Quote:
Originally Posted by Mason Malmuth
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