Probability of rolling $15$ with $3$ dice vs. $5$ dice Similar problem from my textbook to the last question I asked on MSE: Probability of rolling $14$ with $3$ dice vs. $5$ dice

The chance of throwing $15$ is less with three dice than with five as $36$ to $55$.

As expected, I didn't get the given ratio again. So am I wrong, or is the book wrong? Here's what I did:
Let's first do three dice:

*

*$663$: $3$ possibilities

*$654$: $6$ possibilities

*$555$: $1$ possibility

Adding them up, we have$${{3 + 6 + 1}\over{6^3}} = {5\over{108}} = {{120}\over{2592}}$$Let's next do five dice:

*

*$66111$: $10$ possibilities

*$65211$: $60$ possibilities

*$64311$: $60$ possibilities

*$64221$: $60$ possibilities

*$63321$: $60$ possibilities

*$63222$: $20$ possibilities

*$55311$: $30$ possibilities

*$55221$: $30$ possibilities

*$54411$: $30$ possibilities

*$54321$: $120$ possibilities

*$54222$: $20$ possibilities

*$53331$: $20$ possibilities

*$53322$: $30$ possibilities

*$44421$: $20$ possibilities

*$44331$: $30$ possibilities

*$44322$: $30$ possibilities

*$43332$: $20$ possibilities

*$33333$: $1$ possibility

Adding them up, we have$${{217}\over{2592}},$$which is different from the problem statement. So where did I go wrong? Or is there a mistake in the problem statement?
 A: Why don't you just use stars and bars with inclusion-exclusion, which you can do by hand for such small problems.
Basically, part 2, which is very error prone by your procedure, is $x_1+x_2+x_3+x_4+x_5 = 15, 1\leq x_i\leq 6$
And by Theorem 1 of "stars and bars", applying inclusion-exclusion,
you get $\binom{15-1}{5-1} - \binom51\binom{9-1}{5-1} = 651$
You have already counted case $1$ by hand, by stars and bars it would be $\binom{14}2 -\binom31\binom82 +\binom32\binom22 = 10$
You can now proceed further
A: Here is a code I wrote in Python that can help check results with counting in dice sum problems. It basically is an application of generating functions. We can use the following generating function,
$$\left( \sum_{i=1}^{m} x^{i} \right)^n$$
where $n$ is the number of dice and $m$ is the number of faces of each die ($m=6$ for standard dice). If we want to know the number of ways to make a sum of $p$, we only check the coefficient of $x^p$ in the expansion.
from sympy import expand, Symbol, Poly

n = 5  # how many dice?
r = 6  # how many faces for each die (for normal dice r=6)?

x = Symbol('x')
listecik = [x**i for i in range(1, r + 1)]
p1 = expand(((sum(listecik)))**n)
# generating function

a = Poly((p1), x)

print("Using " + str(n) + " dice with " + str(r) + " faces")
print("Sum".ljust(8), "No. of ways")

coefficients = [a.as_poly().coeff_monomial(x**d) for d in range(n, r * n + 1)]
for i, item in enumerate(coefficients, start=n):
    # numbering all the sums (starting with the minimum, n)
    print(str(i).ljust(8), item)

Sample output:
Using 5 dice with 6 faces
Sum      No. of ways
5        1
6        5
7        15
8        35
9        70
10       126
11       205
12       305
13       420
14       540
15       651
16       735
17       780
18       780
19       735
20       651
21       540
22       420
23       305
24       205
25       126
26       70
27       35
28       15
29       5
30       1

