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DogOnMySide

03-28-2006, 08:37 PM Post subject: Help with Programming/Maths!

#1 (permalink)

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Join Date: Apr 2005

Location: Manchester, UK

Posts: 458

Ok... here's an iterative formula which works in my program

Code:

for i = 0 to t
  B=B+xy(a^i)
next

basically I am trying to work out B.

I KNOW x, y and a... they're constants. It's fairly simple when you replace them with numbers.

It's the same as saying

Code:

B = xy(a^0) + xy(a^1) + xy(a^2)... and so on

What i need to do is write a similar loop or function or whatever which allows me to calculate x, assuming i know B.

It might be something like

Code:

x = b/y + b/ya ...

Do you see what I mean? I can't figure it out! this is for that bankroll calculator, by the way!

I am 27. I last did this shit when i was 21. That is a long time

If anyone knows what a for..next loop is and can do algebra/math but doesnt understand the question then please post because i might be able to explain it better

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shysti

03-28-2006, 09:09 PM

#2 (permalink)

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Join Date: Sep 2005

Location: The Internetz

Posts: 838

ugh my head hurts gg you pwned me

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DogOnMySide

03-28-2006, 09:13 PM

#3 (permalink)

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Join Date: Apr 2005

Location: Manchester, UK

Posts: 458

lol

its not that difficult. someone who has done a maths or science degree (and wasn't continually stoned for all 3 years like I was) will eat it for lunch

i'm hoping for demiparadigm or someone

"The best blog you'll ever read. Because after you read it I will poke your fucking eyes out"
- Martha Farqhar
http://mattspokerbankroll.blogspot.com/

shysti

03-28-2006, 09:22 PM

#4 (permalink)

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Join Date: Sep 2005

Location: The Internetz

Posts: 838

well I havent done a degree in that and I hate math IM so pwned

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Hate

03-29-2006, 12:13 PM

#5 (permalink)

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Join Date: Jul 2005

Posts: 322

unless i didn't misunderstand, that would mean that:

B=xy [ (a^0) + (a^1) + (a^2) + ... + (a^t) ]

Thus,

1/x = y/b [ (a^0) + (a^1) + (a^2) + ... + (a^t) ]

=> x = b/y [ (a^0) + (a^1) + (a^2) + ... + (a^t) ]

=> FOR 0<=i<=t x = b/y (SUM** a^i)

** Sum -> think excel sum formula...I'm sure you can handle that in programming.

gabe

03-29-2006, 01:19 PM

#6 (permalink)

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Join Date: Dec 2004

Location: trying to live

Posts: 7,964

gabe is an unknown quantity at this point

B=B+xy(a^i)

B=xya + xya^2 + xya^3 + xya^t
B=xy(a + a^2 + a^3 + a^t)
B/y(a + a^2 + a^3 + a^t) = x

so the code is:

for i = 0 to t
C=C+(a^t)
next
x=B/(yC)

am i close?

DogOnMySide

03-30-2006, 07:52 AM

#7 (permalink)

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Join Date: Apr 2005

Location: Manchester, UK

Posts: 458

Whoa some serious responses. Thanks. They're very similar to my attempts but subtly different... going to give them a go...

To explain all the variables...

x = table buyin e.g 25 for $25NL
y = "caution factor" e.g. 20, if you want 20 * buyins for your bankroll
B = bankroll
t = number of tables you're playing...
a = multi tabling factor... the lower this is the less effect multitabling has on bankroll requirements.

If you are 3 tabling 25NL with a standard "caution factor" of 20... and we call the a factor 0.25(Thru trial and error I think this is reasonable)
then i work out that your required bankroll B is....

B = (25 * 20 * 1) + (25 * 20 *0.25) + (25 * 20 * 0.0625)
B = 656.25

So if i plug in 656.25 into the other formula, then x should come out at 25, AMIRITE?

gabe yours says...
C = 0.25^0 + 0.25^1 + 0.25^2 = 1.3125
x = 656.25/(20 * 1.3125)
x = 25

B.I.N.G.O!!!!

Hate, thanks for the help but I didn't really understand the SUM bit!!! Looks v similar though.

Thanks very much. New BR calculator will have mad props to you for the help...

"The best blog you'll ever read. Because after you read it I will poke your fucking eyes out"
- Martha Farqhar
http://mattspokerbankroll.blogspot.com/

Hate

03-30-2006, 09:56 AM

#8 (permalink)

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Join Date: Jul 2005

Posts: 322

I'm not sure what language you're programming, but my SUM is the equivalent of the excel SIGMA formula ( http://images.google.com/images?q=tb...lhwd/sigma.gif )

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