Probability of winning a tournament Edited Question:
As I promised I've edited this question.  The previous version was written with the intention of simplifying the real question, but it ended in losing the real significance. Now I'm posting the "whole story". ;)
My purpose is to calculate $n$ players' equity in a poker tournament (their probability of ending the tournament) in every $j$ place (1st, 2nd, and so on).

I've previously solved this problem in 2 different ways you can find
  here:
For the Maths:
https://math.stackexchange.com/questions/92942/applying-a-math-formula-in-a-more-elegant-way-maybe-a-recursive-call-would-do-t
And for the code:
https://stackoverflow.com/questions/8605183/how-to-translate-this-math-formula-in-php

So, when I know every players' number of chips, I can easily apply those formulas and get their equity.

There are 2 problems involved that hopefully can be solved with a statistical method. (I'm not a mathematician so I'm not sure it will be feasible).


*

*First problem, even if I know everyone's stack, when the number of players is high, the code is too slow to be implemented;

*Second problem, this code should work by knowing only a limited number of stacks, belonging to the players of the analyzed table.


Optimistically these 2 problems can both be solved with some kind of approximations.
In particular the formulas mentioned above should be applicable to scenarios with 27,45,90 players who are distributed in tables of 9.
For example in the case of 27 players there would be 3 tables: when there are 18 players left they will be redistributed in 2 tables and when there are only 9 left the final table will be opened.
It's not important to take into account the players' skill since it's a high variance game where its influence is reduced to the minimum, and mostly there are coin flips that eliminate players.
So I'm in a situation where I know:


*

*My number of chips.

*The number of chips of the other 8 players of my table.

*The total number of chips.

*The average number of chips.

*The maximum and minimum number of chips.


As I suggested in the previous question, this seems to me (from my humbles math skills) to be a gaussian curve, with a maximum, a minimum and an average number of chips.
I think that's all. If you need additional details please leave a comment, and I will add them as soon as possible.
I wanna thank you for your interest, and for all the previous comments and answers. I hope your statistics can help me solve this. :D 
Best Regards,
Giorgio.


Old Question: 
I'd like to calculate or approximate the probability that I have to win a tournament where every player has a determined amount of chips.
Let's consider a scenario where there are 9 players and I know everyone's number of chips, to calculate my probability of winning I would do: my chips/(tot chips - my chips).
Now imagine those 9 players are put on 3 different tables of 3 players each, and I know the chips only of the 2 players of my table and mine obviously. I also know the total number of chips, the max and min amount chips the players have and the average stack.
Is it possible to make an approximation of my probability of winning?
I have only basic math skills but I think players' stack could be approximated to a gaussian curve,  then use some "statistical trick" to calculate my probability.
Thanks in advance for any hint!
Best regards,
Giorgio
 A: Do players keep playing until they run out of chips? 
Suppose we assume players A B C each have some number of chips and keep playing until they have 0. Suppose we further assume odds of winning a hand are either 1/3, 1/3, 1/3 or some other value (per Michelle's comment about skill). Suppose we further assume each pot consists of either win (+2) or loss (-1) -- an unrealistic but simplifying assumption.   Then doesn't this become a question of the relative probability of hitting 0 first? There's literature I'm not really familiar with that deals with this (Martingale wagers, but without the increasing bets).  With multiple tables, equal chip pools per table and 1 winner per table 2nd round would be p=1/3.
More: I did some quick simulations along these lines. Each player is equally skilled. A always starts with 50% of the chips, B 1/3 of the chips and C 1/6 of the chips. I just varied the total number of chips.  I did 3 runs at 3 levels, each consisting of 2000 simulations until only 1 player was left (18,000 simulations total). This suggests that it might indeed be as simple as p(I win) = my proportion of chips.
 
A: I suspect you are making this more complex than it needs to be.  If there are 27 million players (or even 27 thousand - I think you have a typo) and the game is pure chance and we can reasonably assume that no player has a significant number of chips compared to the total number of chips...  
Then your chance of winning is: $\frac{my chips} {total chips}$
Your change of coming second, given you didn't win, is: $\frac{mychips}{totalchips-winnerchips}$ which is effectively the same as your chance of winning.  And so on.  
Given the vast numbers involved and if as you say it is all chance I doubt it is worthwhile doing any more sophisticated calculation.  The secret of "statistical tricks" is to know when to use an approximation :-)
