Re: The Mathematics of poker by Bill Chen & some dude...
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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]
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