

On the regular 8x8 chessboard, we are taught there are three kinds of material: pieces, pawns, and the kings.
Pieces are given their value in terms of pawns. For example, most people recognize that a bishop is "worth" 3 pawns.
We generally do not include a value for the king, since it is never traded during the course of play.
We are also taught there there are two types of pieces: "major" and "minor" pieces. A major piece is one that can deliver
checkmate, even if only its friendly king is on the board. Since it is possible to mate with king and rook vs. a lone king,
we say that a rook is a major piece. A queen is likewise a major piece. But, since a bishop cannot deliver such a checkmate,
it is designated as a minor piece. The same is true of a knight: it cannot mate a lone enemy king by itself. Therefore,
it, too, is considered a minor piece. Now it is true that a bishop and knight can mate an enemy king if they are both
on the board, but this is not what we were talking about.
Before going any further, a short table of relative piece values is presented below. Please note: The data is not a
"universal truth" of any kind, just a guideline which can be used to help you determine if you should engage in a
particular trade of some kind.
Trice's Chess Piece Values 
Piece 
Value 
Piece 
Value 
Pawn 
1.00 
Archbishop 
6.50 
Knight 
2.50 
Chancellor 
8.25 
Bishop 
3.00 
Queen 
8.75 
Rook 
4.75 
King 



Examine the table shown above. How can it help you play Trice's Chess? Let's say you see a combination that
can be constructed, and in the end you can win two of your opponent's knights, but you will be required to
trade your bishop and one pawn. The two knights are worth 2.50 + 2.50 = 5.00 pawns. The bishop + pawn is worth
3.00 + 1.00 = 4.00 pawns. As a result of this trade, you would "win a pawn" in terms of material distribution.
Now instead, suppose you can win an archbishop and pawn if you exchange your chancellor? According to the table, the
archbishop and pawn would be worth 6.50 + 1.00 = 7.50 pawns, but your chancellor is worth 8.25 pawns. In this case,
you should try to keep your chancellor and NOT execute that trade.


In 1876, a mathematician named Taylor published a paper where he conjectured that the value of a chess piece is proportional to its ability to safely
check an enemy king on another otherwise vacant board. This concept is referred to as a safe check.
Taylor's equations were generalized for a square board of side length n, but in Trice's Chess, we have a rectangular board, so we cannot
use these equations! In 1999, Ed Trice recomputed the safe check values for all of the chess pieces, plus the new pieces, on a rectangular
board that is f files wide and r ranks in height. For the purpose of demonstrating this, the diagrams here will contain varying
numbers of files and ranks (and the formulas work for any size board, even the 8by8 chessboard.)
To begin, examine the diagrams shown above. An aura of Rook checks is shown in (A), illustrating the horizontal and vertical components of its power.
This allows an enemy King that is placed in check the potential to make one of three different types of capturing responses. If the King is in any one
of the four corners of the board when it is in check (B), you can see that there are two squares marked with an "X" showing where it could make a
trivial capture. These are the unsafe checking squares, so what remain would be the squares of safe check. If the King is on the edge
of the board (C) but not in one of the four corners, then it has three squares available to capture the Rook. Closer to the center of the board (D)
the King has four capturing moves that can snare the Rook.
For each of the checking geometries, there will be a different number of King arrangements. In scenario (B) there will always
be 4 corners, no matter what the dimensions of the board are. In scenario (C), the number of edges on the board is given by
2(r2) + 2(f2) = 2r  4 + 2f  4 = 2r + 2f  8 which will vary depending on the dimensions of the board. In scenario (D) there
are (r2)(f2) = rf  2r 2f + 4 center arrangements. Each of these will be used in determining the safe check probability
for the Rook on a board f files wide and r ranks high.
 Corner Squares: 4
 Edge Squares: 2r + 2f  8
 Center Squares: rf  2r 2f + 4
 Total Squares = (4) + (2r + 2f  8) + (rf  2r 2f + 4) = rf
First, place the King in one corner of the board, and enumerate all of the safe checks that a Rook can issue. From the diagram above it is
easy to see that (r  2) safe checks are able to be delivered vertically and (f  2) safe checks can be made horizontally. There are
four corner squares, so the total number of safe checks is given by:
4(r  2 + f  2) = 4(r + f  4)
We note that there are rf squares available to place the King, and rf  1 squares remaining for the Rook once the King has taken up one square on the board. Then, we express
the probability for safe check as the quotient of the safe check square count and the total number of possible arrangements.
P(safe check, corner) = 4(r + f  4)/[(rf)(rf  1)]
Next, place the King on each edge square and repeat the safe checks count that a Rook can issue. From the diagram above there are (r  2) safe vertical checks but now only (f  3) safe checks can be made horizontally. Since the King is out of the corner, it can make a capture to either side
of the Rook when checked horizontally. If the King resided on a vertical edge, a similar observation is made. There would be (f  2) safe checks along horizon and (r  2) safe checks vertically.
But now, there is no longer a "constant" number of these arrangements. There are 2(r  2) instances where the king makes
(r  3) + (f  2) captures and 2(f  2) instances where the king makes (f  3) + (r  2) captures.
The total number of safe checks is given by:
2(r  2)[(r  3 + f  2)] + 2(f  2)[(f  3 + r  2)] =
2(r  2)(r + f  5) + 2(f  2)(r + f  5) =
(2r  4)(r + f  5) + (2f  4)(r + f  5) =
(r + f  5)[(2r  4) + (2f  4)] =
(r + f  5)(2r + 2f  8) =
(r + f  5)2(r + f  4) =
2(r + f  4)(r + f  5)
Expressing this probability for safe check as the quotient of this count and the total number of possible arrangements:
P(safe check, edge) = 2(r + f  4)(r + f  5)/[(rf)(rf  1)]
Finally, place the King on every center square. There are (r  3 + f  3) safe checks for each king arrangement that is inset
one square from the outermost portion of the board. This rectangle is (r  2) squares high and (f  2) squares
wide. The safe check count for this scenario is just a straight multiplication, so the probability is given by:
P(safe check, center) = (r  2)(f  2)(r + f + 6)/[(rf)(rf  1)]
Expressing the total probability for safely checking with the Rook is just the sum of all of these components:
 P(safe check, corner) = 4(r + f  4)/[(rf)(rf  1)]
 P(safe check, edge) = 2(r + f  4)(r + f  5)/[(rf)(rf  1)]
 P(safe check, center) = (r  2)(f  2)(r + f + 6)/[(rf)(rf  1)]
 Total Probability = (r + f  6)/(rf  1) + 2(r + f)/[(rf)(rf  1)]
If you are still reading at this point, then you probably want to do the final simplification of the above formula yourself.
Translating A Safe Check Probability Into A Relative Piece Value
So what does (r + f  6)/(rf1) + 2(r+f)/[(rf)(rf  1)] do to help us determine what a Rook is worth in Trice's Chess? First,
we can check our formula with the Taylor result from 1876. Taylor discovered a Rook could safely check a King onesixth of the time on
the 64square chessboard. So, we set r = f = 8 and see what we get:
(r + f  6)/(rf  1) + 2(r + f)/[(rf)(rf  1)] =
(8 + 8  6)/(64  1) + 2(8 + 8)/[(64)(64  1)] =
10/63 + 32/4032 =
(10x64)/4032 + 32/4032 =
672/4032 = 1/6. Our formula therefore predicts the Taylor result for 8x8 chess.
As a next step, we compute the Rook safe check probability for Trice's Chess using f =10 and r = 8.
(r + f  6)/(rf  1) + 2(r + f)/[(rf)(rf  1)] =
(8 + 10  6)/(80  1) + 2(8 + 10)/[(80)(80  1)] =
12/79 + 36/6320 =
(12x80)/6320 + 36/6320 =
996/6320 = 249/1580 = 0.15759493671
Comparing 1/6 = 0.16666666667 to 0.15759493671 indicates that a Rook on the 10x8 Trice's Chess board is about 94.5% of the strength of the Rook on an 8x8 board. So, the Trice's Chess Rook is about 5% weaker than its chess counterpart.
This is an interesting result, but it does not answer the question: What are the pieces worth on the Trice's Chess board?
Translating A Relative Piece Value Into An Absolute Piece Value
Knowing how one piece on the 10x8 board compares in strength to the 8x8 board is a great start. But in order to determine the
values of the pieces when compared to one another, the formula computed above must be generalized for the Knight and the Bishop
as well. To spare everyone the math, I will present the Knight formula without all of the steps:
Knight Safe Check Probability =
8/(rf  1) + (16  12r  12f)/[(rf)(rf  1)]
The equation for the Bishop is quite a disaster. This formula changes when your rank/file count
goes from odd to even or even to odd. It is populated with conditional clauses that are
not easy to represent in a concise mathematical form. It was easier to just compute the 10x8 result
directly using the graphical method specific to the 10x8 board. Here is a table
of safe check probabilities precomputed for both 8x8 chess and 10x8 Trice's Chess:
Safe Check Probabilities 
Piece 
8x8 board 
10x8 board 
% difference 
Knight 
12/144 
110/1580 
16.67% 
Bishop 
13/144 
133/1580 
6.77% 
Rook 
24/144 
249/1580 
5.50% 
Archbishop 
25/144 
243/1580 
11.41% 
Chancellor 
36/144 
359/1580 
9.11% 
Queen 
37/144 
382/1580 
5.95% 
There are a few things worthy to note at this point. These are not exact values universally accepted by the chess playing public. The
work of Taylor in 1876 provided a foundation upon which players could experiment and "adjust" the relative merits of the pieces over the years.
No strong chess player considers a Rook to be worth exactly two knights worth of material, as the 8x8 safe check entries indicate.
A Rook is much stronger than two Knights in the ending, since it can force a checkmate of the enemy King, whereas a pair of Knights cannot
force a checkmate (they can mate if the opponent commits a grave error).
A Rook also participates in castling so it has a special function to perform. To discourage trading a Rook for two minor pieces, most
masters took the approach of making it less valuable than a pair of minors. The Queen was likewise "tweaked" by the masters. It is not
merely the "sum of its parts" of Bishop + Rook. Since it can do on one square what these piece do on two squares, the modern master has given
it an additional bonus.
The table below shows what a modern chessplayer would consider an approximate trading value for the pieces on an 8x8 chessboard.
8x8 Chess Piece Values 
Piece 
Value in Pawns 
Knight 
3 
Bishop 
3 
Rook 
5 
Archbishop 
N/A 
Chancellor 
N/A 
Queen 
9 
The general consensus was to add one pawn to the value of the Queen, and subtract one pawn from the value of the Rook, once setting
the Knight and Bishop to be exactly equal (a difference of 1/144 is not too big on the 8x8 chessboard.)
Using such generalization as a basis, if we use the 8x8 chess knight as a baseline valued at 3.00 pawns, not
including the Rook "downgrade" and the Queen "upgrade" from "chess master logic", we get these equivalent values:
Values With 8x8 Knight = 3.0 
Piece 
8x8 board 
10x8 board 
Pawn 
1.00 
1.00 
Knight 
3.00 
2.50 
Bishop 
3.25 
3.03 
Archbishop 
N/A 
5.53 
Rook 
6.00 
5.67 
Chancellor 
N/A 
8.17 
Queen 
9.25 
8.70 
Now, if we apply the "Rook downgrade" to the 8x8 chess values, the Rook will be set to 5.00, which would
change the Queen's value to 8.25 instead of 9.25. The chessmasters basically performed the operations...
Rook = Rook  1 Pawn
Queen = Queen x (9/8)
...to get a more "sensible" value system. If we do something similar, we should set our Rook value to 94.5% of the Rook value in chess,
rather than subtract a full pawn from its merit. We should then scale the Queen by 9/8, as well as the Chancellor and Archbishop, since
these pieces are also "two fused into one." Perform all of these functions, without setting the Knight and Bishops to the same value (they
are much more separated in strength on the Trice's chess board) we will get the following table of values:
Quasiadjusted Values 
Piece 
8x8 board 
10x8 board 
Pawn 
1.00 
1.00 
Knight 
3.00 
2.50 
Bishop 
3.25 
3.03 
Rook 
5.00 
4.73 
Archbishop 
N/A 
6.22 
Chancellor 
N/A 
8.13 
Queen 
9.00 
8.73 
One of the results that was "hoped for" was the Queen and Chancellor values remaining relatively constant, which they did do, and the
Rook and Archbishop swapping places in terms of value, which happened as well. The numbers above look about correct both comparitively
from within the Trice's Chess setting, and when contrasted to the 8x8 chess values. It is curious to note that now the Archbishop + Knight
is valued at 1/100th of pawn less than a Queen. In general, you would not want to trade the Archbishop and the Knight for a Queen, so for this
reason, we make one other small adjustment, then round the numbers off to something reasonable for tournament use:
Trice's Chess Piece Values 
Piece 
10x8 board 
Pawn 
1.00 
Knight 
2.50 
Bishop 
3.00 
Rook 
4.75 
Archbishop 
6.50 
Chancellor 
8.25 
Queen 
8.75 


