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12-01-2004, 07:39 AM
hi guys, i'm sure similar questions have been asked before, but i'm a retard when it comes to the search function. so here it goes:

since oct. 18th, i've played 26,909 hands of 15/30 party, and have lost 231 bb. now i'm not concerned over whether or not i can beat this game or not, because i'm sure i can, but what i am concerned over is just how long it's taking to recover and the incredible variance i'm facing.

for example,
from october 18-21, i faced a massive 450bb downstreak, and recovered nicely by mid november to cut it to 130bb. and several up and downs later, i'm faced with another 140bb swing in the past 2 days, giving me the 231bb total from before.

the question is:
with a 19.9bb/100 standard deviation, and assuming my winrate is anywhere between 1.75-2.75 how long is my road to recovery and how typical will 50bb downswings occur, 100bb? 150bb?

and any other info you think would be helpful, thanks guys.

mannika
12-01-2004, 02:21 PM
Your assumption about your winrate being between 1.75 and 2.75 may be incorrect. If your true winrate was 1.75, there would be only a 7% chance of breaking even over 27k hands, not to mention being down by 200 BB. If your true winrate was 2.75, there would be only be about a 1% chance of this happening.

However, for the sake of argument, lets say that you are indeed a 2BB/100 player. In this case, your average road to recovery is calculated merely by assuming you always make your 2BB/100. In this case, to get back from 231 BB, it will take you 11,550 hands to get back to even (on average). I'm not sure about the frequency of downswings, maybe someone else can give an approximation for these.

12-01-2004, 08:08 PM
[ QUOTE ]
Your assumption about your winrate being between 1.75 and 2.75 may be incorrect. If your true winrate was 1.75, there would be only a 7% chance of breaking even over 27k hands, not to mention being down by 200 BB. If your true winrate was 2.75, there would be only be about a 1% chance of this happening.

[/ QUOTE ]

Hey how do you calculate these percentages? i'm dumb at math.

kelvin474
12-02-2004, 01:38 AM
It's assumed that your results over each set of 100 hands are independent and follow a Normal distribution with mean = win rate and standard deviation around 19.9. Over 27,000 hand sample, the mean result is 270*(winrate) and the standard deviation is sqrt(270)*(winrate). This comes from what happens when you add independent normally distributed random variables together.

So, to get the probability of being down 200 BB (or more) after 27k hands, you need to find the sum of the probabilities of all the outcomes &lt;=-200 of a normally distributed variable with mean (270*winrate) and standard deviation around 320. If your a +2BB/hr player, your expected result for 27k hands is +540 BB. The result of -200BB is 740BB worse than the mean. 740 is a little less than 3 times the standard deviation. Because of the shape of the normal distribution, this translates to a little more than 1 in 400 chance that you will net -200 BB or worse over 27k hands.

Regarding the frequency of having a particular downswing: I have a formula for calculating the probability of reaching a particular "goal" at some time in the next N sets of 100 hands (I'm making the assumption that the result is observed only once every hundred hands). Its from Don Schlesinger's Blackjack Attack You can email me at stacksofblack@verizon.net if you would like to see it, its got hyperbolic sines and exponentials and stuff.

Another way you could do it would be to set up a simulation that creates runs of say 1 million hands calculates the number of times the bankroll dips X BB below a highpoint. However, you'd have to define what a downswing is. (when is the downswing over? do I have to recover back to the original point? If i lose 200 BB have i just experienced a 200 BB downswing AND a 150 BB downswing?) I can write some code in MATLAB for thsi and let you know what i come up with, if your interested.

HTH a guy who I could probably not help as much with a straight-up poker question.

KELVIN

BruceZ
12-02-2004, 02:32 AM
[ QUOTE ]
If your a +2BB/hr player, your expected result for 27k hands is +540 BB. The result of -200BB is 740BB worse than the mean. 740 is a little less than 3 times the standard deviation. Because of the shape of the normal distribution, this translates to a little more than 1 in 400 chance that you will net -200 BB or worse over 27k hands.

[/ QUOTE ]

You are associating 3 standard deviations with 0.25%, but that is the area under both tails of the normal distribution function, and for this application we only want the area under one tail, which is half as much or 0.13%. Also, if the standard deviation were 320, then 740 isn't "a little less than 3 standard deviations", it is 740/320 = 2.31 standard deviations which corresponds to 1.04%.

If you have Excel, you can get this by typing =1-NORMSDIST(740/320). Otherwise you need a table of the standard normal distribution. Usually these tables give you the area under the standard normal curve (bell shaped curve) from -infinity to x, and you would look up x = 740/320 = 2.31, for which you will find the value in the table to be 0.9896. That means there is a 98.96% chance that your result will be better than this, and only a 1.04% that it will be this far below average or worse. If you wanted the probability of being 740 above the average, the procedure and numbers would be identical.

kelvin474
12-02-2004, 03:44 AM
Your right Bruce regarding the # of SD, I was thinking "SD is 270" for some reason.

However I always thought "real close to 99.5% of results lie within 3 SD of mean for gaussians". I got 1 in 400 by taking half the .5% that are outside it. Was I remembering the wrong thing? Of course this isn't relevant to CDC since the numbers are what you correctly pointed out. I was in too much hurry when posting and I put down rough estimate that was incorrect and too rough /images/graemlins/mad.gif

BruceZ
12-02-2004, 04:35 AM
[ QUOTE ]
Your right Bruce regarding the # of SD, I was thinking "SD is 270" for some reason.

However I always thought "real close to 99.5% of results lie within 3 SD of mean for gaussians". I got 1 in 400 by taking half the .5% that are outside it. Was I remembering the wrong thing?

[/ QUOTE ]

99.73% lie within 3 SD. Half of the 0.27% outside it is 0.135%.

CDC might want to check out my introductory post on standard deviation (http://archiveserver.twoplustwo.com/showflat.php?Cat=&amp;Board=genpok&amp;Number=344985). There is some info there about how long it takes to break even.

Vaftrudner
12-02-2004, 08:04 AM
[ QUOTE ]
from october 18-21, i faced a massive 450bb downstreak

[/ QUOTE ]

That's really awful man...how many hands?

Hmmm, I think strip wrote somewhere he played some 40 k hands just breaking even... That's the worst I've heard of irl. /v