Winning Players

[AlanBostick]

[ QUOTE ]
try thinking of it this way;

In a hand of poker 1 person out of nine will win the pot, barring a split. Over the lifetime of the game if the same 9 players play (or 9 million) on average only the percentage that represents 1/9 would show a profit (in a zero sum scenario)

[/ QUOTE ]

This reasoning is completely wrong. Here's a counterexample:

Imagine a nine-handed game where everyone wagers $1 and then a 12-sided die is rolled. If it comes up 10, 11, or 12 it gets rolled again; otherwise, the wagers are awarded to the player in the seat whose number comes up.

After many rolls, the players' wins or losses will be normally distributed around a mean of $0. Half the players will be winners, half will be losers.

[Poker Plan]

[ QUOTE ]
[ QUOTE ]
try thinking of it this way;

In a hand of poker 1 person out of nine will win the pot, barring a split. Over the lifetime of the game if the same 9 players play (or 9 million) on average only the percentage that represents 1/9 would show a profit (in a zero sum scenario)

[/ QUOTE ]

This reasoning is completely wrong. Here's a counterexample:

Imagine a nine-handed game where everyone wagers $1 and then a 12-sided die is rolled. If it comes up 10, 11, or 12 it gets rolled again; otherwise, the wagers are awarded to the player in the seat whose number comes up.

After many rolls, the players' wins or losses will be normally distributed around a mean of $0. Half the players will be winners, half will be losers.

[/ QUOTE ]

But you can't equate this to poker. Your example works on the premise that everyone has an equal chance of winning. (ie your 6 will come up as often as my 3). So it's a slightly -EV proposition. There is no way to alter the odds.

In poker (obviously) every hand is different (in value) from it's neighbour.

[Farfenugen]

I think you people are misunderstanding the question so I will restate it.

[ QUOTE ]

If you had to assign a percentage to the amount of people who play poker/gamble seriously, what would the percentage be of that group who are winning players?


[/ QUOTE ]

I have heard the 10% figure and i would agree with it. However, that ten percent is for all poker players.

We are just talking about the subset of serious players. They are much more likely to win.

I would say the ammount of serious poker players who win is more around 50%.

[VBCurtis]

[ QUOTE ]

I have heard the 10% figure and i would agree with it. However, that ten percent is for all poker players.

We are just talking about the subset of serious players. They are much more likely to win.

I would say the ammount of serious poker players who win is more around 50%.

[/ QUOTE ]

This comes down to defining "serious". If you define it by a number of hours, hands, or months played, it becomes a survivorship problem. If you define it by skill, then your skill cutoff determines in large part the % winners, and the problem is circular. Do you include bonus? My roommate is -0.5 BB/100, but is up over $2000 from whoring. Is she a winner or loser?
-Curtis

[OpenWheel]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
try thinking of it this way;

In a hand of poker 1 person out of nine will win the pot, barring a split. Over the lifetime of the game if the same 9 players play (or 9 million) on average only the percentage that represents 1/9 would show a profit (in a zero sum scenario)

[/ QUOTE ]

This reasoning is completely wrong. Here's a counterexample:

Imagine a nine-handed game where everyone wagers $1 and then a 12-sided die is rolled. If it comes up 10, 11, or 12 it gets rolled again; otherwise, the wagers are awarded to the player in the seat whose number comes up.

After many rolls, the players' wins or losses will be normally distributed around a mean of $0. Half the players will be winners, half will be losers.

[/ QUOTE ]

But you can't equate this to poker. Your example works on the premise that everyone has an equal chance of winning. (ie your 6 will come up as often as my 3). So it's a slightly -EV proposition. There is no way to alter the odds.

In poker (obviously) every hand is different (in value) from it's neighbour.

[/ QUOTE ]

He wasn't equating it to poker. His game had equal opponents (outcome determined by chance) and zero rake. He was showing a simple example to clarify that the poster saying 1/9 would be winners long term was wrong.

I must admit that seeing someone make that 1/9 winners claim in that game, here on twoplustwo, made me sad. The normal distribution was the correct answer for that hypothetical.

If everyone was equally skilled and there was a rake then everyone would be a long term loser in that game, there would be no reason to play, although me email box has offers from people wanting to sell me their system for beating such a game.

I would have thought someone would have answered this for some sort of estimated +-ev/hand distribution including the effect of rake. If nobody has yet I'll tackle it in a couple days to relearn some probability stuff I've long forgotten from high school. As a first cut hypothetical I'd just assume unlimited bankrolls and everyone plays indefinitely, to eliminate survivorship issues. I don't know what to do about ev estimates, except maybe skill is distributed around a normal curve also as one estimate.

[dfan]

Survivorship is not the reason players will tend to find approx 45% winners in pokertracker (even though that is a common misconception). It is just an artifact of the large standard deviation of poker winnings/losses in comparison to the mean differences in players long-term win rates.

Here is an analogy. Say players played a game where you drew numbers from a hat and whatever number was on it you got that many dollars if it was positive and gave that many dollars if negative. 100 players sat down with their hats and numbers. Some of these players had hats with the numbers -100 through +100 in them, evenly distributed. These are the long term break even players. Others had hats with numbers -97 through +103 (long-term winning players) while some had hats with numbers -103 to +97 (the long-run losers, you know who you are).

So these hundred player sit down with their little hats and draw out their numbers. What percentage of the winning players will be winners after 50 draws? What percentage losers? What about the break even players? The losing players? Does the distribution of long-term winners, break-even, and losing players even matter much in the outcome?

If you think about it, no matter which group they belong to (winners, losers, or break even) the % that are winners after 50 draws from the hat is going to be close to 50%. The "edge" the winners have, and the "negative edge" the losers have is so small compared to the variation that it will barely even be detectable in the chaos of the wildly swinging outcomes. Even if you increase the number of draws (hands) to 100 or 200 or even 500, the % that will be winners not be that affected by the small positive or negative edges the players have.

So why doesn't PT show roughly 50% winners? Because of rake. If the rake didn't exist, everyone's poker trackers would say just below 50% were winners.

Now if you redid this experiment but had the players make 50,000 draws from the hat instead of 50, 100, or 500, almost all the long term losers will be in the red and almost all the long term winners will be in the black. At that point things like survivorship would have an influence on your stats.