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 Originally Posted by Lukie
Right, which is basically the same thing I am saying, only I wasn't all that clear.
My point was that some people think that if you have a positive edge, you still have to go broke 100% of the time if you play any non-100% edge infinite times, which is not true.
You are right. The problem you are mentioning is called "risk of ruin" and can be simply described as follows:
- assume a gambler starts with an initial number of buy-ins BI
- assume he endlessly (unless he goes broke) plays a game where his probability of winning is P and his probability of loosing is 1-P, with P > 0.5
- his chance of going broke is ((1-P)/P)^BI. In the other cases he will become infinitely rich.
- as an aside, if P<=0.5, the gambler always eventually goes broke
So for example if a gambler has $5 and he repeatedly plays a $1 game where he has 60% chance to win (P=0.6), then his chance of going broke is:
(0.4/0.6)^5 = (2/3)^5 = 32/243 = ~13.2%
The mathematical proof can be found here:
http://www.columbia.edu/~ks20/FE-Not...7-Notes-GR.pdf
You can do further calculations to apply this to a poker player who has a given winrate, standard deviation and some number of buy-ins. See here: http://archives2.twoplustwo.com/show...st682045683150
and here for calculators based on these formulas:
http://www.castrovalva.com/~la/bank.htm
http://www.reviewpokerrooms.com/poke...uirements.html
(note that the winrate and standard deviation are in bb/100 and the bankroll is in bb, not in BI. A typical poker player's standard deviation is 80bb/100, check yours out in PT or HEM).
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