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Old 05-14-2017, 05:12 AM   #1
Mason Malmuth
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Pokerdope Poker Variance Calculator

Hi Everyone:

I was looking at the Pokerdope Poker Variance Calculator. which can be found at:

http://pokerdope.com/poker-variance-calculator/

In the example he has a winrate in BB / 100 of 2.5BB and a standard deviation per 100 of 100BB. And this gives him a minimum bankroll for less than 5% risk of ruin (»?«) of 5991 BB. Does anyone know how he calculates (or estimates) this number. Also, given that he's showing a 95 percent confidence interval isn't he actually calculating a bankroll for a 2.5 percent risk of ruin? or is he getting the calculated/estimated bankroll another way and is actually getting the 5 percent number?

Thanks and best wishes,

Mason
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Old 05-14-2017, 02:22 PM   #2
nickthegeek
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Re: Pokerdope Poker Variance Calculator

They are using the formula:

R(b) = exp(-2*mu*b/sigma^2), where mu is your expected win, sigma is the standard deviation, b is the bankroll and R is the risk of ruin. For derivation and assumptions, see Mathematics of Poker, pp 274-284.

In the page you linked, they set R(b) = 0.05 (since they want a 95% survival probability) and solve for b.

The 95% confidence interval shown has little to do with the RoR. That interval indicates the range you are expected to finish in if you play all the hands. However, if you start with a finite bankroll, you might not be able to end all the hands, since sometimes you lose it all along the path.
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Old 05-14-2017, 04:32 PM   #3
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Re: Pokerdope Poker Variance Calculator

So the 95% confidence interval that is displayed is not really a 95% confidence interval if you start with a finite bankroll (and cannot 'rebuy')?
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Old 05-14-2017, 05:12 PM   #4
Mason Malmuth
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Re: Pokerdope Poker Variance Calculator

Quote:
Originally Posted by nickthegeek View Post
They are using the formula:

R(b) = exp(-2*mu*b/sigma^2), where mu is your expected win, sigma is the standard deviation, b is the bankroll and R is the risk of ruin. For derivation and assumptions, see Mathematics of Poker, pp 274-284.

In the page you linked, they set R(b) = 0.05 (since they want a 95% survival probability) and solve for b.

The 95% confidence interval shown has little to do with the RoR. That interval indicates the range you are expected to finish in if you play all the hands. However, if you start with a finite bankroll, you might not be able to end all the hands, since sometimes you lose it all along the path.
Hi Nick:

BruceZ pointed this out to me many years ago and in the 2004 edition of my Gambling Theory book the following was added:

Quote:
Special Note: It has been pointed out to me by Bruce Zastera, one of the most knowledgeable posters on our forums at www.twoplustwo.com, that it is also possible to go broke before reaching the number of hours at which we evaluate the probability of a loss, as well as after that time, and this problem is more acute at the 95 percent level than I originally realized. This means that these tables are significantly underestimating by a factor of about 2 the amount of bankroll needed to only have a 5 percent chance of going broke. Fortunately, this problem mitigates as the probability of going broke is reduced. Thus the Bankroll Required to Assure a Win tables do contain solid estimates and produce a risk of ruin of approximately 1.1 percent, but the 95 percent tables should probably not be used.
However, in the PokerDope case, which by the way is an excellent tool for showing players the power of the short term luck factor, it seemed strange to me that he would have a 95 percent confidence interval on his graph which actually cprresponds (without regard to exactly how accurate it is when related to a required bankroll) with a 2.5 percent risk of ruin when he was giving a number for a 5 percent risk of ruin.

However, it has now been explained that his 5 percent risk of ruin number was calculated from a formula (which came from the excellent The Mathematics of Poker by Chen and Ankenman) and has nothing to do with the graph. I suggest that PokerDope make this more clear on his site.
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Old 05-14-2017, 05:19 PM   #5
nickthegeek
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Re: Pokerdope Poker Variance Calculator

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Originally Posted by whosnext View Post
So the 95% confidence interval that is displayed is not really a 95% confidence interval if you start with a finite bankroll (and cannot 'rebuy')?
These are completely different concepts.

The 95% confidence interval (although what shown in the page is actually a 2-sigma, 95.4% interval, and the 70% is a one-sigma 68% interval) says that if many players with same EV play 100000 hands each, 95% of them will finish inside that range. It doesn't consider the initial bankroll: each player is assumed to be able to finish all the hands. So the interval doesn't depend on the bankroll, while it depends on the number of hands played (of course the range will change depending on it, since if you play just a few hands, you can't neither win nor lose big amounts).

The risk of ruin expresses the probability of being able to play forever starting with a given bankroll. Even if you play a +EV game, you might incurr in a bad run and lose all your bankroll, making you unable to continue to play (the "ruin"). If you look at the formula, you'll easily realize that it doesn't depend on the number of hands (the assumption is that you survive forever), since both mu and sigma^2 increase linearly with N. So, RoR depends on the bankroll, but not on the number of hands played.
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Old 05-14-2017, 07:44 PM   #6
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Re: Pokerdope Poker Variance Calculator

I understand the math and I appreciate you describing it (again) for the forum viewers. But I think it goes a little too far to say that the two concepts are "completely different".
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Old 05-15-2017, 01:53 AM   #7
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Re: Pokerdope Poker Variance Calculator

I actually can't see many similarities between the two. Try putting a big number of hands, say 1 billion. The 95% range is [24367544 BB, 25632456 BB], hugely on the positive side. The 95% RoR bankroll stays at 5991 BB.

It has to be stressed that the RoR can't be calculated by simulating an arbitrary big number of hands and seeing how negative are the bad case scenarios.
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Old 05-19-2017, 03:49 AM   #8
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Re: Pokerdope Poker Variance Calculator

Quote:
Originally Posted by Mason Malmuth View Post
Hi Everyone:

I was looking at the Pokerdope Poker Variance Calculator. which can be found at:

http://pokerdope.com/poker-variance-calculator/

In the example he has a winrate in BB / 100 of 2.5BB and a standard deviation per 100 of 100BB. And this gives him a minimum bankroll for less than 5% risk of ruin (»?«) of 5991 BB. Does anyone know how he calculates (or estimates) this number. Also, given that he's showing a 95 percent confidence interval isn't he actually calculating a bankroll for a 2.5 percent risk of ruin? or is he getting the calculated/estimated bankroll another way and is actually getting the 5 percent number?

Thanks and best wishes,

Mason
The correct way to do this is to use a probability density function for the outcome of a typical hand of poker in ones typical game of poker if one had such distribution f(x) available (and was well defined under fixed conditions or well averaged out). Then consider the result of repeated summation of this distribution for a given bankroll and find the first return to 0 distribution of position as function of time. Then integrate that over all times.

That is actually better than seeing poker as a game of winrate m and sd s (especially for small bankrolls where one can easily find themselves all in after a few hands something the standard sd/m treatment doesnt anticipate so fast).

One can even consider the identical Wiener process that has m and s like poker ie Dx=m*Dt+W*s (W standard normal random variable) https://en.wikipedia.org/wiki/Wiener_process

However due to law of large numbers probably it wont be very bad to do the following thing;


Lets say one uses the standard risk of ruin expression for fixed unit risked step d. So p, q=1-p and d and bankroll B define that process.


Now avg is m=p*d-(1-p)*d=(2p-1)*d and s^2 = (p*(d-(2p-1)*d)^2+(1-p)*(-d-(2p-1)d)^2=4 d^2 p*(1-p) so s=2d*(p*(1-p))^(1/2)


Sow fix s to poker sd per hand and m to winrate. This yields d=(m^2+s^2)^(1/2), p=1/2+1/2*m/(m^2+s^2)^(1/2).

if B= bankroll

and z= B/d (how many bet step units left)

Then q(ruin)=((1-p)/p)^(z) from standard risk of ruin expression of position z times the bet risked each time against an infinite wealth adversary (eg see classic book by W. Feller for derivation)

So Risk of Ruin =(s/((s^2+m^2)^(1/2)+m))^(2*B/(m^2+s^2)^(1/2)

Notice that if s>>m (so m/s<<1) the above becomes (but only then)

(1-m/s)^(2*B/s)~ Exp[log(1-m/s)*2*B/s)=Exp[-m/s*2*B/s]=Exp[-2*m*B/s^2]

which is the earlier quoted expression.


This gives the same 5% risk of ruin for the m=0.025 and s=10bb/h for the 5991bb example.


Only posted this because it shows how to get there using the standard risk of ruin result.


For anyone interested see these pages too for more understanding of topics involved.

https://en.wikipedia.org/wiki/Gambler%27s_ruin

http://www.columbia.edu/~ks20/FE-Not...7-Notes-GR.pdf
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Old 05-19-2017, 04:44 AM   #9
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Re: Pokerdope Poker Variance Calculator

By the way see here from the archives too when that result was also derived first time long time ago in a more thorough manner by BruceZ;


http://archives2.twoplustwo.com/show...st682045683150
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