[Jerrod Ankenman]

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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.

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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