A Clustering That Knows Nothing Scores 0.5033. One That Gets Half the Items Right Scores 0.3118.

Days 4, 15, 22, 25 and 49 of this series built clustering algorithms, and every one of them ended by printing a number. This page is about the number.

Score two labellings of the same items against each other and the whole input is a contingency table. That means the chance level is a finite sum — something you compute, not something you shuffle for — and it is emphatically not zero.

See it: https://dev48.infy.uk/ml/day73-clustering-metric-chance.html

The comparison that matters

240 items. Under the fixed-margin permutation null, the expected mutual information has a closed form, and a labelling that knows nothing at all about the truth scores an exact 0.5637 on homogeneity at k′ = 100.

labelling	homogeneity	chance-corrected
uniformly random, 93 clusters	0.5033	0.0003
genuinely gets half the items right	0.3118	0.2919

The raw score ranks them backwards. The chance-corrected score ranks them correctly, and by 0.29 of the index's whole range.

Nothing is wrong with the random labelling except that it has a lot of clusters. Homogeneity asks "is each cluster pure?", and a cluster of size 2 is almost always pure. Push k′ up and you approach 1.0 while learning nothing.

Where the crossing happens

The exact k′ at which pure chance overtakes the half-right clustering:

index	homogeneity	geometric	arithmetic
chance overtakes at k′ =	50	92	116

Three normalisations of the same quantity, three different answers to "how many clusters before this metric is lying to me". None of them is far enough away to be safe: 50 clusters on 240 items is not an unusual configuration.

# the fix is not a bigger sample, it is a subtraction
adjusted = (observed - expected_under_null) / (max_possible - expected_under_null)
# expected_under_null has a closed form for the fixed-margin case.
# Compute it. Do not assume it is zero.

The rule

If you report a clustering score without subtracting its chance level, the number is a function of your k as much as your algorithm — and comparisons across different k are meaningless. Report the adjusted variant, or report k alongside and accept that the comparison is not one.

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