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5
Crazy Dice
at z 1 the denominator has a k-fold zero and so there will be a
term c k . All the other zeros are roots of unity and, because we
(1−zð
assumed no common divisiors, all will be of order lower than k.
n+k−1
c
Thus, although the coefficient of the term k is c , the
(1−zð k−1

coefficients of all other terms a j will be aω j n+j . Since all of
(1−ωz) j−1
these j are less than k, the sum total of all of these terms is negligible
n+k−1 n+k−1
compared to our heavy term c . In short C(n) ∼ c ,or
k−1 k−1
even simpler,
n k−1
C(n) ∼ c .
(k − 1)!
But, what is c? Although we have deftly avoided the necessity of
finding all of the other terms, we cannot avoid this one (it’s the whole
story!). So let us write
1 c
+ other terms,
(1 − z )(1 − z ) ··· (1 − z ) (1 − zð k
a k
a 2
a 1
multiply by (1 − zð k to get
1 − z 1 − z 1 − z k
··· c + (1 − zð × other terms,
1 − z a 1 1 − z a 2 1 − z a k

and finally let z → 1. By L’Hˆ opital’s rule, for example, 1−z a → 1
1−z i a i
whereas each of the other terms times (1 − zð k goes to 0. The final
1
result is c , and our final asymptotic formula reads
a 1 a 2 ···a k
n k−1
C(n) ∼ . à 5)
a 1 a 2 ··· a k (k − 1)!

Crazy Dice

An ordinary pair of dice consist of two cubes each numbered 1
through 6. When tossed together there are altogether 36 (equally
likely) outcomes. Thus the sums go from 2 to 12 with varied
numbers of repeats for these possibilities. In terms of our ana-
lytic representation, each die is associated with the polynomial
5
2
3
4
6
z + z + z + z + z + z . The combined possibilities for the