feat!: N-team outcome prediction with draw mass, replacing the 2-team panic
`predict_outcome` asserted `teams.len() == 2` and returned `[p, 1 - p]`, allocating no probability to a draw even with `p_draw > 0`. For a draw-enabled model the numbers were simply wrong, at any team count. It now returns `Result<Prediction, InferenceError>` and supports N teams. Two algorithms, both deterministic: - Who finishes first. Performances are independent Gaussians, so this separates into a one-dimensional integral per team rather than a multivariate orthant probability. Adaptive Gauss-Kronrod evaluates it to ~1e-15, matching the exact two-team closed form. - A specific finishing order. The factor graph only constrains rank-adjacent teams, so a full order is a chain of local constraints, not a general orthant integral. That chain collapses into a sequential recursion over cumulative integrals: O(teams * grid) per order. Fixed-node Gauss-Hermite is the obvious tool for the first and is a trap: when a rival's sigma is small the CDF product becomes a step narrower than the node spacing, and the nodes step over it. Measured 4.4e-4 off the closed form on a mildly skewed matchup and 1.7e-2 on a small-sigma one, while still returning something that looks like a probability. Adaptive refinement is what makes that case safe, and `win_probabilities_survive_a_rival_with_a_tiny_sigma` pins it down. The acceptance test is an identity rather than a golden: the outcome space is exhaustive and disjoint, so the probabilities sum to one. Any drift is integration error and nothing else. Gauss-Hermite failed it at 4.4e-4; this holds to ~1e-9. Also from #21: unknown keys are now reported rather than dropped, so a team of strangers can no longer produce a confident-looking prediction. `predict_quality` returns `Result` for the same reason. BREAKING CHANGE: `predict_outcome` returns `Result<Prediction, _>` instead of `Vec<f64>`; `predict_quality` returns `Result<f64, _>`. Refs #21, #39 Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_011hcFjNDmHXZF8URGLku5zZ
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//! Prediction API: N-team outcomes, draw mass, and the error paths that used
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//! to be panics or silent wrong answers.
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use trueskill_tt::{History, InferenceError, MAX_PREDICTED_TEAMS};
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fn history_with(names: &[&'static str], p_draw: f64) -> History {
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let mut h = History::builder().p_draw(p_draw).build();
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// Give every competitor a recorded skill by playing a small round robin.
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for pair in names.windows(2) {
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h.record_winner(&pair[0], &pair[1], 1).unwrap();
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}
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h.converge().unwrap();
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h
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}
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#[test]
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fn unknown_keys_are_reported_not_silently_dropped() {
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let h = history_with(&["a", "b"], 0.0);
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let err = h
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.predict_outcome(&[&[&"a"], &[&"ghost"]])
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.expect_err("an unknown key must not yield a confident prediction");
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assert_eq!(err, InferenceError::UnknownKey { team: 1, member: 0 });
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// Every prediction entry point, not just one.
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assert!(
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h.predict_win_probabilities(&[&[&"a"], &[&"ghost"]])
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.is_err()
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);
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assert!(h.predict_quality(&[&[&"a"], &[&"ghost"]]).is_err());
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assert!(h.predict_ranking(&[&[&"a"], &[&"ghost"]], &[0, 1]).is_err());
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}
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#[test]
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fn an_entirely_unknown_team_is_an_error() {
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let h = history_with(&["a", "b"], 0.0);
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let err = h.predict_outcome(&[&[&"a"], &[&"x", &"y"]]).unwrap_err();
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assert_eq!(err, InferenceError::UnknownKey { team: 1, member: 0 });
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}
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#[test]
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fn degenerate_team_shapes_are_errors_rather_than_panics() {
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let h = history_with(&["a", "b"], 0.0);
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assert_eq!(
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h.predict_outcome(&[&[&"a"]]).unwrap_err(),
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InferenceError::NotEnoughTeams { got: 1 }
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);
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assert_eq!(
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h.predict_outcome(&[]).unwrap_err(),
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InferenceError::NotEnoughTeams { got: 0 }
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);
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assert_eq!(
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h.predict_outcome(&[&[&"a"], &[]]).unwrap_err(),
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InferenceError::EmptyTeam { team: 1 }
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);
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}
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#[test]
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fn more_than_two_teams_no_longer_panics() {
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let h = history_with(&["a", "b", "c"], 0.0);
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let p = h
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.predict_outcome(&[&[&"a"], &[&"b"], &[&"c"]])
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.expect("three teams must be supported");
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assert!((p.total() - 1.0).abs() < 1e-6, "total = {}", p.total());
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// Three teams, no draws possible: exactly the six strict orderings.
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assert_eq!(p.outcomes().len(), 6);
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}
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#[test]
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fn the_outcome_space_is_capped_rather_than_hanging() {
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let names: Vec<&'static str> = vec!["a", "b", "c", "d", "e", "f", "g", "h"];
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let h = history_with(&names, 0.0);
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let teams: Vec<&[&&'static str]> = Vec::new();
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let _ = teams;
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let too_many: Vec<Vec<&&str>> = names.iter().map(|n| vec![n]).collect();
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let refs: Vec<&[&&str]> = too_many.iter().map(Vec::as_slice).collect();
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let err = h.predict_outcome(&refs).unwrap_err();
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assert_eq!(
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err,
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InferenceError::TooManyTeams {
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got: 8,
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max: MAX_PREDICTED_TEAMS
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}
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);
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// The cheap paths stay available at any size.
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let wins = h.predict_win_probabilities(&refs).unwrap();
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assert_eq!(wins.len(), 8);
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assert!(
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(wins.iter().sum::<f64>() - 1.0).abs() < 1e-6,
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"win probabilities must still sum to one: {wins:?}"
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);
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}
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/// The defect that made every draw-enabled prediction wrong: `[p, 1 - p]`
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/// allocated no mass to a draw even with `p_draw > 0`.
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#[test]
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fn a_draw_carries_probability_mass_when_p_draw_is_positive() {
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let h = history_with(&["a", "b"], 0.25);
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let p = h.predict_outcome(&[&[&"a"], &[&"b"]]).unwrap();
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let draw = p.probability_of(&[0, 0]);
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assert!(draw > 0.0, "a draw-enabled model must give draws mass");
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assert!((p.total() - 1.0).abs() < 1e-6, "total = {}", p.total());
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let wins = p.win_probabilities();
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assert!(
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(wins.iter().sum::<f64>() + draw - 1.0).abs() < 1e-6,
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"wins {wins:?} plus draw {draw} must be the whole space"
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);
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assert!(
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(p.shared_first_place() - draw).abs() < 1e-12,
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"a two-team draw is a shared first place"
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);
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}
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#[test]
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fn a_zero_draw_probability_admits_no_ties() {
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let h = history_with(&["a", "b"], 0.0);
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let p = h.predict_outcome(&[&[&"a"], &[&"b"]]).unwrap();
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assert_eq!(p.probability_of(&[0, 0]), 0.0);
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assert!(p.shared_first_place() < 1e-12);
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}
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/// The two routes to a win probability run through entirely different
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/// algorithms — adaptive quadrature versus the enumerated chain recursion —
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/// so agreement between them is a real cross-check, not a tautology.
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#[test]
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fn the_cheap_and_exhaustive_paths_agree() {
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for p_draw in [0.0, 0.1] {
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let h = history_with(&["a", "b", "c"], p_draw);
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let teams: &[&[&&str]] = &[&[&"a"], &[&"b"], &[&"c"]];
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let cheap = h.predict_win_probabilities(teams).unwrap();
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let exhaustive = h.predict_outcome(teams).unwrap().win_probabilities();
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for (i, (a, b)) in cheap.iter().zip(&exhaustive).enumerate() {
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assert!(
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(a - b).abs() < 1e-6,
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"p_draw={p_draw} team {i}: quadrature {a} vs enumeration {b}"
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);
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}
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}
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}
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#[test]
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fn predict_ranking_agrees_with_the_distribution() {
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let h = history_with(&["a", "b", "c"], 0.1);
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let teams: &[&[&&str]] = &[&[&"a"], &[&"b"], &[&"c"]];
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let dist = h.predict_outcome(teams).unwrap();
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for (ranks, expected) in dist.outcomes() {
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let direct = h.predict_ranking(teams, ranks).unwrap();
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assert!(
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(direct - expected).abs() < 1e-9,
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"ranks {ranks:?}: {direct} vs {expected}"
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);
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}
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}
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#[test]
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fn predict_ranking_checks_its_shape() {
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let h = history_with(&["a", "b"], 0.0);
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let err = h
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.predict_ranking(&[&[&"a"], &[&"b"]], &[0, 1, 2])
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.unwrap_err();
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assert!(matches!(
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err,
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InferenceError::MismatchedShape {
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expected: 2,
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got: 3,
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..
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}
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));
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}
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#[test]
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fn the_stronger_competitor_is_favoured() {
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let mut h = History::builder().build();
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for t in 1..=10 {
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h.record_winner(&"strong", &"weak", t).unwrap();
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}
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h.converge().unwrap();
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let p = h.predict_outcome(&[&[&"strong"], &[&"weak"]]).unwrap();
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let (best, _) = p.most_likely().expect("a most likely outcome");
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assert_eq!(best, &[0, 1], "the winner should be favoured");
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let wins = p.win_probabilities();
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assert!(wins[0] > wins[1], "{wins:?}");
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}
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/// Unequal team sizes change the draw margin, because inference derives it
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/// from the teams' betas. Prediction has to follow, or it describes a
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/// different model than the one that will be fitted.
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#[test]
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fn team_size_affects_the_prediction() {
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let mut h = History::builder().p_draw(0.2).build();
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h.event(1)
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.team(["a", "b"])
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.team(["c"])
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.winner(0)
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.commit()
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.unwrap();
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h.converge().unwrap();
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let p = h.predict_outcome(&[&[&"a", &"b"], &[&"c"]]).unwrap();
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assert!((p.total() - 1.0).abs() < 1e-6, "total = {}", p.total());
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assert!(p.probability_of(&[0, 0]) > 0.0);
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}
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