Probability you are beating a NL game

[XChamp]

Is it possible to make some rigorous statements about whether a player is a *winner* at a NL game (with a max buy-in) based on how many buy-ins he is UP? By 'winner' I mean that you are winning based on skill, not random chance.

Example: Say Player A is up 30 buy-ins in 10,000 hands and Player B is up 30 buy-ins in 100,000 hands. They both play in the same game. What is the difference between these two in terms of what we can say about both players' ability to beat the game?

Which player would you rather be, and why?

At what point can we say that these players have equal likelihoods of being "winners". That is, B has won 30 buy-ins over 100,000 hands. Player A has played 10,000. How many buy-ins must A have won in order for you to declare that A is beating the game with the same likelihood that B is?

[sthief09]

to know this, you need to know your standard deviation

[Siegmund]

Well. Yes. You've not given us enough information to assign any fixed probability to either player's chance of being a winner, though.

The odds point to player A being the better player, on the evidence we have so far. How likely to be better, again, not enough information to say.

Your last question is the only one we can answer: I would rate a player who was sqrt(10) times as much after 10 times as many hands as equally likely to be a winning player. (For instance, 30 buyins / 10,000 and 95 buyins / 100,000.) As the other poster said, we need to know what the SD is to assign any guess as to how much better we think Player A is - all we know is that whatever the SD is, it scales with the square root of number of hands played in the long run.