# Understanding elementary rate laws from a probability stand point

I am learning about chemical kinetics and dynamics and as I understand for a general chemical reaction $$\ce{aA + bB -> cC + dD}$$ whose reaction rate, r, can be described by an elementary rate law can be written as follows: $$r= k[\ce{A]^a[\ce{B}]^b}$$

What I do not seem to conceptually understand is if you look at a chemical reaction from a collision theory standpoint, would it not be true that as one reactant that is in a higher stoichiometric ratio relative to the other, the probabilty that the reactants collide (assuming any collision gives a reaction) would be inversely proportional to the stoichiometry of the reactant. For example in the following: $$\ce {A + 2B -> C}$$

You basically need 2 mols of B for every mol of A for a successful reaction to produce C, so if the probability that A successfully collides with a mol of B is $\propto$ to $1/z$, then the probability that A successfully collides with TWO mols of B $\propto 1/z^2$ which is lower, so why does a reaction rate, which is $\propto$ to the probability of a collision, with an elementary rate law $\propto$ to a concentration raised to an exponent?

Let's divide the reaction volume into V little boxes and randomly throw ${\rm N}_A$ molecules of A into it and ${\rm N}_B$ molecules of B. The probability $p_A$ that a given box has a molecule of A in it is ${\rm N}_A$/V, in the limit that $V >> {\rm N}_A$, i.e. an ideal dilute solution, and likewise the probability $p_B$ that it has a molecule of B in it is ${\rm N}_B/V$. The probability it has two molecules of B in it is ${p_B}^2$, again in the limit that $V >> {\rm N}_B$ and ${\rm N}_B >> 1$.
Consequently, the probability all three atoms are close to one another is $({\rm N}_A/V)({\rm N}_B/V)^2$, which is proportional to the terms in your rate law. (This isn't a derivation, by the way, but only a rationalization I hope is appealing to intuition. I've swept all kinds of detail and qualification under the rug.)