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Old 09-20-2005, 05:23 PM
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Default Re: Online Poker : 75% of loosers

Assuming a 5% rake online (??) That means that every dollar wagered pays out .95. So, to beat this, one must be 1.05263 times better than "average".

Where "average" is 1.0, "winner" is (> or =) 1.053.

Let's consider a "crushing" full-ring limit player is a +3BB/100 player. After 100 hands, a break even (or 1.05) player would have to win 7.5BB just to be at even (you pay 7.5BB every 100 hands). A "crusher" would have to earn 10.5BB over this period.

3/7.5 = .4, which means a crusher is 40% better than a break even player.

So, where "X" is average skill (slight loser), and equals 1X. Break-even is 1.05X, crusher is 1.47X

Among 1000 players, if the 1000th player was a 1.47 player and 500th was a 1.0 player, then about 565 would be a 1.05 player, (if it's a direct relationship ??), which means about 435/1000 players are showing a profit*, or 43.5%, while 56.5% are losing.

*This assumes nobody plays more than 1 table at a time. Since we know that the more likely you are to win, the more likely you are to multi-table, we can easily assume that the winners are much more rare than this. Guessing that the "average winner" plays 3 tables at once, I imagine this makes the winners 3 times as rare, which would bring the number down to 14.5%, which would render 85.5% of online players losers.

This is all quite fuzzy math of course, but hey...whatever. I think the number is probably around 90% in all honesty. No way it's lower than 85%.
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