Converting monthly probability to yearly

If this data is from one company, then I'm not at all sure that is correct.

If the company has 1000 employees in January:  10 are fired, X quit, Y retire, and 262+X+Y are hired.
And so on for Feb, Mar, ... Dec.

In this case the correct answer would be the number people fired during the year divided by the average of employees.

     (10 + 21 + 4 + ... )
   ----------------------------------------
    (1000 + 1252 + 1333 + ...)/12


There is a factor 12 difference in the answers.

Maybe it should be the maximum number of employees rather than the average.


Say you have 1000 employees, and 100 are terminated every month.  
After 10 months there is nobody left.


So the answer would be:

  ( 100 + 100 + 100 + .... + 0 + 0)
--------------------------------------------   ==> 100% probability of getting fired     which clearly correct.
             1000

Monospace attached

Say you have 1000 employees, and 50 are terminated every month.  No other changes 

The method in the first post would give: 

  ( 50 + 50 + 50 + .... )                600
  --------------------------------   =  ----- =  0.071 = 7.1%
  (1000 + 950 + 900 + ...)               8400 

But the answer is clearly   600/1000 = 0.6 = 60%
                                              

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ah, sorry, I was calculating the average monthly termination probability,  not the yearly.

This is kind of a slippery problem.  Too many things changing at once.
You might want to look at the sum of the monthly averages weighted by number of employees.

actually, I think it's just a series of independent probabilities.

Jan and Feb are unrelated, so you multiply the probabilities of rentention to get the combined retention probability of those 2 months.  Subtract that from 1 to get the termination probability.

10      21      4 
---- x ---- x ----
1000   1252   1333 
 
or for the 100 per month example...
 
    100    100  100       100
1 - ---- x --- x--- ... x --- x 0 x 0 = 1
    1000   900  800       100
 
     50     50   50        50    
1 - ---- x --- x--- ... x --- x  = 0.6
    1000   950  900       450
                                              

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ooops I wrote those wrong

I left out the 1-  to get the monthly retention probability

     10           21          4 
(1- ---- )x (1- ----) x (1- ----) 
   1000         1252        1333 
 
or for the 100 per month example...
 
         100        100        100            100
1 - (1- ----) x (1- ---) x (1- --- )... x (1- ---) x 0 x 0 = 1
        1000        900        800            100
 
          50          50          50              50    
1 - (1 - ----) x (1- ---) x ( 1- ---)  ... x (1- ---)  = 0.6
         1000        950         900             450

                                              

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I'm having difficulty seeing the multiplication one intuitively... If it is P(JanTerm) x ... x P(DecTerm)... this would mean regardless of the increase in the probability of terminations, the probability will only be more and more unlikely. For example, (25/100) x (50/100) = 0.125... Give it another (90/100) and it will be 0.1125 even though next month's probability increased.

I originally thought of using the sum of the numerator divide it by the sum of the denominator because i'm trying to simulate rolling a dice... no matter how many times you roll it, it is still 1/6 chances to get anything. However for my case, both the numerator and denominator are changing.

I did consider a weighted average method, but I don't know how to do it for this scenario. Technically, all I have is the frequency of terminations happening. But the frequency itself does not have a value. So I can't really do [(10 * 1000) + (21 * 1252) + (4 * 1333)] / (10 + 21 + 4) = 1189.. Even if that was possible, I'm not sure what that 1189 represents or how to convert it to a sensible probability.

it's not P(jan) x ... x p(dec)


that was my error in 25754628  but I corrected it in 25754749


  1 - (   (1 - p(jan)) x (1-p(feb)) x ... (1-p(nov)) x (1-p(dec)))

you multiply the complimentary probabilities  (1 - p(term) )  to give you a cumulative retention probability for the year.

then subtract that from 1 to get the complimentary probability of cumulative termination probability for the year

Please confirm my understanding of this... For example
[1 - [(1-25/100) x (1-50/100) x (1-90/100)] ] = [1 - (0.75 x 0.5 x 0.1)]
= 1- 0.0375 = 0.9625 which is the cumulative probability of termination according to you.

Now if you assume next month there were 10 terminations out of 100...  it would be
1 - 0.03375 = 0.96625 cumulative probability of termination

Given that next month's probability of termination did dropped, the cumulative probability still increase... And even if the next 7 months had 0 terminations, the cumulative probability will not change. Following this scenario, there is no possible adjustment to that probability and it can only continuously approach 1 as time passes.

To me, that should not be the case. Given that there were terminations in the first 4 months only, and 0 for the next 8 months, the cumulative probability of terminations should not be the same as if I only did the calculation for the first 4 months and ignored the next 8.

Yes,  it doesn't matter how small the probability is in a month, if it's non-zero, it will INCREASE the cumulative probability.  It's not an average.  The probability from month to month will always be greater than or equal to the previous month.  equal if the current month probability is 0.


let's make it even simpler

In January I flip a coin  1/2  heads, cumulative 1/2 all are heads
In February I flip a coin 1/2 heads,  cumulative 1/4 all are heads
In March I flip a coin 1/2 heads, cumulative 1/8 all are heads
April-December I don't flip a coin.   cumulative probability that all coins for the year are heads is still 1/8.

That may be true for your coin.... but if we go back to terminations, isn't it unrealistic? If no one can get fired later on, how can the first month's probability of being terminated be used to determine the entire year's? That's like saying if 10/10 was terminated in january... people were rehired consequently... those people would have 100% chance of being terminated at any point in time even if no one has been terminated for the next 10 years or so.

you're right, declining to flip isn't the same as not-terminating people.


are you just looking for an average probability over time?   sum(p[1..n])/ n

well i'm looking for anything that makes sense or anything that models it more realistically. from my first post, my guess was [(sum of monthly terminations) / (sum of monthly employee count)] to get the probability. If that is not feasible, I would like to know the correct way.

In your response with sum(p[1...n])/ n... your n is the number of months which is 12?

Please tell me which way models the situation more realistically or correctly.