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

[chucky]

If you never take advantage of the book, just copy down the hu jamming chart. It is useful book for thinking about playing against tough players and playing balanced poker. It is NOT a how to book.

[baztalkspoker]

I've only had the book a few days and haven't had a chance to go through it properly yet. I know about ROR formula already but the very useful looking RoRU(risk of ruin with uncertain win rate) formula I can't figure out yet. Anyone have some practical example(s) using pt based data?

[WiltOnTilt]

I just started reading this book. I consider myself above average at "poker math" -- specifically EV calc type stuff. I graduated with a comp sci degree so i'm not too foreign to formal math...

I think it's hilarious how many greek symbols and equations they have in first section marked "Basics". I'm understanding the book so far, although I had to re-read a few pages when i got momentarily lost, but I can't imagine how joe-average poker player could possibly understand the concepts starting off in the "Basics" section.

They should have done a better job dumbing it down and/or removing many of the equations/greek symbols/math lingo until at least later in the book. I know being math guys they feel the need to build the "foundation" of ideas so they can build on it later, but if you want to reach a wide audience it's simply not a good strategy to keep people interested. I have one friend who tried to read this book and threw it out 2 chapters in because of all the graphs/equations/and "foreign letters" that people w/o a math background don't understand.

That said I'm looking forward to reading the rest of the book for myself so I can translate it to others.

WoT

[MicroBob]

"if you want to reach a wide audience"


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

[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]

[ QUOTE ]
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]

[/ 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

[SunyD]

[ QUOTE ]

-- some dude


[/ QUOTE ]

lol classic

[baztalkspoker]

[ 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

[/ QUOTE ]

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]

[ QUOTE ]
[ 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

[/ QUOTE ]

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.

[/ 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