100K variance

[poker1O1]

Sorry if this is the wrong forum, but I feel the people that monitor this one will be more knowledgable than others. So, my question, let's say you're a 16/9/2 player (my point is that you are tight are there is not an overwhelming room for variance) and you make 1.5 BB/100 at 15/30 over 100K hands, all 10 handed. What (in terms of variance if possible) could u expect ur next BB/100 to be over the next 100K hands? Could it really differ by as much as .5 or is <.25 more reasonable (also assume you play exactly the same)?

[SmileyEH]

95% of your results will fall within .5BB/100 and 2.5BB/100 for the next 100k hands assuming you are playing the same game.

-SmileyEH

[CardSharpCook]

[ QUOTE ]
95% of your results will fall within .5BB/100 and 2.5BB/100 for the next 100k hands assuming you are playing the same game.

-SmileyEH

[/ QUOTE ]

that is a huge range.

[pfkaok]

[ QUOTE ]
95% of your results will fall within .5BB/100 and 2.5BB/100

[/ QUOTE ]

Is this correct? i thought that from these 100k hands that we could say with 95% certainty that his TRUE WR is between .5 and 2.5/100. I'm not sure that you can say his next 100k will be in that range 95% of time, but i'm not exactly sure how the stats would work out on that one... maybe it is correct, or at least close though.

[JasonP530]

That is not correct. You need to use the standard deviation for the confidence interval around the WR. Since he didnt give it, you cant figure out how certain you are that it would fall in a certain range.

[poker1O1]

[ QUOTE ]
That is not correct. You need to use the standard deviation for the confidence interval around the WR. Since he didnt give it, you cant figure out how certain you are that it would fall in a certain range.

[/ QUOTE ]
thanks jason, i have not played 100K hands so I'm not really sure what a reasonable s.d. is for that many hands, and also, that type of player. however, u guys said it should be between .5 and 2.5. With that much difference, isn't 2.5+ possible as well then?

[JasonP530]

If you want to be 95% certain that your next 100k hands will be within a certain range, you take the example given from your current data, and add 2 Standard deviations and subtract 2 standard deviations. 95% of the time, your next 100k hands will fall into that range.

[DpR]

Sigh...that is what SMiley did. 100k hands, 16 SD/100 (about right for PP15/30) gives you a 95% confidence interval of +/-1BB. So 0.5 to 2.5.

[DpR]

Yes, 95% CI for 100k hands 16sd is about 1BB.

[poker1O1]

[ QUOTE ]
Sigh...that is what SMiley did. 100k hands, 16 SD/100 (about right for PP15/30) gives you a 95% confidence interval of +/-1BB. So 0.5 to 2.5.

[/ QUOTE ]
ok thanks

[meanjean]

Maybe I'm a moron (well, screw the maybe), but does poker tracker give you standard deviation? or do I have to compute it in some way?

[pfkaok]

[ QUOTE ]

Sigh...that is what SMiley did. 100k hands, 16 SD/100 (about right for PP15/30) gives you a 95% confidence interval of +/-1BB. So 0.5 to 2.5.

[/ QUOTE ]

So doesn't this just mean that its a confidence Int for his TRUE WR? Maybe i'm missing something, but to me, the fact that you have a 95% CI within that range doesn't necesarrily imply that you have a 95% chance of winning in that range for your next 100k hands.

[pfkaok]

[ QUOTE ]
Maybe I'm a moron (well, screw the maybe), but does poker tracker give you standard deviation? or do I have to compute it in some way?

[/ QUOTE ]

yes it does, but i think its been argued that PT figures it in such a way that it might not be an increadibly accurate measure of your real statistical SD

[DpR]

[ QUOTE ]
[ QUOTE ]

Sigh...that is what SMiley did. 100k hands, 16 SD/100 (about right for PP15/30) gives you a 95% confidence interval of +/-1BB. So 0.5 to 2.5.

[/ QUOTE ]

So doesn't this just mean that its a confidence Int for his TRUE WR? Maybe i'm missing something, but to me, the fact that you have a 95% CI within that range doesn't necesarrily imply that you have a 95% chance of winning in that range for your next 100k hands.

[/ QUOTE ]

Yes, that is true. He can be 95% sure that his true win rate is 0.5-2.5. The confidence interval for the results of the next 100k hands is his TRUE win rate (which we are pretty sure is between .5 to 2.5) +/- 1BB.

[pfkaok]

[ QUOTE ]
Yes, that is true. He can be 95% sure that his true win rate is 0.5-2.5. The confidence interval for the results of the next 100k hands is his TRUE win rate (which we are pretty sure is between .5 to 2.5) +/- 1BB.

[/ QUOTE ]

OK, thats what I thought... the initial ? was about how to predict how the next 100k hands will go, and it seems like that would be slightly more complex to do than just a CI.

[DpR]

Yes, for example, there is a 2.5% chance that the lower bound of the 95% CI for his results for the next 100k hands is less than negative 0.5BB/100.

[SmileyEH]

Maybe there should be a sticky about confidence intervals, winrate and the like in general holdem or something.

-SmileyEH

[pfkaok]

[ QUOTE ]
Maybe there should be a sticky about confidence intervals, winrate and the like in general holdem or something.

-SmileyEH

[/ QUOTE ]

This is a good idea, as these ?'s come up quite often in every holdem forum.

[goodguy_1]

I dont know what your sd is but assuming you sd per 100 hands is 18 at 95% confidence level on 100K hands w/ a 1.50bb/100 winrate you should be in a range of 0.38bb/100-2.61bb/100 for your next 100K assuming quality of the games are static.
Svens Winrate Accuracy App. (http://www.svenskpoker.com/math.php)

[pfkaok]

thanks. i thought that the formula for that was different.

do you have any idea of what the actual formula is?

[goodguy_1]

[ QUOTE ]
do you have any idea of what the actual formula is?

[/ QUOTE ] sorry no but I'm sure you could get it in the Probabilty Forum very quickly.

[SmileyEH]

Your standard deviation in winrate over X x 100 hands is.

= standard dev/sqrt(X)

So if your standard deviation is 15BB/100 and your sample is 50k hands, then your standard deviation (corresponding to 68% of all results) is:

= 0.67BB/100

So if your winrate is 2BB/100 assuming constant game conditions your true winrate is 2BB/100 +- 0.67BB/100 68% of the time.

With 95% certainty your true winrate is 2BB/100 +- 1.34BB/100.

Sorry if this is unclear. I've had a bit to drink. Math.com has good info if I recall.

-SmileyEH

[pfkaok]

[ QUOTE ]
Your standard deviation in winrate over X x 100 hands is.

= standard dev/sqrt(X)

So if your standard deviation is 15BB/100 and your sample is 50k hands, then your standard deviation (corresponding to 68% of all results) is:

= 0.67BB/100

So if your winrate is 2BB/100 assuming constant game conditions your true winrate is 2BB/100 +- 0.67BB/100 68% of the time.

With 95% certainty your true winrate is 2BB/100 +- 1.34BB/100.

Sorry if this is unclear. I've had a bit to drink. Math.com has good info if I recall.


[/ QUOTE ]

SOrry, i knew that one, i was looking for the formula that would predict your next 50k hands... like after i have this CI of +/- 1.34 in your example, and my WR over 50k hands, what can i predict to win my next 50k hands (with 95% certainty).

from the example goodguy gave it looks like it'll be a little bit tighter, but maybe it was just in that specific one???

[SmileyEH]

[ QUOTE ]
[ QUOTE ]
Your standard deviation in winrate over X x 100 hands is.

= standard dev/sqrt(X)

So if your standard deviation is 15BB/100 and your sample is 50k hands, then your standard deviation (corresponding to 68% of all results) is:

= 0.67BB/100

So if your winrate is 2BB/100 assuming constant game conditions your true winrate is 2BB/100 +- 0.67BB/100 68% of the time.

With 95% certainty your true winrate is 2BB/100 +- 1.34BB/100.

Sorry if this is unclear. I've had a bit to drink. Math.com has good info if I recall.


[/ QUOTE ]

SOrry, i knew that one, i was looking for the formula that would predict your next 50k hands... like after i have this CI of +/- 1.34 in your example, and my WR over 50k hands, what can i predict to win my next 50k hands (with 95% certainty).

from the example goodguy gave it looks like it'll be a little bit tighter, but maybe it was just in that specific one???

[/ QUOTE ]

Its the same thing conceptually as far as I know. Unless you want your average winrate over the 200k hand sample. Then you would just divide everything by 2.

-SmileyEH

[CallMeIshmael]

[ QUOTE ]
OK, thats what I thought... the initial ? was about how to predict how the next 100k hands will go, and it seems like that would be slightly more complex to do than just a CI.

[/ QUOTE ]

They are the same.

Its hard to 'prove' it in a post but...

lets after 100k hands, say we have a wr = 2, and a sd = 16.

There is an x% chance that his true winrate is 2. And a y% chance that his winrate is 2 +- z (ie. 2.05 and 1.95 are equally likely).

What you would then do...

assume that his true winrate is 2, then calcuate the probability of all winrates over the next 100k hands

then assume that his winrate is actualy 1.95. Then 2.05.
And calculate the probability of all winrates over the next 100k hands..... and so on

In the end, you take take the weighted average of all of these calculations, and you get the original normal curve (normal is symetrical, afterall)

[JasonP530]

You would use the Z statistic (I believe) to find out the confidence intervals other than 67/95/99% since those are more well known. A decent statistics textbook will give you all the info you need.

[poker1O1]

ok, im not claiming to know a lot about statistics, but a decent ammount. I used a Z test, 95% confidence interval, so u have this equation, mean (2.5 BB/100) +- Z * (s.d./(sqrt(n))), and we are assuming s.d. = 16 and n = 100,000, and Z = 1.96. So my results show the next 100K hands should be between about 2.4 and 2.6 BB/100. Please correct me if I'm wrong.

[whitelime]

'n' should be 100k/100 since we are measuring SD and BB/100 in terms of every 100 hands.