Algebraic solution for combinatorics

Question

Let $P(x)$ be the probability that the sum of the faces of 5 fair and independent dice is x

We are all aware that we can find $P(x)$ manually. However, is there an algebraic proof for $P(5) < P(6) <....<P(17)$?

Answer 1

I am thinking an "algebraic" proof using the generating function... $P(i)$ is the coefficient of $x^i$ of $(\frac{x+x^2+x^3+x^4+x^5+x^6}{6})^5$. Now we expand this (we forget the $1/6^5$ on the denominator), either by hand or by a calculator

$$(x+x^2+x^3+x^4+x^5+x^6)^5=x^{30}+5x^{29}+15x^{28}+35x^{27}+70x^{26}+126x^{25}+205x^{24}+305x^{23}+420x^{22}+540x^{21}+651x^{20}+735x^{19}+780x^{18}+780x^{17}+735x^{16}+651x^{15}+540x^{14}+420x^{13}+305x^{12}+205x^{11}+126x^{10}+70x^9+35x^8+15x^7+5x^6+x^5$$

Which, by looking at the coefficients, can yield your result.

Answer 2

Suppose we have $n$ dice. The idea is to create a create a mapping from rolling $ X < \frac{7}{2} n $ to rolling $X+1$, and show that this is a injective map. If so, this implies $P(X) \leq P(X+1)$.
If we can further show that the map is not surjective, then the inequality is strict.

Setting up the map
Order the dice $ D_1, D_2, \ldots D_n$. Let $d_i$ represent the value of dice $D_i$. For any roll where $ \sum d_i = X < \frac{7}{2} n$, Consider the set $\mathbb{I}$ where $i \in \mathbb{I}$ iff $D_1 + D_2 + \ldots D_i < \frac{7}{2} i$. From the conditions, $n \in I$ so the set is non-empty.
Let $I$ be the smallest index in $\mathbb{I}$. Clearly, $d_i < 6$. (In fact, $d_i < 4$.)
We create the map $(d_1, d_2, \ldots, d_{I-1}, d_I , d_{I+1}, \ldots, d_n ) \rightarrow ( d_1, d_2, \ldots , d_{I-1}, d_I + 1 , d_{I+1} , \ldots, d_n)$.
Because of the definition of $I$, the value is uniquely determined for each roll, and hence we have an injection.

I leave it to you to show that this is not a surjection, thus establishing that $P(X) < P(X+1)$.
Hint: Find a class of rolls with value $X+1$ that clearly is not mapped from a roll of $X$.