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

[Adman]

It is probably a great book but unless you have a strong math background you will find it frustrating. I know I was not able to get much out of it at all. I think the guys who are saying that it's "fairly basic" are much stronger in math than they realize. Also, I would add that I don't think understanding the complex material in that book is necessary to be a winning poker player, I'm sure it wouldn't hurt and could only help but certainly is not necessary.

[baztalkspoker]

From what I have seen of if so far it is defintely a book that is very daunting for the non mathematically minded. I have a mathematics and statistics degree - from a long time ago now and I am rusty, but even for someone like me I think it will need very careful reading and involve some relearning to grasp the equations. At the minute I'm still trying to work out the RoRU equation. I wish Chen had given some examples of it in this book [img]/images/graemlins/frown.gif[/img]
Early days yet as have only looked at a small bit of the book but I think an even slightly gentler introduiction and lengthier explanations and examples of formulas would have been a good idea. I get the impression that they have made the mistake of many intelligent math people of not adequately explaining the math assuming too readily that the reader is soaking it up, whereas in reality even people like me are having to do double takes to understand the stuff.

[Jerrod Ankenman]

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From what I have seen of if so far it is defintely a book that is very daunting for the non mathematically minded. I have a mathematics and statistics degree - from a long time ago now and I am rusty, but even for someone like me I think it will need very careful reading and involve some relearning to grasp the equations. At the minute I'm still trying to work out the RoRU equation. I wish Chen had given some examples of it in this book [img]/images/graemlins/frown.gif[/img]

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Example:
Suppose you have played 22,500 hands of 9-handed limit holdem, and you've won 337.5 bets, with a std deviation of
17 bets/100. So your win rate w is 1.5 bb/100, and your standard deviation s is 17.

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.

But you don't really know your true win rate, and you don't have enough of a sample to conclude that it's really super close to those numbers. Instead, we can approximate your win rate by another normal distribution. Suppose your win rate is a random variable with mean w and standard deviation that we can calculate:

s_w = 17/(sqrt(225)) = 1.1333

If we assume your win rate is distributed normally (it isn't, but it's a better approximation than assuming your win rate is exact), then we can do better by calculating your RoR for all the points in that distribution and summing them by their weights. To do this we use calculus, as in our book, and arrive at the RoRU formula:

roru = r(w,b)*(exp(2b^2(s_w^2/s^4)))(phi(w - 2b(s_w^2/s^2)) + phi(-w/s_w)

which if I've done the math right, comes out to 17.89%.

This reflects the uncertainty about our win rate -- the base RoR is our risk of ruin if we are 100% accurate on our win rate and standard deviation. But here we might have just overperformed our expectation - we're only about 1.5 standard deviations away from zero. And at w=zero, risk of ruin is 100%, while at w = 3.0 (the other side), risk of ruin only goes down by, say, 3%.

-- some dude

[SunyD]

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-- some dude


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lol classic

[baztalkspoker]

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Example:
Suppose you have played 22,500 hands of 9-handed limit holdem, and you've won 337.5 bets, with a std deviation of
17 bets/100. So your win rate w is 1.5 bb/100, and your standard deviation s is 17.

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.

But you don't really know your true win rate, and you don't have enough of a sample to conclude that it's really super close to those numbers. Instead, we can approximate your win rate by another normal distribution. Suppose your win rate is a random variable with mean w and standard deviation that we can calculate:

s_w = 17/(sqrt(225)) = 1.1333

If we assume your win rate is distributed normally (it isn't, but it's a better approximation than assuming your win rate is exact), then we can do better by calculating your RoR for all the points in that distribution and summing them by their weights. To do this we use calculus, as in our book, and arrive at the RoRU formula:

roru = r(w,b)*(exp(2b^2(s_w^2/s^4)))(phi(w - 2b(s_w^2/s^2)) + phi(-w/s_w)

which if I've done the math right, comes out to 17.89%.

This reflects the uncertainty about our win rate -- the base RoR is our risk of ruin if we are 100% accurate on our win rate and standard deviation. But here we might have just overperformed our expectation - we're only about 1.5 standard deviations away from zero. And at w=zero, risk of ruin is 100%, while at w = 3.0 (the other side), risk of ruin only goes down by, say, 3%.

-- some dude

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lol cheers 'some dude', a pretty cool alias to have actually. I'm still at a bit of a loss for what phi is representing - I gather it's a function, do I need to look up a log table to utilise this. Any chance you can explain it further here and exactly show how you got the answer 17.89%.

btw thanks for taking the time to respond here. It's appreciated. hmm I realise I have to get busy with going through your book and maybe I won't be posing these questions then anymore.

[Jerrod Ankenman]

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[ QUOTE ]


Example:
Suppose you have played 22,500 hands of 9-handed limit holdem, and you've won 337.5 bets, with a std deviation of
17 bets/100. So your win rate w is 1.5 bb/100, and your standard deviation s is 17.

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.

But you don't really know your true win rate, and you don't have enough of a sample to conclude that it's really super close to those numbers. Instead, we can approximate your win rate by another normal distribution. Suppose your win rate is a random variable with mean w and standard deviation that we can calculate:

s_w = 17/(sqrt(225)) = 1.1333

If we assume your win rate is distributed normally (it isn't, but it's a better approximation than assuming your win rate is exact), then we can do better by calculating your RoR for all the points in that distribution and summing them by their weights. To do this we use calculus, as in our book, and arrive at the RoRU formula:

roru = r(w,b)*(exp(2b^2(s_w^2/s^4)))(phi(w - 2b(s_w^2/s^2)) + phi(-w/s_w)

which if I've done the math right, comes out to 17.89%.

This reflects the uncertainty about our win rate -- the base RoR is our risk of ruin if we are 100% accurate on our win rate and standard deviation. But here we might have just overperformed our expectation - we're only about 1.5 standard deviations away from zero. And at w=zero, risk of ruin is 100%, while at w = 3.0 (the other side), risk of ruin only goes down by, say, 3%.

-- some dude

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lol cheers 'some dude', a pretty cool alias to have actually. I'm still at a bit of a loss for what phi is representing - I gather it's a function, do I need to look up a log table to utilise this. Any chance you can explain it further here and exactly show how you got the answer 17.89%.

btw thanks for taking the time to respond here. It's appreciated. hmm I realise I have to get busy with going through your book and maybe I won't be posing these questions then anymore.

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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

[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

[johnnyrocket]

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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]

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he may have taken the natural log and u worded this a little weird? i dont have a calc in front of me but natural log is the inverse of exp so it would be much smaller, i'll try it on a calc later if i find mine

[baztalkspoker]

[ QUOTE ]


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]