Question for statisticians

[Aisthesis]

Supposing that one has a "true" win-rate of n bb/100. This is not actual win-rate but just an assumed number based on one's own (fluctuating) quality of play relative to the field.

Now, the law of large numbers says that your actual win-rate will approach n the more hands you play.

It seems like there should be some statistical formula involving standard deviation and number of hands played for just how fast actual win-rate (call it a) approachs n.

I would think this would actually be more like a probability function of the form: if n is your true win-rate and s is your standard deviation, and you've played h hands (in 100s), then the probability that a lies in the range [n-s/10, n+s/10] is p.


Anyone have a good formula (or formulas) for this?

[DiamondDog]

is this what you're after?

Not sure I understand it myself, but it seems to be talking about something similar.

[Aisthesis]

I think so, actually, but I'm going to have to digest it... [img]/images/graemlins/smile.gif[/img]

[BruceZ]

[ QUOTE ]
Supposing that one has a "true" win-rate of n bb/100. This is not actual win-rate but just an assumed number based on one's own (fluctuating) quality of play relative to the field.

Now, the law of large numbers says that your actual win-rate will approach n the more hands you play.

It seems like there should be some statistical formula involving standard deviation and number of hands played for just how fast actual win-rate (call it a) approachs n.

I would think this would actually be more like a probability function of the form: if n is your true win-rate and s is your standard deviation, and you've played h hands (in 100s), then the probability that a lies in the range [n-s/10, n+s/10] is p.


Anyone have a good formula (or formulas) for this?

[/ QUOTE ]

You want the standard error, which is simply s/sqrt(h), where s has units of bb for 100 hands. Then your actual win rate will lie within [n-s/sqrt(h), n+s/sqrt(h)] about 68% of the time, [n-2*s/sqrt(h), n+2*s/sqrt(h)] about 95% of the time, etc.

[uDevil]

BruceZ,

Is there a 2+2-like discussion forum for statistics? I've got some questions that require both specialized knowledge and patience with an ignorant newb.

[BruceZ]

[ QUOTE ]
BruceZ,

Is there a 2+2-like discussion forum for statistics? I've got some questions that require both specialized knowledge and patience with an ignorant newb.

[/ QUOTE ]

That would be this forum which you may regard as the "Probability and Statistics" forum.

[uDevil]

[ QUOTE ]
[ QUOTE ]
BruceZ,

Is there a 2+2-like discussion forum for statistics? I've got some questions that require both specialized knowledge and patience with an ignorant newb.

[/ QUOTE ]

That would be this forum which you may regard as the "Probability and Statistics" forum.

[/ QUOTE ]

Uh-huh. I meant like 2+2 but not 2+2. But ok, I'll try to compose a sensible question and start a new thread. [img]/images/graemlins/smile.gif[/img]

[Aisthesis]

I like this and will have to plug it in when I'm finished playing tonight. The "normsinv" thing on the other one threw me for a loop, and I still don't actually understand at all how it's calculated--but, conveniently, one can just plug it in to Excel ...

[BruceZ]

[ QUOTE ]
I like this and will have to plug it in when I'm finished playing tonight. The "normsinv" thing on the other one threw me for a loop, and I still don't actually understand at all how it's calculated--but, conveniently, one can just plug it in to Excel ...

[/ QUOTE ]

You use NORMSINV to get the number of standard errors N corresponding to a given probability P, and NORMSDIST to get the probability P corresponding to a given number of standard errors N. Use these formulas in Excel to do this:

N = NORMSINV(0.5+P/2)

P = 2*NORMSDIST(N)-1

This is what tells us that a 95% probability corresponds to about +/- 2 standard errors (actually 1.96).

These functions cannot be computed manually since they involve integrals which cannot be evaluated in closed form, so these have been computed numerically, and they are evaluated with the use of tables or Excel.

[Silent A]

[ QUOTE ]
[ QUOTE ]
Supposing that one has a "true" win-rate of n bb/100. This is not actual win-rate but just an assumed number based on one's own (fluctuating) quality of play relative to the field.

Now, the law of large numbers says that your actual win-rate will approach n the more hands you play.

It seems like there should be some statistical formula involving standard deviation and number of hands played for just how fast actual win-rate (call it a) approachs n.

I would think this would actually be more like a probability function of the form: if n is your true win-rate and s is your standard deviation, and you've played h hands (in 100s), then the probability that a lies in the range [n-s/10, n+s/10] is p.


Anyone have a good formula (or formulas) for this?

[/ QUOTE ]

You want the standard error, which is simply s/sqrt(h), where s has units of bb for 100 hands. Then your actual win rate will lie within [n-s/sqrt(h), n+s/sqrt(h)] about 68% of the time, [n-2*s/sqrt(h), n+2*s/sqrt(h)] about 95% of the time, etc.

[/ QUOTE ]

OK, I'm taking a risk trying to correct Bruce here, but ...

let,
S1 = SD per hand
s = SD/100 hands
Sh = SD per "h" hands

general equation:

(Sh)^2 = h*S1^2

Therefore,

s^2 = 100*S1^2
(Sh)^2 = h*S1^2

so,

s^2/100 = (Sh)^2/h

Sh = sqrt(h)*s/10

if,

WR = your win rate/100 hands after "h" hands

then,

Total winnings = h*WR/100

95% confidence interval is,

2*Sh = 2*sqrt(h)*s/10

Total winnings with confidence interval is:

h*WR/100 +/- 2*sqrt(h)*s/10

multiply this by 100/h to get it in terms of per 100 hands,

WR +/- 2*s/sqrt(h/100) = WR +/- 20*s/sqrt(h)