Homemade Henderson-Hasselbalch: The recurrence relation

vivianimbriotis | March 19, 2025, 9:40 p.m.

If we have k particles of B around a single particle of A, what is the chance of a reaction occurring?

If there's one particle of B near A, then let's say it has a fixed probability of reacting, called p.

If there's multiple particles, then let's check if they react one at a time. The probability that each one reacts, GIVEN that no reaction has occurred so far, is p.

$$P(\text{A and B1 react}) = p$$

$$P(\text{A and B2 react | A and B1 did not react}) = p$$

$$P(\text{A and B3 | (A and B1 did not react) AND (A and B2 did not react)}) = p$$

and so on. We also know some other things - there's only one particle of B, right? So

$$P(A \text{ and } B_i \text{ react} | A \text{ and }B_j \text{ react }) = 0\text{ for }i \neq j$$

What's the probability that ANY of the Bs react with A (i.e. that ANY reaction occurs)? With n total particles of B available, we'll denote this probability \(P_n\).

$$P_1 = p$$

Clearly. Then for each particle we add, the total probability of ANY reaction is the chance of a reaction already having happened with N-1 particles...

$$P_{n-1}$$

...plus the chance that a reaction hasn't happened yet, times the chance it happens when we introduce the Nth particle...

$$+ (1-P_{n-1}) \cdot p$$

...so the grand probability is

$$P_n = P_{n-1} + (1-P_{n-1}) \cdot p$$

$$P_n = P_{n-1} + p - P_{n-1} \cdot p$$

$$P_n = P_{n-1}(1-p) + p$$

Now we have a recurrence relation. To solve this we first need to find a fixed point...

$$P^* = P^*(1-p) + p$$

$$P^* - P^*(1-p) = p$$

$$P^*p = p$$

$$P^* = 1$$

This makes perfect sense. If the probability of a reaction having already happened after N-1 particles is 1, then adding another particle doesn't change anything. Now we have a fixed point, we can define a dummy variable to convert this to homogenous form:

$$x_n = P_n - P^*$$

$$x_n = P_n - 1$$

$$P_n = x_n + 1$$

(this works for all n, so we also have immediately that)

$$P_{n-1} = x_{n-1} + 1$$

Substituting this into out recurrence relation:

$$x_n + 1 = (x_{n-1} + 1)(1-p) + p$$

$$x_n + 1 = (x_{n-1} + 1) - (p \cdot x_{n-1} + p) + p$$

$$x_n + 1 = x_{n-1} + 1 - p \cdot x_{n-1} - p + p$$

$$x_n = x_{n-1} - p \cdot x_{n-1}$$

$$x_n = (1 - p) \cdot x_{n-1}$$

Recognizing a geometric form (every iteration multiplying by (1-p)), we can solve for \(x_n\):

$$x_n = A\cdot(1 - p)^{n}$$

$$\text{and }x_1 = P_1 -1 = p-1$$

$$\text{so }A = -1$$

Therefore

$$P_n = 1 - (1 - p)^{n}$$

This satisfies our intiution that if we take the limit as N grows large, the probability of a reaction approaches unity.

If we do a first order approximation and assume that p^2 and all higher terms are negligible, then we can go via a binomial expansion...

$$P_n = 1 - {n\choose 0}(-p)^0 - {n\choose 1}(-p)^1 - {n\choose 2}(-p)^2 - ... - {n\choose n}(-p)^n$$

$$P_n = 1 - 1 + np - ...\text{higher-order terms}$$

$$ P_n \approx np$$

Ta-da!