The Mathematics of poker by Bill Chen & some dude...

[baztalkspoker]

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exp(2*(300^2)*(1.13333^2/1.5^4)) = 15.929


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Thanks again senor Jerrod.
I understand phi now thanks. And to give a little something back to 2+2 community to calculate the phi of a value in Excel use NORMSDIST function.

I'm still baffled though in at least one spot. I worked out that exp(2.768141347) = 15.929, but the figure in brackets here appears to be equal to 45668.86716, the exp(45668.867169) is massive of course. Am I reading the equation wrong or missing out on a bracket somewhere? [img]/images/graemlins/confused.gif[/img]

[Inf1n1tY]

Its a good book. I'm not far away from title it a "must read". it really helps you analyse your own hands and think diferently about situations.

btw: i spend about 2,5 month of getting the math down in the first ca. 50 pages

[baztalkspoker]

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phi is the cumulative normal distribution function. Suppose you have a normal distribution with mean mu and standard deviation s. For any value, you can make a "z-score," which is essentially the number of standard deviations away from the mean that you are.

z(x) = (x - mu)/s

So if your distribution has a mean of 10 and a standard deviation of 5, then 2.5 has a z-score of -1.5.

Phi(z) is the probability that if you randomly select a point from your distribution, it will lie to the left of the z-score z.

So take the familiar example that 68% of points lie between +1 and -1 standard deviations. This implies that phi(-1) is 16%, phi(0) is 50%, and phi(1) is 84%.

I got 17.89% by using the following variables:

w = 1.5
s = 17
n = 225
s_w = 1.13333
b = 300

ror(w,b) = exp(-2*1.5*300/17^2) = .0444
(that's term 1 in the roru formula)

exp(2*(300^2)*(1.13333^2/1.5^4)) = 15.929
(thats the second term)

phi(1.5 - 2*300*(1.13333^2/1.5^2)) = .121673
(that's the third term)

phi(-1.5/1.13333)
(that's the fourth term)

Multiplying terms 1,2, and 3 together and adding term 4 gives 17.89%.

-- still some dude

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Ah I spotted a little error you made that caused my confusion. It should have read exp(2*(300^2)*(1.13333^2/ 17 ^4)) = 15.929
(thats the second term)

phi(1.5 - 2*300*(1.13333^2/ 17 ^2)) = .121673
(that's the third term)

You had entered the win rate in to the formula instead of the standard deviation.Easily done. [img]/images/graemlins/wink.gif[/img]

[baztalkspoker]

I substituted 400 big bets for 300 big bets in Jerrod 'Some dude' Ankenman's formula, the result for RoRU that I got were rorU = 35.53% almost twice the rate for having 300 big bets, obviously this can't be correct.

I also did a check of the example given on page 302. Implementing the formula as described by Jerrod and I got a slightly different answer of 3.566% RoRU.

Both of my calculations were done in excel with formulas that correctly worked out Jerrod's example given earlier on this thread!!

I can't see any mistake that I might have made. Is it possible that there is a problem with this formula or with the way the dude descibed it here?

[baztalkspoker]

Oops spotted the mistake that I made with the 400 big bets [img]/images/graemlins/blush.gif[/img].

My 2nd more monor observation is right though I still think.

[Troll_Inc]

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Traditional risk of ruin says, for a 300 bet bankroll:

ror = exp(-2*w*b/s^2) = exp(-2*1.5*300/289) = 4.44%

This is the risk of ruin if your TRUE win rate is 1.5 bb/100 and your TRUE standard deviation is 17 bb/100.

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On what basis do you use standard deviation to predict what will happen a future sittings at a poker table?

[Doc T River]

Chen just wrote it so he would get invited to High Stakes Poker. [img]/images/graemlins/wink.gif[/img]

[Jerrod Ankenman]

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Traditional risk of ruin says, for a 300 bet bankroll:

ror = exp(-2*w*b/s^2) = exp(-2*1.5*300/289) = 4.44%

This is the risk of ruin if your TRUE win rate is 1.5 bb/100 and your TRUE standard deviation is 17 bb/100.

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On what basis do you use standard deviation to predict what will happen a future sittings at a poker table?

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<montypython>It's only a model.</montypython>

[Troll_Inc]

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Traditional risk of ruin says, for a 300 bet bankroll:

ror = exp(-2*w*b/s^2) = exp(-2*1.5*300/289) = 4.44%

This is the risk of ruin if your TRUE win rate is 1.5 bb/100 and your TRUE standard deviation is 17 bb/100.

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On what basis do you use standard deviation to predict what will happen a future sittings at a poker table?

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<montypython>It's only a model.</montypython>

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How about next time you guys come up with a model you do so for poker, and not some imaginary game?

[Barfunkel]

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"if you want to reach a wide audience"


I never got the impression they wanted to reach a wide audience.

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They definitely should write more books for the narrow audience. I could easily read a whole book about valuebetting the river or somesuch topic.