Why RRF beats score mixing for hybrid search

Your vector store returns cosine similarities between 0 and 1. Your BM25 index returns scores that might be 4.2 or 47.8 depending on term frequency and corpus statistics. Averaging them is meaningless — yet "weighted score mixing" is still the first thing most teams try when they add hybrid search.

The scale problem#

BM25 scores are unbounded and corpus-dependent: add documents and the same query's scores shift. Cosine similarity is bounded but its useful range is model-dependent — some embedding models cluster everything between 0.6 and 0.9. Any fixed mixing weight you tune today silently degrades as your corpus grows.

Rank fusion sidesteps it#

Reciprocal Rank Fusion ignores scores entirely and fuses positions:

def rrf(rankings: list[list[str]], k: int = 60) -> dict[str, float]:
    scores: dict[str, float] = {}
    for ranking in rankings:
        for rank, doc_id in enumerate(ranking):
            scores[doc_id] = scores.get(doc_id, 0.0) + 1.0 / (k + rank + 1)
    return scores

A document ranked #1 by dense search and #8 by BM25 gets 1/61 + 1/69 — no scale mismatch, no tuning drift, no re-calibration when the corpus doubles.

Why k = 60 works#

The constant dampens the head of each ranking. With small k, rank #1 dominates and fusion barely differs from taking one ranking. With large k, everything flattens toward equal weight. The literature's default of 60 is remarkably robust — we've moved it across corpora from 10k to 10M chunks without meaningful nDCG changes.

When score mixing is defensible#

If both retrievers emit calibrated probabilities — say, two dense indexes sharing an embedding model — score mixing preserves more information than ranks. In every mixed dense+sparse setup we've measured, RRF matched or beat tuned score mixing, while requiring zero maintenance.

The takeaway: fuse by rank, spend your tuning budget on the reranker.