I previously asked:
Need help in calculating the average life of an online gambling player.
Here is some data:
Group A: 100 joined in month 1 to the site, 90 stayed in month 2, 50
stayed in month 3.

Group B: 75 joined in month 2, 60 stayed in month 3.

Group C: 120 joined in month 3.

So in total, Month 1 has 100 players, month 2 has (90+75)=165 players,
month 3 has (50+60+120)=230 players.

One answer i got was that the average life of a player can be calculated as:
Then, at the end of month 1, there were 100 players, all of whom had
been playing for 1 month, so the average age of the player would have
been 1 month.

At the end of month 2, there were 75 players who had been playing for
1 month, and 90 who had been playing for 2 months.  The average "age"
would then be (75*1 + 90*2)/(70+90) = 1.55 months.

At the end of month 3, there were 120 players who had been playing for
1 month, 60 who had been playing for 2 months, and 50 who had been
playing for 3 months.  The average "age" would then be (120*1 + 60*2 +
50*3)/(120+60+50) = 1.70 months.


NEW QUESTION:
Is it also possible (or better) to use the original number of players
in the denominator?
i.e. 
In month 1 its the same as before.
In month 2 it will be (90*2+75*1)/(100+75)=1.45
In month 3 it will be (50*3+60*2+120*1)/(100+75+120)= 1.32

Is this a valid calculation, or it is not mathematically correct?
Thank you.

As I understand your goals, using the sum of the original number of
players who joined in each month in the denominator will not give you
a useful result.  Doing this is equivalent to assigning a "zero age"
to all the people who have dropped out of the game, and including them
in the weighted average you are calculating.  That is, for month 3,
the calculation you propose is really equivalent to:

    (50*3 + 50*0 + 60*2 + 15*0 + 120*1)/(100 + 75 + 120)= 1.32



Although your followup question expired before I had a chance to
comment on it, I did provide a comment on that question in the comment
section of your original question.  See
http://answers.google.com/answers/threadview?id=568774.