In this paper, Majority rule, hierarchical structures, and democratic totalitarianism: A statistical approach, I read:
A social system is considered in which people are supporters of either a $G$-orientation or a $H$-orientation. The proportions of $G$ and $H$ persons are respectively $p_0$ and $\left(1-p_0\right)$.
We now constitute a cell of three persons selected randomly without replacement from the population of the social system. Eight configuration can be obtained through the sampling. They are classified into four types with respect to the numbers of $G$ and $H$. The types are $3 H, 2 H-1 G, 1 H-2 G$, and $3 G$ assuming a practically infinite population the probabilities of the four types are respectively $\left(1-p_0\right)^3, 3\left(1-p_0\right)^2 p_0, 3\left(1-p_0\right) p_0^2$, and $p_0^3$.
Once a cell is constituted it elects a representative either a $H$ or a $G$ using the principle of majority rule. Types $3 H$ and $2 H-1 G$ elect a $H$-person while types $1 H-2 G$ and $3 G$ elect a $G$-person. The probability to have a $G$-person elected is therefore $$ p_1=p_0^3+3 p_0^2\left(1-p_0\right) $$
I have trouble explaining these calculations but maybe it's just because I don't know how to think well with probabilities. For example, if I have 5 cells, respectively GGH,HGG,HHH,GGG,GGG, the elections will then give the result, respectively, G,G,H,G,G. The new probability to have G is then 4/5 but how can I relate this value to the formula above?
As stated in the comments, we can re-cast this as the usual coin flipping experiment, with $P(G)=p_0$ so the number of $G$-oriented people in a cell of $3$ is $Binomial(3,p_0)$.
The probability mass function for this Binomial Distribution is:
$$P(G=k)={3 \choose k}p_0^k(1-p_0)^{n-k}$$
From this, we can get all the probabilities in the article.
To get the probability of electing a $G$-oriented leader, we just add up the probability we get $2$ or $3$ $G$-oriented people in the cell:
$$P(G=3) = p_0^3,\;P(G=2)=3p_0^2(1-p_0) \implies P(H < 2)=p_0^3+3p_0^2(1-p_0)$$