# Mathematical Biology Models?

This question is a follow-up to a previous question I posted Probability of Seeing "X" % of Balls in "Y" Turns?

Set up:

• Suppose we have integers 1,2,3...99, 100

• Each integer has an equal probability of being selected Game:

• In round=1, we pick 5 numbers randomly without replacement and then put them back

• In round=2 we again pick 5 numbers randomly without replacement and then put them back

• We do this until round = 100

I wrote an R program to simulate this situation:

 round   numbers_picked         cumulative_unique_numbers_seen     percent_of_new_numbers
1  31, 79, 51, 14, 67                              5                    100
2  42, 50, 43, 14, 25                              9                     80
3  90, 91, 69, 99, 57                             14                    100
4   92, 9, 93, 72, 26                             19                    100
5    7, 42, 9, 83, 36                             22                     60
6  78, 81, 43, 76, 15                             26                     80
7    32, 7, 9, 41, 74                             29                     60
8   23, 27, 60, 53, 7                             33                     80
9  53, 27, 96, 38, 89                             36                     60
10  34, 93, 69, 72, 76                             37                     20
11  63, 13, 82, 97, 91                             41                     80
12  25, 38, 21, 79, 41                             42                     20
13  47, 90, 60, 95, 16                             45                     60
14   94, 6, 72, 86, 97                             48                     60
15  39, 31, 81, 50, 34                             49                     20


I am wondering if there is some probability distribution that can be used to answer the following question:

Suppose we are currently at round = n and we have seen "m" unique numbers. If we DO NOT know that there are 100 total numbers - what is the probability we will have seen 99% of all numbers by round = k? (k>n)

I have been beginning to learn about mathematical biology models and their use in these kinds of problems. For example:

Can someone please suggest if there are some mathematical biology models that can be used in this problem?

Answering your first question, let us consider the following natural and simplified model.

Suppose that for each natural number $$N\ge m$$, $$A_N$$ is the hypothesis that there are $$N$$ numbers in total and $$p_N$$ is the a priori probability of $$A_N$$. To simplify the model, suppose that at each round we pick only one number and all choices have equal probabilities. I expect that the probability required to calculate depends on probabilities $$(p_N)_{N\ge m}$$ even for $$k=n$$.

Indeed, assuming $$A_N$$, for each natural $$r$$ let $$q_{N,r}$$ be the vector of dimension $$N$$ such that for each natural $$k\le N$$ the $$k$$th entry $$q_{N,r,k}$$ of the vector is equal to the probability that up to $$r$$ rounds we have seen exactly $$k$$ unique numbers. Then $$q_{N,r,1}=\frac 1{N^{r-1}}$$, $$q_{N,1,k}=0$$ for any natural $$k$$ with $$2\le k\le N$$ and for each natural $$r$$ we have $$q_{N,r+1,k}=q_{N,r,k}\cdot\frac kN+q_{N,r,k-1}\cdot \frac {N-k}N.$$ Let $$M_N=\|m_{N,i,j}\|$$ be the $$N\times N$$ matrix such that $$m_{N,1,1}=\frac 1N$$ and for each natural $$k$$ with $$2\le k\le N$$ we have $$m_{N,k,k}=\frac kN$$ and $$m_{N,k,k-1}=\frac {N-k}N$$, and all others entries of $$M$$ are zeroes. By the recurrence, $$q_{N,r}=M_N^{r-1}q_{N,1}$$.

Then given that up to $$n$$ rounds we have seen exactly $$m$$ unique numbers, Bayes' formula suggests that for each natural $$N\ge m$$, the a posteriori probability $$p_N'$$ of $$A_N$$ is $$\frac {p_N\cdot q_{S,n,m}}{\sum_{S=m}^\infty p_S\cdot q_{S,n,m}}.$$

Let $$\bar m=\left\lfloor\frac m{0.99}\right\rfloor$$. Then the probability that up to the round $$n$$ he have seen at least $$99\%$$ of all numbers should be equal to $$\sum_{S=m}^{\bar m} p'_S=\frac {\sum_{S=m}^{\bar m} p_S\cdot q_{S,n,m}}{\sum_{S=m}^\infty p_S\cdot q_{S,n,m}}.$$

This value depends on the probabilities $$(p_N)_{N\ge m}$$ provided there exist natural numbers $$S,T>\bar m$$ such that $$q_{S,n,m}\ne q_{T,n,m}$$. This clearly holds, for instance, when $$n>1$$ and $$m=1$$ because then for distinct natural $$S,T$$ we have $$q_{S,n,m}=\frac 1{S^{n-1}} \ne \frac 1{T^{n-1}} =q_{T,n,m}.$$