How do we convert a generating function that counts to a generating function of probabilities?

Question

I am learning about the binomial coefficient and counting. We define: $$ (1+x)^m = \sum^m_{n=0} \begin{pmatrix} m \\ n \end{pmatrix} x^n $$ the coefficients of the powers of $x^n$ represent the number of ways of choosing $n$ objects from a set of $m$. Is there a way to convert this into a probability distribution by dividing each coefficient by $$ \sum^m_{n=0} \begin{pmatrix} m \\ n \end{pmatrix}. $$

Example

We have a set of three objects, $\{a, b , c\}$, we can draw 1 object three ways $(a + b + c)$, two objects three ways $(ab + ac + bc)$ and three objects one way $(abc)$. Then,

$$ (1+x)^3 = 1x^0 + 3x^1 + 3x^2 + 1x^3 $$ If we divide each coefficient by the sum of coefficients (in this case 8) then they represent probabilities(?). Can we then say that $(1+x)^n$ is the generator for a distribution function $$ p_n = \frac{1}{\sum^m_{n=0} \begin{pmatrix}m \\ n \end{pmatrix}} \begin{pmatrix} m \\ n \end{pmatrix} $$ Such that

$$ (1+x)^n = \sum_n p_n x^n $$

Answer 1

Note that $$ p_n =\binom{m}{n}/\sum_{n=0}^{m}\binom{m}{n}=\binom{m}{n}/2^{m}. $$ So $$ \sum_{n=0}^{m} p_n x^n=\frac{(1+x)^m}{2^{m}}=\left(\frac{1}{2}+\frac{x}{2}\right)^m=\sum_{n=0}^m\binom{m}{n}\left(\frac{1}{2}\right)^n \left(\frac{1}{2}\right)^{m-n}x^n.\tag{1} $$ You can think of $p_n$ as the probability of getting $n$ heads when we toss a fair coin $m$ times.

More precisely if $X\sim\text{Binom}(m,1/2)$, i.e. $X$ is a random variable that follows a binomial distribution with $m$ trials and probability of success $1/2$, then $p_n=P(X=n)$.

We say that the series in (1) is the probability generating function of $X$ as it encodes the probabilities of $X$.

Answer 2

What you are observing is on the right track - indeed it makes polynomials exceptionally helpful for combinatorial problems that may be otherwise extremely cumbersome.

Suppose that we are to perform $n$ independent experiments. For each $k$, we have that event $A$ happens with probability $p_k$, and it does not occur with probability $1-p_k := q_k$. Then, consider the product $$(p_1 x + q_1 ) ( p_2 x + q_2 ) \dots (p_n x + q_n )$$ The coefficient of $x^m$ in the above is precisely the probability that $A$ occurs exactly $m$ times in the $n$ experiments.

In your question, you would be computing the case where the $p_k = q_k = 1/2$.