This is revealed to be more difficult than it looks but we can at
least provide an algorithm to compute the exact values rather than
rely on simulations.
Suppose that the die has $m$ faces and is rolled $n$ times. Rolling
the die with the least occured value being $p$ and the most occured
value being $q$ and both of them marked yields the species
$$\mathfrak{S}_{=m}
(\mathfrak{P}_{=0}(\mathcal{Z})
+ \mathcal{U}\mathfrak{P}_{=p}(\mathcal{Z})
+ \mathfrak{P}_{=p+1}(\mathcal{Z})
+ \cdots
+ \mathfrak{P}_{=q-1}(\mathcal{Z})
+ \mathcal{V}\mathfrak{P}_{=q}(\mathcal{Z})).$$
This has generating function
$$G(z,u,v) =
\left(1+u\frac{z^p}{p!} + \sum_{r=p+1}^{q-1} \frac{z^r}{r!}
+ v\frac{z^q}{q!}\right)^m.$$
Subtracting the values where sets of size $p$ and $q$ did not occur we
obtain the generating function
$$H_{p,q}(z) =
\left(1+ \sum_{r=p}^{q} \frac{z^r}{r!}\right)^m \\
- \left(1+ \sum_{r=p+1}^{q} \frac{z^r}{r!}\right)^m
- \left(1+ \sum_{r=p}^{q-1} \frac{z^r}{r!}\right)^m
+ \left(1+ \sum_{r=p+1}^{q-1} \frac{z^r}{r!}\right)^m.$$
We then obtain for the desired quantity the closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{n!}{m^n}
[z^n] \sum_{p=1}^n \sum_{q=p}^n \frac{q}{p} H_{p,q}(z).}$$
Introducing
$$L_{p,q}(z) =
\left(1+ \sum_{r=p}^{q} \frac{z^r}{r!}\right)^m$$
we thus have
$$\frac{n!}{m^n}
[z^n] \sum_{p=1}^n \sum_{q=p}^n
\frac{q}{p} (L_{p,q}(z)
-L_{p+1,q}(z)-L_{p,q-1}(z)+L_{p+1, q-1}(z)).$$
This is
$$\frac{n!}{m^n}
[z^n] \sum_{p=1}^n \sum_{q=p}^n
L_{p,q}(z)\left(\frac{q}{p}
-\frac{q}{p-1}-\frac{q+1}{p}+\frac{q+1}{p-1}\right)^*$$
where the star indicates that those terms with $p-1=0$ and $q+1=n+1$
do not contribute. We also have for $p\lt q$
$$[z^n] L_{p,q}(z)
= [z^n] \sum_{k=0}^m {m\choose k}
\left(1+ \sum_{r=p+1}^{q} \frac{z^r}{r!}\right)^{m-k}
\left(\frac{z^p}{p!}\right)^k
\\ = \sum_{k=0}^m {m\choose k} [z^{n-pk}] \frac{1}{p!^k}
\left(1+ \sum_{r=p+1}^{q} \frac{z^r}{r!}\right)^{m-k}.$$
Furthermore we obtain for $p=q$
$$[z^n] L_{q,q}(z)
= [z^n] \left(1+ \frac{z^q}{q!}\right)^m
= [[n \bmod q\equiv 0]]
\times {m\choose n/q} \frac{1}{q!^{n/q}}.$$
With these we can implement a recursion, which in fact is not all that
much faster than working with $H_{p,q}(z).$ This yields the following
graph where we have used a six-sided die. We reach resource limits
fairly quickly (esp. space) but we do have the exact values for up
to $120$ rolls, which is completely impossible by enumeration ($94$
digits total case count). The convergence is very slow which tells us
that we must deploy probabilistic methods to make progress with this
problem.
4.5+
| HH
+ H HH
+ H H
+ H H
| H H
+ H
4+ H H
| H
+ H H
+ H
+ H H
| H
+ H
3.5+ H H
+ HH
| H
+ H H
+ H
+ HH
| H H
3+ HH
+ HH
+ H HH
| HH
+ HH
+ H HH
2.5+ HHH
| HH
+ HHH
+ H HHH
| HHHH
+ HHHH
+ H HHHHHH
2+ HHHHHHH
| HHHHHHHH
+ HHHHHHHHHH
+ H
+ H
|
+
1.5+
+ H
|
+
+
+
|
1+HH
-+--+---+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--
0 20 40 60 80 100 120
This was the Maple code.
with(combinat);
ENUM :=
proc(n, m)
option remember;
local rolls, res, ind, counts;
res := 0;
for ind from m^n to 2*m^n-1 do
rolls := convert(ind, base, m);
counts := map(mel->op(2, mel),
convert(rolls[1..n], `multiset`));
res := res + max(counts)/min(counts);
od;
res/m^n;
end;
L := (m, rmin, rmax) -> (1 + add(z^r/r!, r=rmin..rmax))^m;
X :=
proc(n, m)
option remember;
local H;
H := (p, q) ->
expand(L(m, p,q)
- L(m, p+1,q) - L(m, p,q-1)
+ L(m, p+1,q-1));
n!/m^n*
coeff(add(add(q/p*H(p,q), q=p..n), p=1..n),
z, n);
end;
LCF :=
proc(n,m,p,q)
option remember;
if n < 0 then return 0 fi;
if p = q then
if n mod q <> 0 then return 0 fi;
return binomial(m,n/q)/q!^(n/q);
fi;
add(binomial(m,k)*1/p!^k*LCF(n-p*k, m-k, p+1, q),
k=0..m);
end;
LVERIF :=
(m, p,q) -> add(LCF(n, m, p, q)*z^n, n=0..q*m);
XX :=
proc(n, m)
option remember;
local res, p, q, cf;
res := 0;
for p to n do
for q from p to n do
cf := q/p
- `if`(p>1, q/(p-1), 0)
- `if`(q1 and q
Addendum. A rather fascinating sequence appears when we compute
the value $n$ of the number of rolls of a die with $m$ sides that
maximizes the average ratio between most and least. We obtain a
sequence that might well be linear, or it might not, here it is:
$$1, 5, 9, 13, 16, 20, 24, 28, 33, 37, 41, 46, 50,
\\ 55, 60, 64, 69, 74, \ldots $$
A linear fit to this sequence is given by
$$- 4.82352941176471806+ 4.27966976264189913\,m.$$
This pattern does seem to suggest the problem merits additional
investigation. I hope these data and the conjecture as to suspected
linearity will be a start.
MX :=
proc(m)
option remember;
local n, cur, nxt;
if m = 1 then return 1 fi;
n := 1;
cur := XX(1, m);
do
nxt := XX(n+1, m);
if cur > nxt then
break;
fi;
n := n+1;
cur := nxt;
od;
n;
end;
Remark. There is a better recurrence at the following MSE link.
