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which givesP(err|F) = 1− |S|N + 1 N(d)=1− F −∑ FIn the above (a) is due to Bayes rule, (b) considers thatand thatP(S 0 , ̂φ = 0)P(err|S 0 , ̂φ = 0)NF∑ ( )|Sφ |−1+P φφ=1φ=1 P φP(err|S 0 , ̂φ = 0) = P(err|S 0 , ̂φ ≠ 0) = 1+P(S 0 , ̂φ ≠ 0)P(err|S 0 , ̂φ ≠ 0)(A.3). (A.4)= P(S 0 , ̂φ = 0)·1+P(S 0 , ̂φ ≠ 0)·1 = P(S 0 ) = N −|S|N ,(c) considers that P(S φ ) = |S|N , that P(̂φ = φ|S φ ) = 1−P φ , that P(err|S φ , ̂φ ≠ φ) = 1, and thatP(err|S φ , ̂φ = φ) = |S φ|−1,|S φ |and finally (d) considers that ∑ Fφ=1 |S φ| = |S|.□Proof of Lemma 3 LetC F be the total number ofN-tuples v that introduce F effective featurecategories. Thenρ! N!C F =(ρ−F)!(N −F)! FN−F (A.5)where the first term ρ!(ρ−F)!describes the total number of ways F categories can be chosen to hostsubjects, the second term N!(N−F)!describes the total number of ways F initial people, out of Npeople, can be chosen to fill theseF categories, and where the third termF N−F describes the totalnumber of ways the F effective categories can be freely associated to the rest N − F subjects.Finally we note thatP(F) = C F∑ Ni=1 C ,iwhich completes the proof.□Example 14 Consider the case where ρ = 9,N = 5,F = 3. Then the cardinality of the set of allpossible N-tuples that span F = 3 effective categories, is given by the product of the followingthree terms.– The first term is (ρ·(ρ−1)···(ρ−F +1)) = ρ!(ρ−F)!= 9·8·7 = 504 which describesthe number of ways one can pick whichF = 3 categories will be filled.– Having picked theseF = 3 categories, the second term is(N ·(N −1)···(N −F +1)) =N!(N−F)!= 5 · 4 · 3 = 60, which describes the number of ways one can place exactly onesubject in each of these picked categories.