my goal is pretty self serving but i would also like help in putting all of these
formulas on the same page so no more searching, looking, and wondering if someday
it may appear. i am sure if you are in this forum you are probably looking for or
may someday be looking for the answers to these Qs. well i think it would be a
lot easier if we had one master webpage.

a webpage to explain standard deviation for our (as poker players) use:
http://www.internettexasholdem.com/p...n-vt14280.html


my numbers:
(all very small sample sizes but will work for most Qs)

cash:
9772 hands
standard deviation/100 hands=81.082 BB
WR/100 hands=8.45

SNGs:
3592 games at given level
12.07 ROI

confidence intervals:
75%=.68
90%=1.29
95%=1.65
99%=2.33

formula's i have:

key:
SD=standard deviation
WR=winrate in BBs per 100 hands
H=hands
CI=confidence interval

A.
WR range at %

WR +/- 2 SD/((#H/100)^1/2)

so for me:
at a 95% confidence interval this formula will tell me that 95% of the time
through any given set of 100 hands my winrate will be between 24.85BBs and -7.95BBs.

or easier version:
95% of the time my winrate will be 8.45 +/- 16.404

a more in depth explanation here:
http://archives2.twoplustwo.com/show...b=5&o=0&fpart=

B.
The number of hands needed to be confident you are breakeven or better

100*(((CI*SD)/WR))^2)

so for me at a 95% confidence interval:
i would need 25067 hands to tell if i am a break even player or not.

a more in depth explanation here:

http://www.toxicpoker.com/forum/just...598-page1.html



answers to these questions are great but i am really looking for the formula's.
some questions i have about standard deviations:

1.
i would like to know with a 95% confidence interval how many hands it would take for me to know that i am
5BB/100 hands winning player?
you will notice this is just a fraction different from formula A.

2.
with a 95% confidence interval what is the probability that i will go broke with a 10000BB bankroll
buying in for 100BB each time (also 100 buy ins). i believe this formula will either directly be able to
be applied to SNGs or have a version that will apply to SNGs. (the SNGs will once again have 100 buy ins).
i would like to be able to input my own number of buy ins into the formula as well.

3.
since it is really hard for me to apply this one to me i gave the idea to find the formula.
i am a 10BB/100 hands winner over 40k+ hands, i have just run at 3BB/100 hands for 3000 hands.
what is the probability?


i cant say thank you enough to anyone who can help with these and any other formula's you may have or know!
also anyone who can put together a webpage with these (if we get enough) or already knows of a webpage with all of these would be my hero!

bump

For 1. I don't think you have enough information.

Say you were a 10Buyins/100 games winner then you would have a p (chance of winning each game) of 0.55. Then if 5Buyins/100 games represents your actual result and that is at the 5% lower end of the binomial curve you would need:

1.96*sqrt(N*p*(1-p))/N = 0.025 so N = 1521 games


If your p is different then N will be different too. If your win rate was really 5Buyins/100 games it would take an infinite number of games to prove it because the right hand side of the equation becomes zero.

2. You need to know your actual win rate and the payout structure of the SNG. Also, the 95% level is not really relevant to the question. It's easiest to use a montecarlo to see how many times you bust out.

For a 5 seater winner-takes-all you find a breakeven player will go bust 65% of the time after 10K games. A player with an EV of 0.05 buyins goes bust 13.6% of the time after 10K games and a player with EV of 0.1 buyins goes bust 2.1% of the time. You just need to set up the simulation according to the parameters you need.

3. I am not quite sure if it is appropriate to use the binomial formula for the SD here. It might be if you are just playing one type of D2N's or SNG's , but not otherwise.

If it was appropriate you would reason:

if 10Buyins/100 games is equivalent to a p of 0.55 then the chance of running at p = 0.515 for 3000 games is:

Binomdist(0.515*3000,3000,0.55, true) = 0.006%

Apologies if this is all wrong. I just had a go at it because nobody else had responded. Also I realise that he is sometimes talking about hands and I am am talking about games.

Good effort, but it was also a thread that was troll bumped, so I didn't bother.

Sorry sir, but that was a sincere bump.

Fair enough, but I wouldn't expect someone to dig through the forums that deeply for an apparently old and ignored thread. I'll try to answer them sufficiently then.


1. It's an algebra exercise.


WR +/- 2 SD/((#H/100)^1/2) for 95% CI

If we want to know we'll be above a certain winrate with 97.5% certainty (97.5% is used here because a -2SD event or worse will only happen 2.5% of the time while a +2SD event or better will occur 2.5% of the time. So a 95% CI means the probability of both extreme events don't occur. Now we are just concerned with the low end extreme event, hence using 97.5% certainty.) If you rather truly use 95% certainty here you need to change 2SD to 1.645SD in the math here.

MWR = minimum desired win rate to 97.5% certainty

MWR = WR - 2 SD/((#H/100)^1/2)

Now solve for hands played:

H = 100*[2SD/(WR - MWR)]^2

(note how similar this is to OP's formula B )

So given OP's stats, the number of hands he needs with a WR of 8.45bb/100 to yield a long term WR of 5bb/100 with 97.5% certainty given a SD of 81.082bb/100 is:

H = 100*[2*81.082/(8.45-5)]^2 = 220,938 hands

for 95% certainty MWR > 5bb/100:

H = 100*[1.645*81.082/(8.45-5)]^2 = 149,066 hands

2. for cash games, these are typical RoR calculators found at sites like this:

http://www.evplusplus.com/poker_tools/risk_of_ruin/

For tourneys, this method could also be used, but it's less robust, and simulation of tourney results given expected ROIs/finish distributions are more accurate. Info can be found in this thread:

http://forumserver.twoplustwo.com/25...20/index2.html

3. This is much more difficult, imo. I have to give it more thought to deal with it properly. But if we just naively assume that the WR of the larger 40k sample is TRUE WR, then you can estimate the probability fairly easily if given the SD. In reality, this is a stretch because the test sample is not significantly smaller than the sample we are assuming to be the long-term winrate.

TWR = True Win Rate/100
SWR = Sample Win Rate/100
SD = Standard Deviation/100
H = hands in sample
x = number of deviations from TWR

SWR*(H/100) = TWR*(H/100) + x*SD*(H/100)^0.5

x = (H/100)^0.5*(SWR-TWR)/SD

Probability of x event occurring (or worse) = NormDist(x)
(where NormDist(x) determines the probability a normal distribution has a z-score the value of x or lower)

A Normal Distribution calculator can be found here:

http://stattrek.com/tables/normal.aspx

So for the example given:
TWR = 10bb/100
SWR = 3bb/100
SD = 82.085bb/100
H = 3000

x = (H/100)^0.5*(SWR-TWR)/SD
x = (3000/100)^0.5*(3 - 10)/82.085 = -0.4671

NormDist(-0.4671) = 0.3202

So having a WR of 3bb/100 or lower is 32% likely over 3k hands if the player has a true win rate of 10bb.

Cliffnotes: Variance is a bitch.

Thanks a lot.

btw, I didnt dig that deep, I just googled for articles and this was the first hit.

Here's a useful bump. Never again form a plural using an apostrophe. Ever.

bump for general utility