# probability distribution of total number of cells with 1/2 probability of dividing.

There is a cell that has a 1/2 probability of dividing into two daughter cells (the parent cell disappears) and a 1/2 probability of stop dividing. And each daughter cell has a 1/2 probability of dividing again.

The probability of division of a cell, Y, obeys a binomial distribution, $$Y\sim B(2,1/2)$$, and the divisions of different cells are independent of each other.

Cell death, cell nutrient depletion, is not considered. If there is only one cell at the beginning. Is it possible to find the probability distribution function for the total number of cells X?

To make sure there is no ambiguity, the python code to simulate one event is shown below

from random import random
def get():
""" calculate total number of cells """
if random() < 1/2:
return get() + get()
return 1


My efforts: I try to find all distinct divide way when $$X=n$$. Then divide the number by $$2^n$$. Intuitively this is equivalent to variety of trees with $$n$$ mode. Starting from a node, each dividing is equivalent to find a node then plus a new edge. And I found the sequence of distinct tree numbers with $$n$$ nodes in https://oeis.org/A000055. But this series grows faster than $$2^n$$ and is therefore wrong.

• When a cell has two daughters, does that make 3 cells (plus any descendents of the daughters?) Commented Dec 3, 2023 at 16:53
• After the cell divides, it becomes two daughter cells. The total number is 2 if daughter cells don't divide. The final total cell number count any possible descendents of daughters. Commented Dec 3, 2023 at 17:01
• What method are you supposed to use? Probability generating functions? Commented Dec 3, 2023 at 17:06
• With your "2 if daughters don't divide", you get the probability generating function $G(x)$ satisfying $G = \frac12 x +\frac12 G^2$ which gives $G(x)= 1-\sqrt{1-x} = \frac12 x+\frac18x^2+\frac{2}{32}x^3+\frac{5}{128}x^4 + \frac{14}{512}x^5+ \frac{42}{2048}x^6+\cdots$ showing the relevant probabilities and the Catalan numbers as numerators. Commented Dec 3, 2023 at 17:27
• Similarly with "3 if daughters don't divide", you get the probability generating function $G(x)$ satisfying $G = \frac12 x +\frac12 x G^2$ which gives $G(x)= \frac{1-\sqrt{1-x}}{x} = \frac12 x+\frac18x^3+\frac{2}{32}x^5+\frac{5}{128}x^7 + \frac{14}{512}x^9+ \frac{42}{2048}x^{11}+\cdots$ showing the same probabilities but only for odd numbers in the population. Commented Dec 3, 2023 at 17:27

Let $$N$$ be the final number of cells. We have $$P(N=1)=1/2$$. Next, $$P(N=2)=1/8$$, because the initial cell needs to divide, and then both children must cease dividing.
For $$P(N=3)$$, there are two ways this can happen. The first cell divides into a left and right, so you can either have the left cell re-divide and the right cell die, or vice versa. Both of these variants have a probability of $$(1/2)^5$$, so $$P(N=3)=2\cdot (1/2)^5$$. Both of the possible divisions are illustrated below:
     .          .

In general, to find $$P(N=n)$$, you need to count up the number of ways the cells can divide to have a final population of $$n$$. The number of ways corresponds exactly to the number of full binary trees with $$n$$ leaves. It is well known that full binary trees with $$n$$ leaves are enumerated by the Catalan numbers, $$C_{n-1}$$. Finally, for each of these trees, the probability of that tree occuring is $$(1/2)^{2n-1}$$. This is because a full binary tree with $$n$$ leaves will have $$n-1$$ internal nodes, and each of the $$n+(n-1)=2n-1$$ nodes of the tree is associated with an event with probability $$1/2$$. All in all, we conclude $$P(N=n)=C_{n-1}(1/2)^{2n-1}=\frac{1}{n}\binom{2n-2}{n-1}(1/2)^{2n-1}.$$