\nblack line in Figure 2<\/a>) will prefer the right candidate, and
\nwe\u2019ll call this region $r_4$. In the remaining
\nregion between $r_1$
\nand $r_4$, voters will
\nalways prefer the center candidate, since they are the closest one. But
\nfor reasons that will become clear below we\u2019ll split this into two
\nregions, $r_2$ and
\n$r_3$ divided at the
\nmidpoint between the left and right candidates (shown as a solid black
\nline in Figure 2<\/a>). Now on to the election!<\/p>\n

\n\"Vertical
Left, center, and right candidates are shown as blue,
\nyellow, and red triangles, respectively, placed along the political
\nspectrum. The electorate is divided into four regions, $r_1, r_2, r_3$
\nand $r_4$.<\/figcaption><\/figure>\n

Plurality: The Simplest
\nSystem<\/h1>\n

Under plurality voting, each voter gets to vote for one candidate and
\nthe candidate with the most votes wins the election. Because of the
\ncareful work we did above, a candidate\u2019s vote share can be easily
\ncomputed as a sum of definite integrals. The left candidate vote share
\nas a function of each candidate\u2019s position is
\n$$\\begin{aligned}
\nS_{\\text{L}}^\\text{Plurality}(\\ell, c, r) & = & \\int_{R_1}f(x)
\n\\,\\,dx,
\n\\end{aligned}$$ where $f(x)$ is the probability
\ndensity function of the voter distribution, possibly the Gaussian
\nmixture model as described above. The vote shares for the center and
\nright candidates can be computed similarly as
\n$$\\begin{aligned}
\n S_C^\\text{Plurality}(\\ell, c, r) = \\int_{R_2\\cup R_3}f(x) \\,\\,dx
\n\\,\\,\\,\\,\\text{ and } \\,\\,\\,\\,S_R^\\text{Plurality}(\\ell, c, r) =
\n\\int_{R_4} f(x) \\,\\,dx.
\n\\end{aligned}$$
\nUsing the values of $\\ell, c,$ and $r$ from the previous section, we can
\nsee that the left candidate wins the election (see Figure 3<\/a>. Despite the fact that the voter
\ndistribution skews right and the right candidate has even moved slightly
\ntowards center to appeal to more voters, the presence of the slightly
\nright-of-center candidate has taken a big enough bite from the right
\ncandidate\u2019s base to cost them the election. In this way, the center
\ncandidate is acting as the spoiler<\/a>,
\nsince they lost the election and their removal from the election would
\nhave changed the outcome.<\/p>\n

\n\"Bar
Under plurality voting the left candidate wins the
\nelection.<\/figcaption><\/figure>\n

The Borda Count:
\nAsking Voters to Say More<\/h1>\n

Another well-known voting method is the Borda count which requires
\nvoters to order candidates from most-preferred to least-preferred. For a
\nrace with $n$ candidates, votes
\nare tabulated by assigning $n \u2212 1$ points to a first place
\nranking, $n \u2212 2$ to a second
\nplace ranking and so on, until the last place candidate who receives 0
\npoints. Any candidates not ranked will receive 0 points, the so-called
\npessimistic Borda count (although there are other
\noptions<\/a> for what to do with incomplete ballots). Assuming again that
\na voter will order their candidates according to their proximity on the
\npolitical spectrum, we can compute the left candidate\u2019s vote share as
\n$$\\begin{aligned}
\nS_L^\\text{Borda}(\\ell, c, r) & = & 2 \\cdot \\int_{R_1}f(x) \\,\\,dx
\n+ 1 \\cdot \\int_{R_2}f(x) \\,\\,dx.
\n\\end{aligned}$$
\nThe left candidate gets first choice ranks from
\nvoters in $r_1$, and
\nsecond choice ranks in $r_2$, but as soon as we
\npass over the midpoint line between $\\ell$ and $r$, the left candidate becomes the
\nthird choice. For the remaining candidates, the vote shares are computed
\nas
\n$$\\begin{aligned}
\nS_R^\\text{Borda}(\\ell, c, r) & = & 2 \\cdot \\int_{R_4}f(x) \\,\\,dx
\n+ 1 \\cdot \\int_{R_3}f(x) \\,\\,dx,
\n\\end{aligned}$$ and
\n$$\\begin{aligned}
\nS_C^\\text{Borda}(\\ell, c, r) & = & 2 \\cdot \\int_{R_2\\cup
\nR_3}f(x) \\,\\,dx + 1 \\cdot \\int_{R_1\\cup R_4}f(x) \\,\\,dx.
\n\\end{aligned}$$ Once again using the values of $\\ell, c,$ and $r$ from above, we now see that the
\ncenter candidate wins (see Figure 4<\/a>). Even though
\nthe center candidate is sitting in the valley of the distribution and
\nhas fewer first-choice votes than the other candidates, the fact that
\nthey are nobody\u2019s least favorite candidate gives them the boost they
\nneed to win handily. Who\u2019s the spoiler now?<\/p>\n

\n\"Bar
Under the Borda count method the center candidate wins the
\nelection.<\/figcaption><\/figure>\n

Instant Runoff
\nVoting: Elimination as a Strategy<\/h1>\n

In instant runoff voting, each voter casts a ballot ranking
\ncandidates from first choice to last choice (as in the Borda count), but
\nnow votes are tabulated in multiple rounds. In the first round, each
\ncandidate\u2019s share of first-choice votes are counted; if any candidate
\ncrosses the $50\\%$ threshold, they are
\ndeclared the winner and the election is over. In the more likely
\nscenario that no single candidate crosses the threshold, the candidate
\nwith the lowest share of first-choice votes is eliminated, and their
\nballots are reassigned to their second-choice candidates. At this point,
\nfirst-choice vote shares are counted, and in the case of our
\nthree-candidate election, the candidate with the highest number of votes
\nis elected. For each individual round, votes shares are computed exactly
\nas in plurality voting, with just the limits of integration depending on
\nwhich candidates are still in the race.<\/p>\n

Under instant runoff voting, the center candidate is eliminated after
\nthe first round (all of those second choice votes do nothing to help
\nthem here). In the second round of voting, the left candidate gets all
\nof the votes from regions $r_1$ and $r_2$ and the right
\ncandidate gets all of the votes from regions $r_3$ and $r_4$. Now the fact that the
\nright candidate is slightly closer towards center means that they are
\nable to capture enough of the votes from the eliminated candidate to
\nbring them to victory (see the figure below).<\/p>\n

\n\"Bar
Under instant runoff voting the right candidate wins the
\nelection.<\/figcaption><\/figure>\n

A Quick Detour Into
\nCandidate Strategy<\/h1>\n

This is an interesting story in the context of Hotelling\u2019s
\nLaw<\/a> from economics which says that in any market it is rational for
\nproducers to make their products as similar as possible. In the world of
\npolitics, this law specializes to the Median Voter
\nTheorem<\/a> which says that for a single-peaked
\nelectorate<\/a>, candidates will converge to the median voter. At first
\nit looks like we\u2019ve spotted a violation of the Median Voter Theorem
\nsince the left and right candidates are able to win in some cases, but
\nof course in a real-life scenario neither the candidates nor the
\nelectorate are completely stable. You can imagine (and we\u2019ve
\nmodeled!<\/a>) what happens when candidates move in opportunistic ways.
\nThere\u2019s also the question of whether voters actually turn out to vote,
\nwhich can be impacted by all kinds of things including strategic
\nvoting<\/a>. Put another way, the Median Voter Theorem tells us where the
\nequilibrium should<\/em> be, but the dynamics of getting
\nthere (or not) becomes more interesting with the addition of movement
\nand a little bit of reality.<\/p>\n

Conclusion:
\nMathematics as Democratic Infrastructure<\/h1>\n

Returning now to the Alaska special election. Begich was the
\ncenter-most candidate and seemed like the good compromise and likely
\nwinner. While he didn\u2019t have the strength of first-choice votes, he also
\ndidn\u2019t have the anchor of last-choice votes. Functionally, he was a lot
\nlike the center candidate in the previous sections. Under a voting
\nmethod like Borda count, Begich almost certainly would have won, but
\nunder plurality or instant runoff, he would (and did) lose. The
\nspecifics of the left and right candidates in the worked-out example
\nabove are slightly different than what happened in Alaska (notably,
\nPeltola would have won both plurality and instant runoff), but they
\ndemonstrate the brittleness of the electoral process.<\/p>\n

Mathematics is an important piece of the democracy, and this is not a
\npurely academic exercise. All over the world governments have been
\nthinking about what it would look like to adopt different voting
\nmethods. In the immediate aftermath of the Alaska special election,
\nstates were all over the place in terms of ranked choice voting. Some
\nstates like Florida
\nand Tennessee<\/a> banned ranked choice voting, while others had active
\ncampaigns to adopt it. In Alaska, polls
\nshowed that $54\\%$ of Alaskans<\/a> supported repealing ranked choice
\nvoting, although when put to a vote the
\nrepeal ultimately failed<\/a>. As of March 2026, nineteen
\nstates have banned ranked-choice voting<\/a>.<\/p>\n

While mathematics can\u2019t tell us which system is right, or which
\ncandidate is right, it can tell us what a voting system prioritizes.
\nEvery ballot quietly asks a question, and depending on the voting method
\nwe use we choose to listen to or discard little bits of the answer. That
\nis why the voting system, which does so much to shape our democracy and
\ngovernance, deserves at least as much scrutiny as the candidates
\nthemselves.<\/p>\n

\n\"The
The author getting ready to vote in Madison, Wisconsin.<\/figcaption><\/figure>\n","protected":false},"excerpt":{"rendered":"

Every ballot quietly asks a question, and depending on the voting method we use we choose to listen to or discard little bits of the answer… The Story Your Ballot Doesn\u2019t Tell: The Mathematics of Single-Winner Elections Anna Haensch University of Wisconsin – Madison Introduction: Who\u2019s the Spoiler? The greatRead More →<\/a><\/span><\/p>\n","protected":false},"author":2,"featured_media":1599,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"advanced_seo_description":"","jetpack_seo_html_title":"","jetpack_seo_noindex":false,"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,232,137,25],"tags":[141],"class_list":["entry","author-uwhitcher","post-2732","post","type-post","status-publish","format-standard","has-post-thumbnail","category-213","category-anna-haensch","category-math-and-social-sciences","category-probability-and-statistics","tag-voting"],"jetpack_featured_media_url":"https:\/\/i0.wp.com\/mathvoices.ams.org\/featurecolumn\/wp-content\/uploads\/sites\/2\/2023\/03\/mathvoices-banner-feat-col.png?fit=2760%2C580&ssl=1","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/posts\/2732","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/comments?post=2732"}],"version-history":[{"count":8,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/posts\/2732\/revisions"}],"predecessor-version":[{"id":2746,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/posts\/2732\/revisions\/2746"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/media\/1599"}],"wp:attachment":[{"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/media?parent=2732"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/categories?post=2732"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/tags?post=2732"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}