So if it's that, why do you use = symbol when I and Hirosi Kano use ≈ ?!
Well, foremost because the latter symbol is not on my keyboard. :-) But empirical piece values are like measurements in physics; they are never precise, there is always a margin of error. So equality in the mathematical sense never exists, the meaning of '=' here always is "equal to within the precision of the measurement". There also exists the problem that the entire concept of piece values that can be added to get the strength of the army is only an approximation. Q loses to R+B more often than not in Capablanca Chess when all Chancellors and Archbishops are still on the board, while without the latter two, Q usually beats R+B (as in orthodox Chess). So presence of other material affects the value, even when nothing else is known from the position than the material that is present. (Which of course was already known from the Bishop as well, that it gets much better if the complementary Bishop is present.)
But this has nothing to do with the variation being large or small; in statistics average and variation are independent quantitities, an a set of wildly varying values can still have a very precise average. It just means you have to realize that a piece value is an average, with all limitations that follow from that. As the proverb goes: "A statician waded trough a river that was on average 1m deep. He drowned!".
This is why chess players employ the 'rule of thumb' that a simple advantage is better than an equal complex advantage. With Q vs R+B (part of a pair) you can have exactly the same advantage (piece-value-wise) as with R+B+P vs R+B. But the latter imbalance (a plain P ahead) gives a much more predictable outcome than Q vs R+B, which the Q player might very well lose. As far as the average is concerned 50% win + 50% draw is the same result as 75% win + 25% loss.
Why I count A≈7 (if accurately, 7.44) Archbishop has a less color-switching ability than Centaur or Chancellor.
Well, it is also possible to 'count' Q=1 and P=3. But if you use that as a guide for which pieces to trade for which other, you won't win many games. Likewise with 'counting' A=7.44. You will voluntarily trade into badly lost positions, just because you though that A was worth less than Rook + Knight, while in practice the Rook and Knight you are left with then lose most of the time. What you can imagine and what would really work are entirely different things.
Well, foremost because the latter symbol is not on my keyboard. :-) But empirical piece values are like measurements in physics; they are never precise, there is always a margin of error. So equality in the mathematical sense never exists, the meaning of '=' here always is "equal to within the precision of the measurement". There also exists the problem that the entire concept of piece values that can be added to get the strength of the army is only an approximation. Q loses to R+B more often than not in Capablanca Chess when all Chancellors and Archbishops are still on the board, while without the latter two, Q usually beats R+B (as in orthodox Chess). So presence of other material affects the value, even when nothing else is known from the position than the material that is present. (Which of course was already known from the Bishop as well, that it gets much better if the complementary Bishop is present.)
But this has nothing to do with the variation being large or small; in statistics average and variation are independent quantitities, an a set of wildly varying values can still have a very precise average. It just means you have to realize that a piece value is an average, with all limitations that follow from that. As the proverb goes: "A statician waded trough a river that was on average 1m deep. He drowned!".
This is why chess players employ the 'rule of thumb' that a simple advantage is better than an equal complex advantage. With Q vs R+B (part of a pair) you can have exactly the same advantage (piece-value-wise) as with R+B+P vs R+B. But the latter imbalance (a plain P ahead) gives a much more predictable outcome than Q vs R+B, which the Q player might very well lose. As far as the average is concerned 50% win + 50% draw is the same result as 75% win + 25% loss.
Well, it is also possible to 'count' Q=1 and P=3. But if you use that as a guide for which pieces to trade for which other, you won't win many games. Likewise with 'counting' A=7.44. You will voluntarily trade into badly lost positions, just because you though that A was worth less than Rook + Knight, while in practice the Rook and Knight you are left with then lose most of the time. What you can imagine and what would really work are entirely different things.