The "Average" is derived by setting down a piece on each square of an empty board, counting up its mobility, and taking the average for all squares as the result.
In the mid 1970s, I thought of a way to improve this calculation.
When is the board empty? Doesn't it make more sense to calculate the mobility allowing for the presence of other pieces on the board?
In the opening position, the board is half-full and half-empty. What is the Rook's mobility in the opening position?
If you said "zero", you're wrong. The right answer is two. The Rook at a1 does useful work by defending the Pawn at a2, even though the Rook can't move to a2.
Should the answer be more than two? Does the Rook's support of the Pawn on a2 have any effect on a3 or a4? For now, we'll assume it doesn't ( and, really, it doesn't, but what if there were another Rook on a2 instead of a Pawn on a2? ), and get back to that question much, much later.With the board half-full, then, the Rook on a1 is certain of being able to reach a2 but has only a 50% chance of getting to a3, one chance in 4 to reach a4, and an 0.125 probability of hitting a5.
Of course, when all the pieces are still on the board, the Rook's chance of hitting a3 from a1 is 31/63, not 0.5, because there are 31 pieces on the board other than the Rook, and 63 squares they can be on, since the Rook itself is on a square. I'm simplifying things for the sake of discussion.If we combine the use of these probabilities with the "Average" method, we get values of 6.0 for the Rook, 4.75 for the Bishop, and 10.75 for the Queen. These values are much too low.
To simply say "combine the use of these probabilities..." may not explain well enough how this calculation is done.The reason these values are much too low is that the board doesn't stay 50% full for long; soon some pieces are captured, and the probability of getting from one place to another increases -- and at the same time, the value of the long-range pieces increases.
Let's start with a Rook on a1. It is certain to reach a2, so our working total begins with 1.0; its probability of reaching a3 is 0.5, so we add that to the total and get 1.5; its probability of reaching a4 is 0.25, so we add that to the total and get 1.75; repeat for a5, a6, a7; repeat for vertical retreats ( there are none from a1 ); repeat for right and left horizontal moves; now repeat for every possible starting square ( 64 of them ) and divide by 64 to get the average.
There are shortcuts you can use if you do this by hand...
When there are only 16 pieces on the board, the Rook on a1 has a 75% chance of reaching a3, 9 chances out of 16 for hitting a4, and so on.
The actual value of the Rook at any moment in the game ( up to the late endgame ) must be greater than its current mobility would indicate, because you have to take account of the fact that its mobility will increase as pieces are traded off.Could this explain the difference between the Standard values of the pieces and the Spielmann values? Spielmann played for mate, not for endgames.
This hypothesis fails, in my experience. The Spielmann values are correct in the endgame, at least until things get very very empty.When there are 20.3 pieces on the board, the probability that any square is occupied is 0.68275, and the mobility of the Rook is calculated as 7.88; the mobility of a Bishop is then 5.81, roughly a third of a Pawn more than that of a Knight.
We managed to get just the right answers. Does that mean we're done? Is the "Magic Number" 0.68275?
Now I have you all excited with the thought that it will be simple to calculate the exact and correct values of the chess pieces.