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Old Aug 27, 2005, 12:13am   #41
nofuture
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Quote:
Originally Posted by NorthView
Quote:
Originally Posted by nofuture
give me a break... I like to visit other places to be bashed about.
I bet you do. Do you prefer to dress up while you're being bashed about, or do you prefer it au naturelle?
I like to wear poker t-shirts saying: My daddy went all-in last night.
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Old Aug 27, 2005, 12:15am   #42
nofuture
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Quote:
Originally Posted by pstevens101
Quote:
Say you can compute SD(ring) and SD(mtt). Now why shouldn't it be possible to compute SD(ring+mtt)?

For n games that is: (xi the ith game and X the arithmetic average)

SD = sqrt(1/n(sum([xi - X]^2)))

For n ringgames:

SD(ring) = sqrt(1/n(sum([xi - X]^2)))

For m MTTs:

SD(mtt) = sqrt(1/m(sum([xi - X]^2)))

For n+m ring and MTT games: (yi the ith game of ring and MTT and Y the arihmetic average)

SD(ring+mtt) = sqrt(1/(n+m)(sum([yi - Y]^2)))
As someone who is about to start a PhD in Statistics, this looks very ugly.
Could people please not do this, its not the kind of thing I want to be reading whilst eating my lunch.
Do you know what irony is? The SD of your lunch....
And BTW: I will of course regard everyones utility function of my posts in future - we're discussing sophisticated maths here that will swirl your brain.
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Old Aug 27, 2005, 8:18am   #43
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Originally Posted by nofuture
Don't you know there are people at 1$/2$ who don't know what a flush draw is? From that alone I could make a living and perhaps will do. I didn't say I can play well, I just said I can play better than others.
the best way to disarm a smartarse is to make yourself look even stupider than he could.

congratulations.

You would get on a lot better if you were aware of your station and acted in an appropriate manner. Show the game the respect it deserves.

Kc
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Old Aug 31, 2005, 1:39pm   #44
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Originally Posted by nofuture - posted 20 Aug
First I went to 10$ with .01$/.02$ Limit, then I won 30$ in a MTT and lost 20$ in .10$/.20$ NL. Then I went to 60$ .10$/.20$ Limit and lost 20$ in .50$/1$ Limit. I will probably play .10$/.20$ Limit/NL until I reach 150$ to conquer the .50$/1$ Tables.

What site are you playing at? I play at Everstpoker.

Quote:
Originally Posted by nofuture - posted 25 Aug
I play the highest limits possible at my site
...which would be 5$/10$ (1000NL)! I think you're talking BS...

Quote:
Originally Posted by nofuture
From that alone I could make a living and perhaps will do
I wish good luck! but just in case you end as a dragqueen in a rotten gay-club, please promise that you won't blame pokertips.org....
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Old Sep 03, 2005, 6:53pm   #45
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You guys are funny. I play Limit so 1$/2$ is indeed the highest Limit you can play, because the 2/4 Tables are hardly ever open.

I will blame Pokertips only for wasting my time with stupid accusations.
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Old Sep 03, 2005, 7:43pm   #46
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Quote:
Originally Posted by pstevens101
Quote:
Say you can compute SD(ring) and SD(mtt). Now why shouldn't it be possible to compute SD(ring+mtt)?

For n games that is: (xi the ith game and X the arithmetic average)

SD = sqrt(1/n(sum([xi - X]^2)))

For n ringgames:

SD(ring) = sqrt(1/n(sum([xi - X]^2)))

For m MTTs:

SD(mtt) = sqrt(1/m(sum([xi - X]^2)))

For n+m ring and MTT games: (yi the ith game of ring and MTT and Y the arihmetic average)

SD(ring+mtt) = sqrt(1/(n+m)(sum([yi - Y]^2)))
As someone who is about to start a PhD in Statistics, this looks very ugly.
Could people please not do this, its not the kind of thing I want to be reading whilst eating my lunch.
Could you write out a cleaner version (or more correct one) for those of us not starting a PhD in Stats?

I would be really interested to see how you tackle this.
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Old Sep 05, 2005, 12:16pm   #47
nofuture
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Originally Posted by XA_kid
Could you write out a cleaner version (or more correct one) for those of us not starting a PhD in Stats?

I would be really interested to see how you tackle this.
I nearly lost hope that there could be a normal post in this thread.

It's just a complicated way for saying: it's possible to add SD(Ring)+SD(MTT).
You take the deviation from the arithmetic average for each game and add it up.
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Old Sep 07, 2005, 9:57pm   #48
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Quote:
Could you write out a cleaner version (or more correct one) for those of us not starting a PhD in Stats?

I would be really interested to see how you tackle this.
If you have a data sample of size n with variance Vx and another sample of size m with variance Vy then th overall variance is

(nVx + mVy)/(n+m).

Quote:
It's just a complicated way for saying: it's possible to add SD(Ring)+SD(MTT).
You take the deviation from the arithmetic average for each game and add it up.
I'm not convinced it is.
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