distribution of tablets
I used to consume nicotine tabs (giving up smoking :). I ate a half at a time. When getting a whole tab out, I cut it into two and put a half back. I experienced a higher and higher probability of getting a half during the lifecycle of such a pack. I was curious how this distribution looks like, so asked help from friends regarding the exact math behind and finally got a solution which is like the one below.
tabs
Notationally, let n1 denote the number of whole, and n2 the number of half tabs.
Number of whole tabs in step t+1 is n1(t+1), namely you can get a whole or a half (in step t+1), this is respectively:
/ \
n | t | / \
/ \ 1 \ / | / \ |
n | t+1 | = ------------------- | n | t | - 1 | +
1 \ / / \ / \ | 1 \ / |
n | t | + n | t | \ /
1 \ / 2 \ /
/ \
n | t |
2 \ / / \
+ ------------------- n | t |
/ \ / \ 1 \ /
n | t | + n | t |
1 \ / 2 \ /
Equation means that you can get a whole, in this case with n1/(n1+n2) probability. The number of wholes decreases by one (you cut that whole tab into half and put one half back). In the other hand you can draw a half tab with n2/(n1+n2) probability and the number of wholes remains the same (you consume the half).
Doing the same with nr. of half tabs as n2:
/ \
n | t | / \
/ \ 1 \ / | / \ |
n | t+1 | = ------------------- | n | t | + 1 | +
2 \ / / \ / \ | 2 \ / |
n | t | + n | t | \ /
1 \ / 2 \ /
/ \
n | t | / \
2 \ / | / \ |
+ ------------------- | n | t | - 1 |
/ \ / \ | 2 \ / |
n | t | + n | t | \ /
1 \ / 2 \ /
When you get a whole, in this case you put back a half (+1 for the halves). Or a half and you consume that (there’ll be one less).
After some simplification, the equations are like this in differential (&dimensionless) form:
n n - n
. 1 . 1 2
n = - ------ , n = -------
1 n + n 2 n + n
1 2 1 2
Making it simpler (N is the number of tabs, P is the probability you getting a whole):
. . .
N := n + n , NP := n , N-NP = n , n = NP+PN
1 2 1 2 1
. . . . . . . . .
n = N-NP-PN , NP+NP=-P , N-NP-NP = (2NP - N) / N = 2P - 1
2
. . .
N = P - 1 , PP-P+NP=-P , P = - PP/N
Finally:
. . N = P - 1 , P = - PP/N
Solving this numerically with below code and plot it with GnuPlot:
1 #include<iostream>
2
3 int main()
4 {
5 double dt = 1e-3;
6 double N = 1;
7 double P = 1;
8
9 for(double t=0; t<2.0; t+=dt)
10 {
11 std::cout<<t<<" "<<N<<" "<<P<<std::endl;
12 double oN = N;
13 N=N+P*dt-dt;
14 P=P-P*P*dt/oN;
15 }
16
17 return 0;
18 }
plot of N and P
