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#11
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what bankroll must he have if he wants his ruin risk to be less than 10%? Assume a SD of 20BB's per hr.
Happy pokering, [img]/forums/images/icons/laugh.gif[/img] SittingBull |
#12
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Just invert above formula. B is bankroll:
B = -(sigma^2/2u)ln(r) For r = 10% = 0.1 and SD = 20*60 = $1200: win rate = $10/hr, B = $165,786 win rate = $20/hr, B = $82,893 win rate = $30/hr, B = $55,262 win rate = $40/hr, B = $41,447 win rate = $50/hr, B = $33,157 win rate = $60/hr, B = $27,631 These bankrolls are very large (450-2700+ BB) because the standard deviation is so large. |
#13
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larger bankroll to win a smaller hourly rate than he would need to win a larger hourly rate. What am i missing?? [img]/forums/images/icons/confused.gif[/img]
Hmmm Just wondering, SittingBull |
#14
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For a given standard deviation, the less you win, the larger bankroll you need, since your winnings are too small to overcome the swings caused by your standard deviation. We are assuming a constant standard deviation independent of winnings. Often, winning less will decrease your standard deviation, and winning more will increase it, so you need to know both to make this calculation. Bankroll and risk of ruin depend more on standard deviation than win rate since they depend on the square of the standard deviation or the variance.
20 bb/hr might be a standard deviation for a shorthanded game, or a very volatile game. People play these games because they can win more money. If you win twice as much money, but at the same time you also double your standard deviation, then you will need twice the bankroll to play this game for the same risk of ruin. You would have to win 4 times as much money in order to have the same risk of ruin with the same bankroll. |
#15
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If your win rate were only $10/hr in a 30-60 game, you would be much better off playing in a 10-20 game if you could win that same $10/hr, because then your standard deviation would be 1/3 as much, and you would only need 1/9 as much bankroll for the same risk of ruin.
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#16
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game is -10BB's to +10BB's.
Is this information sufficient to estimate my SD? Or would U need more information? Just wondering, [img]/forums/images/icons/shocked.gif[/img] SittingBull |
#17
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If your average swing per hour is +/- 10 bb, then your standard deviation should be approximately 12.5 bb. That is, 10 bb is .8*sigma. Note that this is different from saying half of your results lie within +/- 10 bb, and half lie outside. That would mean 10 bb was .67*sigma, and sigma would be approximately 14.9 bb. Your results should lie within +/- 1 sigma 68% of the time. All of these methods will tend to overestimate sigma a little since your actual hourly results are not exactly normally distributed. To get your true standard deviation you would average the square of your hourly swings to get the variance, then take the square root. See Mason's essay on computing your standard deviation in the essay section for variable length sessions.
The .8*sigma comes from: [2/(sqrt(2pi)*sigma)]*integral[0 to infinity]x*exp[-x^2/(2*sigma^2)] = 2*sigma/sqrt(2pi) = .8*sigma. |
#18
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#19
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Make them a proposal: Youhave to choose a winner among 4 hands. After you choose they remove one of the loosing hands and you get to choose again! This way you get 2:1 instead of 3:1 [img]/forums/images/icons/wink.gif[/img]
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