Demonstration 1 of 1
The public leader's advantage, shown and earned, by number of candidates screened
How much of the public leader's visible advantage is still there on private rows as more near-equal candidates are screened?
Each candidate's public RMSE is its true quality plus sampling noise from only 400 rows. Taking the minimum over more candidates picks more noise: the advantage the leaderboard shows keeps growing with the candidate count, while the advantage that exists on private rows or fresh rows changes far less (up or down by about 0.01 across seeds), so the noise in the visible lead keeps growing. Cross-validation uses 1,500 rows, so it is usually a quieter judge here, but it is also one more selection, so its pick loses ground too.
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s_c is candidate c's RMSE (lower is better), the superscripts pub and priv mark public and private rows, median over c is taken across the screened candidates and pick is the candidate with the lowest public RMSE.
Predict first. With 100 candidates screened, the public leader looks 0.062 RMSE better than the median candidate on the public rows. How much of that advantage shows up on private rows?
Choose an example
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Constructed example: seeded synthetic regression data and 3 to 100 ridge candidates per competition, measured by the chapter activity.
Calculated values
- Lead shown on public rows
- 0.052
- Lead earned on private rows
- 0.018
- Lead earned on fresh rows
- 0.017
- Public leader also private leader
- 23% of splits
- Regret: public / CV / random
- 0.017 / 0.009 / 0.042
With 30 candidates the public leader looks 0.052 RMSE better than the median candidate on the public rows, but earns 0.018 on private rows and 0.017 on fresh rows: 0.052 - 0.018 = 0.034 of the visible lead was selection noise, and 35% of it is real. The public leader is also the private leader in 23% of splits. Against the best candidate, picking by public score costs 0.017 RMSE on fresh rows, picking by cross-validation 0.009 and picking at random 0.042.
Worked steps
- Noise in the visible lead: 0.052 - 0.018 = 0.034 (shown minus earned on private rows).
- Share of the visible lead that is real: 0.018 / 0.052 = 0.351.
- Cross-validation against the public leader: 0.017 - 0.009 = 0.008 less regret.
- Public leader against a random pick: 0.042 - 0.017 = 0.025 less regret.
Use the idea
Treat a small public lead as weak evidence, judge each change on paired local folds, and fill the final submission slots with candidates that have supported local evidence rather than only the public leader.
Where the conclusion applies
Constructed data with ridge variants standing in for the many near-equal models of a Playground competition, and one fixed public share (20%). Cross-validation had less regret than the public pick in every state here, by 0.002 to 0.01, but under another seed the two were tied (cross-validation very slightly worse) in one state; it is a property of the training set (1,500 rows) being much larger than the public set (400 rows). With a relatively larger public set or a leaking cross-validation the order could change. Costs here are RMSE units on a target with noise standard deviation 3.
Check your understanding: If the public lead overstates so much, why not ignore the public score and pick at random?
Chapter 61 source: section "Reading the Leaderboard: Signal vs. Noise".