A362633 - OEIS

A362633

Square array read by antidiagonals: Consider n k-sided fair dice, whose faces are numbered 1, ..., n*k (in any order). The outcome of a roll of the dice determines an ordering of them. T(n,k) is the minimum difference of the number of outcomes resulting in the most common ordering and the number of outcomes resulting in the least common ordering, n,k >= 1.

%I #11 May 13 2023 13:00:23

%S 0,0,1,0,0,1,0,1,2,1,0,0,3,2,1,0,1,2,6,2,1,0,0,5,8,6,2,1

%N Square array read by antidiagonals: Consider n k-sided fair dice, whose faces are numbered 1, ..., n*k (in any order). The outcome of a roll of the dice determines an ordering of them. T(n,k) is the minimum difference of the number of outcomes resulting in the most common ordering and the number of outcomes resulting in the least common ordering, n,k >= 1.

%C In other words, T(n,k)/k^n is the minimum difference of the largest and smallest probabilities of the n! possible orderings of the dice.

%C T(n,k) = 0 means that there exists a set of n k-sided dice such that all n! orderings are equally likely, i.e., a permutation-fair set of dice. The smallest k for which T(n,k) = 0 is 1, 2, 6, 12 for n = 1, 2, 3, 4. A necessary condition for T(n,k) = 0 to hold is that k^n be divisible by n!, i.e., that k be a multiple of A034386(n). This condition is not sufficient, because 6^4 is divisible by 4! but T(4,6) > 0.

%H James Grime and Brady Haran, <a href="https://www.numberphile.com/videos/go-first-dice">Go First Dice</a>, Numberphile video, 2023.

%H Eric Harshbarger, <a href="http://www.ericharshbarger.org/dice/go_first_dice.html">Go First Dice</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Go_First_Dice">Go First Dice</a>.

%F T(n,k) = A362634(prime(k)^n).

%F T(1,k) = 0.

%F T(2,k) = (k mod 2).

%F T(n,1) = 1 if n >= 2.

%F T(n,2) = 2 for n >= 3.

%e Array begins:

%e n\k| 1 2 3 4 5 6

%e ---+-----------------

%e 1 | 0 0 0 0 0 0

%e 2 | 1 0 1 0 1 0

%e 3 | 1 2 3 2 5 0

%e For three 2-sided "dice" (or coins), the best we can do is to number the faces (1,5), (2,6), and (3,4), respectively. We then have the following 8 possible outcomes and orderings:

%e d1 d2 d3 | ordering

%e ---------+---------

%e 1 2 3 | d1<d2<d3

%e 1 2 4 | d1<d2<d3

%e 1 6 3 | d1<d3<d2

%e 1 6 4 | d1<d3<d2

%e 5 2 3 | d2<d3<d1

%e 5 2 4 | d2<d3<d1

%e 5 6 3 | d3<d1<d2

%e 5 6 4 | d3<d1<d2

%e Here, the orderings d2<d1<d3 and d3<d2<d1 never occur while all the others occur twice each, so T(3,2) = 2-0 = 2.

%Y Cf. A034386, A362634.

%K nonn,tabl,more

%O 1,9

%A _Pontus von Brömssen_, Apr 29 2023