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.

Can the average life of a player be determined from this data? or is
additional info required?
Thanks.

Ariel

  Since you say 'calculating'  I assume that you are ask about a problem
  in probability theory: How long it takes to lose one's initial capital, 
  given the parameters of a game,

  rather than a problem in Psychology: When do people get discouraged?

That is, you are curious about a problem like this:


"A customer of the Neptune Palace Casino is betting at a Roulette
table. He is following a gambling strategy that is often used by
prudent gamblers. He has dedicated a capital of $200 to this session
with the plan of not winning more than $20. He invariably bets $1 on
RED at each spin and plans to do continue playing until one of the
following three events occurs E1: He loses all of the $200, E2: He is
ahead by $20, E3: He has placed 2000 successive bets. Problem: What is
the probability that E2 takes place before E1 or E3?
http://www.math.ucsd.edu/~anistat/gamblers_ruin.html

and variants on that.

If so, you have formulated a classical problem in probability
called  : Gambler's Ruin

There is lot of literature and even java applets answering that.
 Simply enter this into Google
 SEARCH TERM: Gambler's Ruin
or (same thing) click here 
://www.google.com/search?hl=en&q=gamblers+ruin&btnG=Google+Search


Hedgie

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Request for Answer Clarification by stocks4ariel-ga on 16 Sep 2005 13:02 PDT

Hi.
Thanks for the effort, but i am still at a loss.
What i needed to know from the information i porvided is how to
calculate the average life of a player.

This includes of course some people who left the site, however, after
the 3 months in the example, more new people will come and people will
stop coming.

In some instances, there is a possiblity that those who left in the
first month will comeback in a much later month.

So i am not only considering the fact that they run out of money.

The sites provided were not so useful to me - perhaps i am not reading
them right.
I can send you a simple excel spreadhseet to show you how i calculated
average life, and would appreciate it if you can have a look and let
me know your thoughts.
Regards,
Ariel.

---

Clarification of Answer by hedgie-ga on 16 Sep 2005 20:44 PDT

Answer to your specific question is NO. 
You cannot calculate average life of player from data you gave.

 You can make some guesses, such as hafshaw-ga did in his
comment, but that can be argued with:
For example: 10 players left SOMETIME during the month 1, so 
average life of a player is certainly less than one month.

If you want to have average life without guesses,
you need to have a record for each player, like this:

player  joined    left ,    returned, left again, ...          total in days
  1       date     date      date      date                        ???
  2  etc

 Than, you add (total in days) for all players and divide that by
number of players.

Only alternative to such detailed record keeping is to make Markov Chain
model, of the type used in the Gambler's Ruin. Namely: you ASSUME that
each player has a constant probability to quit (=Pq) if s/he is playing
and to return (=Pr) after s/he did quit. 
Constant probability would mean same every day and same for all players.
 
That's the simplest Markov type model one can have for this problem.
Then you would  need to estimate these two numbers for your particular
audience.  You would still need for each
(or sample = subset off all, selected at random)
player the 'number of day played'  and number of days 'being absent' after
they did quit.
If you add third probability Pn (that a new person joins for the first time
on any given time) (which may depend on level of advertising) - then,
 assuming again that all this is not changing with time, you have a
simple model which would allow you to estimate numbers of players at
the end of each month etc as well as average gambling time  -  total
number of days a person who joined once will spend in the 'gambling
mode'.

  You still need to decide how to define what it means
  'a person did quit'  (does that mean 'no bet in two days', 'in a week'  ? 

Considering this ambiguity it may be a better model, to ask: 

 For a person whp placed at least one bet in [this] casino, 
 what is the probability that s/he will place another bet on any other day. 

 Then you have just one number (Pb) to estimate, which describes
both absences and gambling periods. You would still need the records of the
type described above, just a bit more simple:
player k1 : a bet on day   date1
player k2 : a bet on day   date2
player k3 : a bet on day   date3

which the software shoud be able to collect

Hedgie

Great. Thank you for the clarification.  I may come back with a
additional questions. but for now it seems sufficient.

For simplicity, let's assume that people only join or drop out on the
first day of each month, and calculate the average "age" of the
players present on the last day of each month.

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.


Is this the sort of answer you are looking for?  If so, then there are
ways to analyze the problem by making an analogy to the "residence
time" of a particle in a reservior, a common problem, for example, in
chemical engineering.

Your followup question expired already, but in case you are
interested, here are some comments addressing your new questions:

> ....is it only going to increase every month?

No.  One can easily imagine scenarios in which the average age
decreases, but such a condition cannot persist indefinitely.  For
example, let say that in both the first month and second months of a
game, 10 people join each month, and they all continue to playAt the
end of the second month, the average age will be 1.5 months.  Then
let's say 10 new people join at the beginning of the 3rd month, but
*all* the previous players drop out.  Then, at the end of the 3rd
month, the average age would be only 1 month.  The average age would
have dropped.


> In what cases can it actually decrease?

The change in average age, delta-T(t), from one time period to the
next is given by:

delta-T(t) = 1/N(t) * SUM from i = 1 to t of [delta-n_i(t)*(i*delta-t - T(t-1)]

Where N(t) is the total number of players at the end of time period t

delta-n_i(t) is the change in the number of players who have been
playing for i time periods at the end of time period t.  If you ignore
the complication of people dropping out and then rejoining, then
delta-n_i(t) will be greater than or equal to zero only for i = 1 (new
people joining) and i = t (the oldest cohort).  For 1<i<t,
delta-n_i(t) will be less than or equal to zero.

delta-t is the duration of the time period in question (i.e., months
in your case, but this could be weeks, days, years, etc.)

T(t-1) is the average age of the players at the end of the previous time period

This says that if the weighted sum of the change in the number of
players who have been playing for *less* than the average amount of
time is greater than the weighed sum of the change in the number of
players who have been playing for *more* than the average amount if
time, then the average age will go down.  In both cases, the terms in
the sum are weighted by the difference between the actual "age" and
the agerage "age" in the previous timestep.