Hey Yoodle15.
How did you setup this problem? I can see a way to do it using Markovian modelling (with a lot of absorbing states).
This is the question I have:"Once a year employees at a company are given the opportunity to join one of three pension plans, A, B, or C. Once an employee decides to join one of these plans, the employee cannot drop the plan or switch to another plan. Past records indicate that each year 4% of the employees elect to join plan A, 14% elect to join plan B, 7% elect to join plan C, and the remainder to not join any plan. In the long run, what percentage of employees will elect to join plans A, B, and C? On average, how many years will it take an employee to decide to join a plan?"
From my calculations, I have found that in the long run, 16% of the employees will take insurance A, 56% will take insurance B and 28% will take insurance C. However, I feel suspicious of this answer because it seems so odd/unrealistic that there are no employees left who have not chosen to take any insurance plan. I also calculated that on average it will take an employee 4 years to get an insurance plan, and that answer seems fairly realistic.
I made a matrix with A as people in Insurance Plan A (4%), B as people in Insurance Plan B (14%), C as people in Insurance Plan C (7%), and D as people with no insurance plan (75%). In the matrix, one can see how people in 4% of people in D progress to A, 14% of people in D progress to C, and and 7% of people in D progress to C; 75% are left remaining in D. Here is the matrix:
A B C D A 1 0 0 0 B 0 1 0 0 C 0 0 1 0 D 0.04 0.14 0.07 0.75
Now, to find the behaviour of the Markov chain on the long-run, I found the limiting matrix of the above matrix.
Its limiting matrix is (according to my calculations, which may be faulty):
A B C D A 1 0 0 0 B 0 1 0 0 C 0 0 1 0 D 0.16 0.56 0.28 0
Hence, on the long run, there won't be any employees with no insurance plan (movement of D to D is 0): 16% of the employees will have taken Plan A, 56% will have taken Plan B, and 28% will have taken Plan C.