Tied dice (n = 4 addendum)

We present a very tedious proof of the n = 4 case of the Tied Dice Theorem. Suppose X = (x1; x2; x3; x4); Y = (y1; y2; y3; y4) 2 D(4; a; b; s) are distinct, tied, non-balanced dice. Let denote the number of pairs (i; j) with xi = yj ; as X and Y are tied, there must be (16 )=2 pairs (xi; yj) with xi > yj , and (16 )=2 pairs (xi; yj) with xi < yj . For each value of we list the various possible arrangements of the labels of X and Y by considering the winning con…guration c1c2c3c4 of X, where ci = jfj j xi > yjgj; the winning con…guration is a nondecreasing sequence of non-negative integers whose sum is (16 )=2. We hope that the layout of the addendum is clear, with sub-cases underlined and sub-sub-cases underlined and indented. In each case we provide an example of a Z 2 D(4; a; b; s) which ties neither X nor Y , and we brie‡y explain the nonzero values of fX(Z) and fY (Z). For instance the …rst sub-case considered below includes the statement “Z = (x2; y2+1; y3; y4+y1 x2 1) ties neither X nor Y (fX(Z) fX(Y ) = 0 (1 or 2) and fY (Z) fY (Y ) = (at least 2) 1).”Recalling that fX(X) = fX(Y ) = 0 = fY (X) = fY (Y ) because X and Y are tied, the quoted statement indicates that if we think of Z as having been obtained from Y by reducing y1 to x2, increasing y2 to y2 + 1 and increasing y4 to y4 + y1 x2 1, then we can see that fX(Z) 6= 0 6= fY (Z) because reducing y1 contributes 1 or 2 to fX(Z) and contributes 1 to fY (Z), while increasing y2 and y4 contributes at least 2 to fY (Z) and does not a¤ect fX(Z).