They might get different bankrolls even after a ridiculous number of hands. If we were to talk about infinity (not the vodka) of course they would be equal.
But i think even for millions of hands their results will diverge. That's because one of them might go down limits while the other will move up. So while one plays NL10 the other is at NL200.
Some dude not as lazy as me could make a little program simulating this problem (basic programming knowledge) to put a nail on it.
Here's a java program that can graph this concept with 100's or even 1000's of players:
http://www.pokervariancesimulator.fr/
Just click go and see how vastly the same player can differ over time. Also for you guys like like All-in EV graphs try 5 million hands and 100 players. The top player is still well above the winnings of the bottom.
Originally posted by Tosh5457
they will converge. opposite of diverge.
In theory, they will eventually have the same bankroll in the end, but that might take a few million hands to happen.
But if for example on the case of trading, you win 100% per year on your account. Two traders start with 10K. If by the effect of variance account of trader A goes to 11K and player B's account goes to 10K, since they have the same EV (100% per year) it's most probable in the long run that trader A will have more money than trader B (more 10%), and then they'll tend to diverge (the difference between them gets bigger and bigger, although player A is expected to have 10% more than player B and this percentage is expected to remain constant). It's like trader A started with 11K and trader B started with 10K.
In fact, in the beggining if they have the same edge it's expected that they'll have the same bankroll in the end. But as one of them gets a different bankroll by the effect of variance, that's no longer the most probable scenario, the most probable scenario is that the difference between their bankrolls will rise.
thats because when 1 guy has a bigger bankroll, the guy effectively "plays on another limit". his lot size as in ur example changes to a % of his bankroll.
secondly, you assume no equality in variance. if A loses 10% each trade consecutively in 10trades, and win 10% each trade consecutively in the next 10 trades, his account balance, at the end of these 20 trades, would be the same as B, who makes equal amount of wins/losses within this 20 trades at 10% each trade in any order of wins and losses.
Originally posted by tokyoaces
Here's a java program that can graph this concept with 100's or even 1000's of players:http://www.pokervariancesimulator.fr/
Just click go and see how vastly the same player can differ over time. Also for you guys like like All-in EV graphs try 5 million hands and 100 players. The top player is still well above the winnings of the bottom.
Random walk model?
the absolute difference MAY rise, but the relative difference will get smaller, i.e. convergence? or is that not convergence? i studied maths for a year or two, i should know ♦ i'm so bad
Originally posted by gadget51
Converges, has to.
except that it doesn't, lol
if after some n number of hands the difference between the two players bankrolls is x bucks, it should stay x bucks from that point on with 'neutral luck'
Originally posted by Kruppe
the absolute difference MAY rise, but the relative difference will get smaller, i.e. convergence? or is that not convergence? i studied maths for a year or two, i should know ♦ i'm so bad
what you're thinking of is that with an increasing n, A/B will converge to 1
The reason is doesn't converge is because you're adding pot size into the equation.
ie. What redskwerl said.
Hi guys,
i'm not sure i got the question correctly, but from mathematical point of view i would see it this way: let's consider the following game: you flip a coin every second and if you flip heads you win 1$ and if tails you lose 1$, we assume that you can be up or down any number of $.
if two players play this game and we are interested in the difference between their bankroll it's like only one is playing and flips twice every second co actually the same game. so suppose that one player plays this game for infitite time
for any number of $ given we can find time at which he will have exactly this number of $ (this is a well known mathematical fact, if you are interested try google "random walk"), so in other words difference of bankrolls of our two players will be this number, this means that this difference will have arbitrary big oscillations to both plus and minus so as a sequence cannot converge (at least not in classical sense) so it doesn't converge even in the case of such simple game
of course in poker such math can't be probably even applied. for example if phil ivey and patrick antonius would make a pact to follow first player to all of his tables and the other player would be playing say with me, i would omit convergence in any sense 
Originally posted by Wriggers
I always thought that if you run under EV then there is no reason you should run above EV in the future, or to put it simply: "The dice has no memory".So if you run 10BI under EV for 20k hands, you EV will then be expected to stay 10BI under EV, not for you to run 10BI above EV for the next 20k hands so you will meet up with you long term EV. That wouldn't make sense at all, that's basically saying "I was unlucky before so i've got to be lucky in the future".
So no, they should not diverge or converge if they have ran at different levels of EV.
the dice works both ways. positively and negatively. your current thinking only accounts for negative variance.
Originally posted by conall88
the dice works both ways. positively and negatively. your current thinking only accounts for negative variance.
No, it doesn't. You are thinking of an all-in preflop push/fold game. This is not the game we play.
Originally posted by redskwerl
Originally posted by Kruppe
the absolute difference MAY rise, but the relative difference will get smaller, i.e. convergence? or is that not convergence? i studied maths for a year or two, i should know ♦ i'm so badwhat you're thinking of is that with an increasing n, A/B will converge to 1
exactly. and is that convergence of A and B or not?
If the game is zero sum why can it not convege? Just a question not a criticism, even if I still think it should! 
I also think the problem requires too many assumptions to be posed correctly, but as usual I am likely wrong.
a lot of approximations and non sense in these posts 
To put things clearly: suppose A and B play the same limit indefinitely and are equally skilled, call A$ the winnings of A, B$ the winning of B, then we have:
- the absolute value of A$ - B$ tends to infinity (the longer the experience, the bigger the difference)
- it is the relative difference (A$-B$)/A$ which goes to zero when the sample becomes big. This means that although the difference A$-B$ might be really big, it is negligible with respect to the winnings of each player in the long term.
Hope this clarify things a little bit on this topic
Cheers
