Well, the values that Kaufman found were B=N=3.25, R=5 and Q=9.75. So also there 2 minor > R+P (6.5 vs 6), minor > 3P (3.25 vs 3), 2R > Q (10 vs 9.75). Only 3 minor = Q. Except of course that this ignores the B-pair bonus; 3 minors is bound to involve at least one Bishop, and if that broke the pair... So in almost all cases 3 minors > Q.
You can also see the onset of the leveling effect in the Q-vs-3 case: it is not only bad in the presence of extra Bishops (making sure the Q is opposed by a pair), but also in the presence of extra Rooks. These Rooks would suffer much more from the presence of three opponent minors than they suffer from the presence of an opponent Queen. (But this of course transcends the simple theory of piece values.) So the conclusion would be that he only case where you have equality is Q vs NNB plus Pawns. This could very well be correct without being in contradiction with the claim that 2 minors are in general stronger.
BTW, in his article Kaufman already is skeptical about the Q value he found, and said that he personally would prefer a value 9.50.
If you don't recognize teh B-pair as a separate term, then it is of course no miracle that you find the Bishop on average to be stronger. Because i a large part of the cases it will be part of a pair.
Well, the values that Kaufman found were B=N=3.25, R=5 and Q=9.75. So also there 2 minor > R+P (6.5 vs 6), minor > 3P (3.25 vs 3), 2R > Q (10 vs 9.75). Only 3 minor = Q. Except of course that this ignores the B-pair bonus; 3 minors is bound to involve at least one Bishop, and if that broke the pair... So in almost all cases 3 minors > Q.
You can also see the onset of the leveling effect in the Q-vs-3 case: it is not only bad in the presence of extra Bishops (making sure the Q is opposed by a pair), but also in the presence of extra Rooks. These Rooks would suffer much more from the presence of three opponent minors than they suffer from the presence of an opponent Queen. (But this of course transcends the simple theory of piece values.) So the conclusion would be that he only case where you have equality is Q vs NNB plus Pawns. This could very well be correct without being in contradiction with the claim that 2 minors are in general stronger.
BTW, in his article Kaufman already is skeptical about the Q value he found, and said that he personally would prefer a value 9.50.
If you don't recognize teh B-pair as a separate term, then it is of course no miracle that you find the Bishop on average to be stronger. Because i a large part of the cases it will be part of a pair.