Bankroll Requirements - Recreational Mathematics
Tom Weideman
I tried to post this last night, but apparently my ISP had problems with
its news server, so here goes again...
*****
In case you haven't seen it, there's an interesting thread whose subject
is "Bankroll requirements" over on the 2+2 Forum. It has to do with the
derivation of the minimum bankroll needed by a player with a given win
rate, standard deviation, and tolerance for risk that is found in
Malmuth's _Gambling Theory & Other Topics_. I encourage everyone to go
over and have a look.
Thinking about that topic gave me an outrageous idea for a way to
calculate bankroll needs, and in the Christmas spirit, I'd like to share
with you that bizarre spawn. If you are not the sort of person who
occasionally does a mathematical calculation for fun ("Gee, I wonder how
many ping-pong balls would fit in this room?"), then you may want to
move on to some post that discusses cardroom smoking or starting hand
requirements.
Okay, so Malmuth's derivation for minimum bankroll does not adequately
address the fact that the statistics used require paths that pass
through ruined states (like a random walker temporarily going beyond an
absorbing barrier). [Read the 2+2 thread for more details.] So what
CAN we do about getting a minimum bankroll number? I came up with this
warped model:
Poker Player's Ruin
-------------------
We can calculate the exact bankroll needed for a given risk tolerance
for one simple case: The classic "Gambler's Ruin" problem. In this
problem, you play a very simple recurring game for which you have a
probability p of winning 1 unit, and a probability (1-p) of losing 1
unit. You continue playing until either you are busted, or your
opponent is. Without going into the details, the probability that you
will go bust first ("R" stands for probability of being ruined) is given
by the equation:
R = [A^B - A^b]/[A^B - 1],
where: A = 1/p - 1
B = opponent's bankroll
b = your bankroll
Now if your are looking at playing poker forever, the bankroll of your
opponent is (effectively) infinite. You might think this makes your
probability of busting equal to 1, but it does not. If you play this
game with an edge (p>1/2), then A^B goes to zero as B goes to infinity,
leaving only:
R = [1/p - 1]^b
Solving for b (your minimum required bankroll, we get:
b = log(R)/log(1/p - 1)
This b will be the number of betting units (remember, the game played
requires the win or loss of one of these units) needed to reduce your
chance of going bust down to the value R, given that your probability of
winning a single game is p.
Great! Now all we have to do is somehow mold our poker results into
this form. All the data we have to work with is our average win rate
per hour (or hand), and our variance per hour (or hand), so let's go
from there. We start by computing the expected win per game (E) and the
standard deviation per game (S) of our model game:
E = (p)(+1 unit) + (1-p)(-1 unit)
E = 2p-1 units per game
S^2 = <x^2> - <x>^2 = [(p)(+1 unit)^2 + (1-p)(-1 unit)^2] - (2p-1)^2
S^2 = 1 - (2p-1)^2 = 4p(1-p)
S = 2*sqrt[p(1-p)] units per game
Because we are modeling this game to simulate our poker results, the
expectation equation gives a conversion between game units and big bets
per hour. We get this by just insisting that the expectations are the
same:
(2p-1) units per game = expected big bets per hour
Or:
1 unit = (expected hourly BB's earned)/(2p-1)
We'll need this conversion later, but now we need to be able to plug
something in for p in the bankroll equation. To get this, we use the
fact that the model is also constructed to have the same standard
deviation as our poker results, so the ratio S/E is the same for the
model as for our poker statistics:
S/E(model game) = 2*sqrt[p(1-p)]/(2p-1) = S/E(poker statistics)
Since S/E varies for each individual poker player, I'll leave it as a
variable, and call it just "SE". Now we can solve for p in terms of SE:
[omitting dull algebra, and keeping in mind that 1>p>0.5]
p = 0.5*[1+(SE^2+1)^(-0.5)]
Plugging this p back into the ruin equation gives the bankroll required,
measured in model game units:
b = (0.5)*log(R)/log{SE/[sqrt(SE^2+1)+1]}
Plugging p into the conversion factor from units to big bet gives:
1 unit = (hourly BB earn rate)/(2p-1)
1 unit = (hourly BB earn rate)*sqrt(SE^2+1)
Finally, we use this conversion to give the minimum bankroll in units of
big bets:
min br (big bets) = (0.5)*log(R)*E*sqrt(SE^2+1)/log{SE/[sqrt(SE^2+1)+1]}
where: R = accepted risk (probability of ruin)
E = earn rate (hourly, in big bets)
SE = standard deviation - to - earnings ratio (hourly)
Wow, what a mess. Please feel free to find all the algebra errors, and
report them back to the group with a snicker. Let's try some numbers.
Suppose we are willing to assume very little risk. Using 0.14% (a
probability of 0.0014), an earn rate of 1 BB/hr, and a standard
deviation of 10 BB/hr, we calculate the required bankroll to be:
br = 331 big bets
Compare this with the number found in _Gambling Theory & Other Topics_
(the claim there is that 0.0014% risk equates to a 3-sigma event), which
comes out to be
br (GT&OT) = 225 big bets
Quite a large difference! I'm by no means claiming that my calculation
is correct (because it isn't), but this is a wide discrepancy
nonetheless. Oh, why isn't this method correct? Just because we can
convert mean and standard deviation into this model doesn't mean that
the statistical distribution created by this model is the correct one.
In fact, it clearly is NOT the correct distribution, because according
to the model, each hour you sit at the poker table you either win or
lose the same fixed amount (with probabilities for each of these events
being p and 1-p repectively). For the numbers in the example, you would
either win or lose about 10 big bets per hour, winning about 55% of the
time, and losing 45% of the time. These numbers give the correct earn
rate and standard deviation, but clearly the distribution of results is
significantly different from real life.
Malmuth's calculation surely gives a lower limit on the exact value, and
though I haven't tried to prove it, it's possible my number is an upper
limit. Hope you enjoyed my little jaunt through bizarromathworld. I
think I hear reindeer on the roof... to all a good night!
Tom Weideman
William Chen
Hmm. Well maybe I should post my formula.
The chance of busting out with Bankroll B, std dev s and
win rate w is exp (-2*w*w/(B*s)). This is based on the
probability of busting out with a bankroll of 1 dollar
when you have a (1+w)/2 chance of winning a bet.
Bill
William Chen
Argh that was wrong. For a win rate of w, Bankroll of B
and std deviation of s, your chance of busting is exp(-2Bw/s^2).
Bill
Tom Weideman
Thank you for that enlightening commentary. Will you be joining 2+2's
authoring staff soon?
Hints:
1. Read a post before you comment on it. The part where my post says
explicitly that the results are not correct make your statement, "Argh
that was wrong" particularly annoying.
2. Some details behind your own contribution might be helpful, as some
of us are interesting in understanding WHY a result is what it is,
rather than having it handed down on tablets. For example, you should
mention to what extent your equation is an approximation of the exact
value (i.e. what simplifying assumptions were made, etc.).
3. Are you aware of how close our answers are to one another? [Again,
judging from the "Argh that was wrong", I'd guess you have no clue how
close they are.] Let me demonstrate.
Clearly s>>w for all realistic cases, so I'll take my result and ignore
all terms that appear of order (w/s)^2 and above (note my notation uses
S and E rather than s and w, but I'll adjust):
My result was:
B = (0.5)*log(R)*w*sqrt[(s/w)^2+1]/log{(s/w)/[sqrt((s/w)^2+1)+1]}
Now for math... I'll do simplying in pieces. First the sqrt's:
sqrt[(s/w)^2+1] = (s/w)sqrt[1+(w/s)^2] = (by approx) = s/w
Plug back in:
B = (0.5)*log(R)*w*(s/w)/log{(s/w)/(s/w+1)}
Simplifying:
B = (0.5)*log(R)*s/[log{1/(1+w/s)}]
B = -(0.5)*log(R)*s/log(1+w/s)
Now expand the natural log in a power series in w/s, throwing out the
terms of power greater than or equal to (w/s)^2 to get:
log(1+w/s) = (by approx) = w/s
B = -(0.5)*log(R)*s^2/w
Solving for R (the risk of busting) gives:
R = exp(-2Bw/s^2)
Imagine that.
Tom Weideman
Tom Weideman
I have learned from private email that Mr. Chen is responding to his own
earlier post, rather than mine (this earlier post has still not arrived
on my news server, but when I try to check the reference I find the
referenced article is not mine but some unknown article).
I apologize for the harsh rhetoric.
Tom Weideman
Lone Locust of the Apocalypse
Tom Weideman <zugz...@dcn.davis.ca.us> writes:
>William Chen wrote:
>>
>> Argh that was wrong. For a win rate of w, Bankroll of B
>> and std deviation of s, your chance of busting is exp(-2Bw/s^2).
>>
>> Bill
>
>Thank you for that enlightening commentary. Will you be joining 2+2's
>authoring staff soon?
>
>Hints:
>
>1. Read a post before you comment on it. The part where my post says
>explicitly that the results are not correct make your statement, "Argh
>that was wrong" particularly annoying.
Um, I think Bill was referring to his OWN post when he said "Argh that
was wrong." Now, I may have somewhat of an advantage here because I
know Bill personally so I'm familiar with his quirks and can maybe fill
in blanks that other people can't, but I think that in this case if you
read his other article in this thread, then it's pretty clear that the
"argh" post is intended as a followup to that one.
Why wasn't it posted as such then? Well, maybe Bill's not 100% clear
on Internet conventions, or maybe the article hadn't yet shown up on
his news server so he could follow-up to it so he chose the original
article again, or maybe he's having trouble with his newsreader or
ISP.
Anyway I think you were a bit quick to jump to a conclusion and your
tone was unwarranted.
-- Z.
____
I'm nice... He's nice... and we're both f---ing lunatics. \ /
Can I come in please? [_Flatliners_] \/
Dave Horwitz
Tom Weideman wrote:
> I apologize for the harsh rhetoric.
Not so "recreational" when you risk getting hacked
up by somebody slinging sharp formulas.. :)
-Quick
Andrew Morton
Tom Weideman wrote:
> William Chen wrote:
> >
> > Argh that was wrong. For a win rate of w, Bankroll of B
> > and std deviation of s, your chance of busting is exp(-2Bw/s^2).
> >
> > Bill
>
> Thank you for that enlightening commentary. Will you be joining 2+2's
> authoring staff soon?
Whoa, boy. I could be mistaken, but I think you seriously misread Chen's
post. On my server, he responded twice to you, the first time with a
formula that he briefly described the basis for, and the second time 20
minutes later with a different formula. His "Argh" comment was directed at
his own first post, I believe. Also, while he didn't explain this
thoroughly, I think the derivation of his formula was presented a couple
months ago in a thread I started (or at least participated in, anyway).
Second point, regarding the differences between your model and the MaxLik
formula, for high levels of risk tolerance. If MaxLik suggests you need
only $67 instead of $450 or so, what does your formula say about the risk
for such a player who actually does begin play with a bankroll of only
$67? ie, can you drive your formula backwards? Is the player with only
$67 nearly certain to go broke, as the discrepancy between the two results
might suggest?
Thanks for the interesting post (um... the original one. ;-)
andy
William Chen
In article <34A7EE...@dcn.davis.ca.us>,
Tom Weideman <zugz...@dcn.davis.ca.us> wrote:
>William Chen wrote:
>>
>> Argh that was wrong. For a win rate of w, Bankroll of B
>> and std deviation of s, your chance of busting is exp(-2Bw/s^2).
>>
>> Bill
Well, I couldn't get a hold of my article to follup up to since there
is some delay in posting. I guess I should have said oops I was wrong.
I thought both articles would post at the same time so it would be clear.
>
>Thank you for that enlightening commentary. Will you be joining 2+2's
>authoring staff soon?
>
>Hints:
>
>1. Read a post before you comment on it. The part where my post says
>explicitly that the results are not correct make your statement, "Argh
>that was wrong" particularly annoying.
Hint: follow your own advice, unless only my second article posted to your
reader at the time you responnded.
>
>2. Some details behind your own contribution might be helpful, as some
>of us are interesting in understanding WHY a result is what it is,
>rather than having it handed down on tablets. For example, you should
>mention to what extent your equation is an approximation of the exact
>value (i.e. what simplifying assumptions were made, etc.).
What was annoying was I wrote an explaination of my derivation twice in
this newsgroup already. I also posted it in the 2+2 forum. I'm a little
unsatisfied with my explaination in 2+2, so let me try to clarify it tonight.
>
>3. Are you aware of how close our answers are to one another? [Again,
>judging from the "Argh that was wrong", I'd guess you have no clue how
>close they are.] Let me demonstrate.
Yes, I read your article. I realize you made a mistake (since I like read
both of your followups before posting this message), but this tone really
isn't the right way to get constructive criticism about what you wrote.
>My result was:
>
>B = (0.5)*log(R)*w*sqrt[(s/w)^2+1]/log{(s/w)/[sqrt((s/w)^2+1)+1]}
>
>Now for math... I'll do simplying in pieces. First the sqrt's:
[rest deleted]
>
>Solving for R (the risk of busting) gives:
>
>R = exp(-2Bw/s^2)
>
>Imagine that.
>
Gee, imagine. I also think the way we derived the formulas were similar too.
I'll have some more to say about it tonight.
Bill.
Bill
Tom Weideman
Andrew Morton wrote:
> Whoa, boy. I could be mistaken, but I think you seriously misread Chen's
> post. On my server, he responded twice to you, the first time with a
> formula that he briefly described the basis for, and the second time 20
> minutes later with a different formula. His "Argh" comment was directed at
> his own first post, I believe. Also, while he didn't explain this
> thoroughly, I think the derivation of his formula was presented a couple
> months ago in a thread I started (or at least participated in, anyway).
As I posted, it was all a misunderstanding (Even at the time of this
posting, Bill's first post hasn't appeared on my server, and he didn't
quote his earlier post in his follow-up, so I assumed he was refering to
my post). I hope that this is all behind us now.
> Second point, regarding the differences between your model and the MaxLik
> formula, for high levels of risk tolerance. If MaxLik suggests you need
> only $67 instead of $450 or so, what does your formula say about the risk
> for such a player who actually does begin play with a bankroll of only
> $67? ie, can you drive your formula backwards? Is the player with only
> $67 nearly certain to go broke, as the discrepancy between the two results
> might suggest?
I'm confused. I thought the previous "MaxLik" discussion had to do with
computing hourly standard deviation from session numbers. Maybe the
thread carried on into discussing bankroll requirements, though.
But to answer your question, putting the numbers into the equation in
reverse, I get that the "25% risk bankroll" of $69.36 derived from
Malmuth's method corresponds to a 79.5% risk using my formula.
Oh, and one last note: Because of my not-so-nice tone in my response to
Bill, I think readers may overlook the interesting result presented
therein: His formula and mine are VERY close approximations of each
other, given that s>>w. We must have done very similar things in our
derivations. If I get a chance, I'll go back and look at both to see if
I can make out where we diverged.
Tom Weideman
Abdul Jalib
William Chen <w_c...@ix.netcom.com> writes:
> In article <34A7EE...@dcn.davis.ca.us>,
> Tom Weideman <zugz...@dcn.davis.ca.us> wrote:
>
> >William Chen wrote:
> >>
> >> Argh that was wrong. For a win rate of w, Bankroll of B
> >> and std deviation of s, your chance of busting is exp(-2Bw/s^2).
> >>
> >> Bill
> Well, I couldn't get a hold of my article to follup up to since there
> is some delay in posting. I guess I should have said oops I was wrong.
> I thought both articles would post at the same time so it would be clear.
It was your fault, Bill, because you didn't follow up to your own
article. If you follow up to some other article and don't include
any quoted reference material, anyone's first thought will be that
you are refering to the article you are following up to. It's just
plain rude to reply without anyone knowing what you are replying to -
it becomes noise.
And, because of the distributed and rather flakey system of news
propagation, you cannot even know that your two articles, sent seconds
apart, will even arrive at every news machine within *days* of each other.
Your first article could even be lost from many machines, with only the
second article ever appearing on these machines.
> >My result was:
> >
> >B = (0.5)*log(R)*w*sqrt[(s/w)^2+1]/log{(s/w)/[sqrt((s/w)^2+1)+1]}
> >
> >Now for math... I'll do simplying in pieces. First the sqrt's:
> [rest deleted]
> >
> >Solving for R (the risk of busting) gives:
> >
> >R = exp(-2Bw/s^2)
> >
> >Imagine that.
> >
> Gee, imagine. I also think the way we derived the formulas were similar too.
> I'll have some more to say about it tonight.
The formulas in this thread, flaming or not, are of great interest to me,
and they seem better than most of the published ruin formulas.
Back in 1991, before he was stalked and murdered after playing a 20-40
hold'em game, Michael Hall wrote something about Mason's method of
calculating required bankroll. The full text is available at
ftp://ftp.csua.berkeley.edu/pub/rec.gambling/ruin.txt.gz
Michael Hall wrote:
==========================================================================
VI. Mason Malmuth on the Ruin Formula
From a letter to the editor, by Mason Malmuth, Blackjack Forum,
December 1988...
Mason Malmuth:
"George C's article on `The Ruin Formula' (BJ VIII #3) was
very well done,
[Personally, I think it was a sloppy piece of shit - Abdul]
... but if you use the method give in my
book, `Gambling Theory and Other Topics', you can extend
these ideas even further and produce what I think are
even more interesting numbers. Specifically, consider
the following equation:
LL = (WR)(N)-(3)(SD)sqrt(N)
where LL is your lower limit,
WR is your win rate per hour,
N is the number of hours you play, and
SD is the standard deviation.
For all practical purposes, this is the equation that gives
the lower limit (at three standard deviations) for how you
would do for some period of time. That is, for practical
purposes, this equation is your worst possible result."
Okay, let's apply Mason's equation to George's problem, so
WR=392, N=383, SD=5451, producing LL=-169898.00. Actually,
come to think of it this doesn't really give you much of
an indication as to your risk of ruin during those 383
hours - it is just the worst result you would expect after
383 hours IF YOU HAD AN INFINITE BANKROLL.
But Mason Malmuth continues:
"Going one step further, if we take the above equation and
take the first derivative with respect to N, set it to zero,
and solve for N, we find the number of hours where our bankroll
can by minimized
0 = WR - (3)(1/2)(SD)/(sqrt(N))
2
/ (3)(SD) \
N = | -------- |
\ (2)(WR) /
Now by substituting this value back into the original lower
limit equation, we produce practical bankroll requirements.
In George C's example WR=.783, and SD=10.9. Thus N=436,
2
/ (3)(10.9) \
436 = | --------- |
\ (2)(.783) /
and the required bankroll is 341 units.
-341 = (.783)(436)-(3)(10.9)sqrt(436)"
First, note that 341 units is 341 of George C's $500 units,
which is really 1705 $100 units, or $170,500. Using my version
of the ruin formula, I calculate a 0.7% chance of ruin before
doubling 1705 $100 units, under George C's conditions.
Mason's technique seems to work okay, but I'd trust the ruin
formula more. Part II of this article gave a means of
determining bankroll size for a given risk of ruin.
==========================================================================
End quote.
--
Abdul Jalib
William Chen
This explaination is being posted simutaneously to the 2+2
forum and the Bankroll thread on rgp.
I didn't finish my explaination of my derivation. My claim was
that given a bankroll B, win rate w>0, and standard deviation
s, the risk of bankrupcy is approximately:
R = exp(-2Bw/s^2).
If B = s^2/w, then R = 0.13 for our formula instead of the
5% for the formula in _Gambling Theory_
Explaination. We take the gambler's ruin problemm where
the bets are the same size, which is solvable in closed
form. Suppose the bet size is $1, and your probability of
winning (there are no pushes) is p = (1+w)/2, and w>0. Let
x(B) be the chance of going bankrupt with an integer bankroll
of B. Note that we have:
x(B) = x(1)^B (Losing B dollars is the same as going
bankrupt with $1 B times).
Also, x(1) = (1-p) + p*x(2) = px(1)^2 +(1-p).
Consider the first play. Either he loses and is bankrupt
or he wins and needs to lose $2.
Solving for x(1) we get x(1) = [1, (1-p)/p]. Note if p<=1/2,
the actual root is 1 and for p>1/2 the solution
is x(1) = (1-p)/p. Thus x(B) = [(1-p)/p]^B.
Now, suppose we take the case where w<<1, B>>1, so it
would take lots of trials to go bankrupt. The binomial
disribution approaches the Gaussian by the law of large
numbers. Then:
x(B) = [(1-w)/(1+w)]^B ~ exp (-2wB). Now let's suppose that
in one sufficiently large time unit, N trials are done.
The win rate for the trial period would be W = Nw, and the
standard deviation would be S = sqrt(4Np(1-p))or approx
sqrt(N) since p is close to 1/2.
Then X(B,W,S) = x(B) ~ exp(-2WB/S^S).
The result we have is that if the distribution of results
for a 1-hr session is normal then the formula is exact since
we can make w arbitarily small and B arb. large...
Since this is the assumption made in _Gambling Theory_ anyway
for poker results, I believe this ruin formula is more
accurate.
Bill
Bill
Kim Green
Tom Weideman <zugz...@dcn.davis.ca.us> writes:
>probability p of winning 1 unit, and a probability (1-p) of losing 1
>unit. You continue playing until either you are busted, or your
>opponent is. Without going into the details, the probability that you
>will go bust first ("R" stands for probability of being ruined) is given
>by the equation:
>R = [A^B - A^b]/[A^B - 1],
>where: A = 1/p - 1
> B = opponent's bankroll
> b = your bankroll
So if B=b you have absolutely no risk of going broke?
Now I see why money management is so important...
Tom Weideman
Kim Green wrote:
>
> Tom Weideman <zugz...@dcn.davis.ca.us> writes:
> >probability p of winning 1 unit, and a probability (1-p) of losing 1
> >unit. You continue playing until either you are busted, or your
> >opponent is. Without going into the details, the probability that you
> >will go bust first ("R" stands for probability of being ruined) is given
> >by the equation:
> >R = [A^B - A^b]/[A^B - 1],
> >where: A = 1/p - 1
> > B = opponent's bankroll
> > b = your bankroll
>
> Now I see why money management is so important...
Heh. Sorry. The equation should read:
R = [A^(B+b) - A^b]/[A^(B+b) - 1]
I knew where I was going (B effectively infinite), so I thought of B as
being the combined bankroll, B+b.
I'm glad someone bothered to honor the request I made later in the post:
> Please feel free to find all the algebra errors, and
> report them back to the group with a snicker.
Thank you, Kim.
Tom Weideman