Max ER VS "other" strategies... long

[cdfsrule1]

After months of silence, I feel compelled to put in my $0.02 here.
Frankly, a lot of what I have been reading about EV, variance, best
game, etc, bothers me. Sure, I find much of the vpfree's emails
extremely valuable and interesting, but a lot of the
mathematical "opinions" expressed here are just plain misleading (,
usually because they are incomplete or overly simplified). BTW I
apologize in advance-- I don't mean to be picking on anyone in
particular. In this email, I will focus on just one, oft-misused
term, "lower risk"…

"Lower Risk"
The term "lower risk" has been thrown around a bit lately,
and most
often it has been used in a misleading way. Each of us probably has a
different personal definition of "Risk" and, without
qualification,
the comparative phrase "lower risk" has little meaning. To
me, risk
is the probability of a negative event occurring. So if I ask what
is MY risk of loosing my entire bank roll (for a particular session
and game, etc), I am looking for the probability of the loss
occurring. Mathematically, the probability is equivalent to the
integral (or sum, also known as "the area under the curve")
of the
probability distribution function (PDF) for the game I am actually
playing from minus infinity to size of my back roll (in other words
it is 1-the probability that from my bankroll to positive infinity).
This integral is known as the cumulative distribution function or
CDF. BTW, If you know the CDF (or PDF, though CDF is better defined)
of the game (you are actually playing) them you know everything
(since any 2 vp hands are truly uncorrelated). Given the CDF, you
can compute a lot of useful quantities: the EV (arithmetic mean), the
mode (the most likely outcome), the median (the 50% point), the
bankroll required for a particular likelihood of success (say 90%),
etc.

You can also use the CDF (or PDF) to compare two VP games. For
example, if someone asks me, which game has lower risk [to my bank
roll), I would compare the CDF's for the two games, and
specifically
the cumulative probability of a loss exceeding my bank roll for the
two games.

Important note: The PDF (and CDF) for a session reflects the number
of hands played and how the hands are played. That makes sense. BTW,
it is usually a good approximation to assume that the PDF (or CDF)
can be computed just from the payout table, so long as the number of
hands played is large enough and the play is good enough (correct
strategy with a relatively few number of errors). That is a good
thing, since one can easily compute this kind of PDF without Monte
Carlo simulation (using simple convolution). On the other hand,
sometimes that approximation breaks down, and a Monte Carlo
simulation becomes necessary since the PDF computed using simple
convolution method would significantly differ from the long-term
average PDF of simulated actual play (Monte Carlo)

Ok, so practically speaking, what does this all mean? First,
statements like,

Whenever a player compares a negative game to a positive game, the
positive game will _always_ be lower risk than the negative game.

are wrong. "Will _always_ be" is just too strong of a
phrase. To
understand this, we turn to the trusty ol' CDF or PDF and
consider
two fictitious vp games. Game 1 (vp1) has PDF that looks like a
(truncated) Gaussian (otherwise known the "bell curve" or
"normal
distribution") at least for a very large number of hands.
Gaussians
have the property that the mean (EV), mode (peak or most likely
outcome) and median (50% CDF point) all occur at the same value (how
nice!). [ BTW The CDF of this curve is called the error function or
ERF]. Assume for now that the vp1 game has a 0 EV. Hence the most
likely outcome for this game is 0 as is the EV.

Now consider Game 2 (vp2), a game that will have a different PDF, one
which has a long positive going tale, but whose main area is
concentrated at negative values. Such curves are sometimes referred
to as "log-normal", "chi-squared",
"heavy-tailed", or "long-tailed"
tailed distributions. An interesting property of such distributions
(for vp) is that the mode (or most likely outcome) is always below
(to the left or the negative side) of the mean (or EV). That is, the
most likely outcome for this kind of game might be a loss (negative),
even if the EV is positive! [In case you're wondering, in order
for
all the math to work out, something has to change, and usually that
something is the variance increasing and higher moments becoming non-
zero, which skew the distribution]

So if we compare the PDF's (or CDF's) for vp1 and vp2, we
find that
the probability of (a certain large) bankroll loss may very well be
lower for the EV=0 game than the positive EV game. Wow. {Aside: If we
plotted the two PDF's we would see that the bulk of the vp2 cruve
lies the the negative side of the peak area of vp1 curve. In terms
of CDF's, we would find that the CDF for vp1 crosses the CDF for
vp2. [I imagine that I may confused may people here; I guess
pictures would help a lot]}

On the other hand (since the CDF curves cross), if we allowed the
value of our of bankroll used to decrease (that is, if we ask the
question again, but now consider the risk of losing a smaller and
smaller bankroll) we might eventually find that that vp2 offers lower
risk [Aside:this may seem paradoxical, but it isn't. Recall that
we
are holding the number of hands fixed and that we are considering
relative risk. Hence, the absolute risk of losing the smaller
bankroll or the area under the PDF curve, is always larger for both
games]. We take this concept a little farther and consider the
question `which game has lower risk' as a function of
bankroll (or
better yet just `outcome' where a negative number is a loss
and a
positive number a win). Amazingly, we already have our answer: for a
particular negative outcome, if the CDF of vp2 is above (larger
probability) vp1 is the lower risk game. [Aside: Likewise, for a
particular positive outcome, if the CDF of vp2 is above (larger
probability) the CDF of vp1, then vp1 is the higher risk game, though
I would prefer to use the phrase "vp2 has a higher probability of
a
winning $X"]

When I get a free moment I will share some CDF's of vp games with the
list.

[Steve_Jacobs]

BTW I
apologize in advance-- I don't mean to be picking on anyone in
particular. In this email, I will focus on just one, oft-misused
term, "lower risk"�

I don't think you need to apologize for challenging something
that you feel is incorrect.

"Lower Risk"
The term "lower risk" has been thrown around a bit lately,
and most
often it has been used in a misleading way. Each of us probably has a
different personal definition of "Risk" and, without
qualification,
the comparative phrase "lower risk" has little meaning. To
me, risk
is the probability of a negative event occurring. So if I ask what
is MY risk of loosing my entire bank roll (for a particular session
and game, etc), I am looking for the probability of the loss
occurring.

I agree with the statement above. I would generalize it somewhat
by saying that risk is the probability of going broke before reaching
some specific goal. Any risk calculation can also be framed in terms
of computing the probability of success rather than the probability of
failure. Classical "risk of ruin" uses the goal "play indefinitely" but
the same mathematical formulation works when the goal is a specific
target bankroll. Another example would be the risk of losing your
bankroll before hitting a royal flush -- I prefer to frame this in terms
of "probability of success" and call this "best shot at royal," but it is
the same as doing a certain type of risk calculation. I think it is fair
to claim that any optimal strategy that is based on optimizing a
probability falls under the broad category of "risk."

Mathematically, the probability is equivalent to the
integral (or sum, also known as "the area under the curve")
of the
probability distribution function (PDF) for the game I am actually
playing from minus infinity to size of my back roll (in other words
it is 1-the probability that from my bankroll to positive infinity).

I'm sorry, but that is not how risk of ruin is calculated. If you take
an initial bankroll of N units and compute a PDF from a large number
of trials (say 10*N) then the PDF will have a "tail" on the left which
represents a net loss larger than the initial bankroll. In addition, some
results in the positive side of the PDF will represent cases where the
bankroll went negative but eventually returned to the positive side.
A proper RoR calculation will not include any cases where the
bankroll was negative at any time. Including such cases means
that play continued after the player has gone broke. Clearly, any
calculation which included such bankroll "paths" is flawed.

[text deleted]

Ok, so practically speaking, what does this all mean? First,
statements like,

>> Whenever a player compares a negative game to a positive game, the
>> positive game will _always_ be lower risk than the negative game.

are wrong. "Will _always_ be" is just too strong of a
phrase.

My statement is true. A negative game ALWAYS yields 100% RoR (based
on the "classical" definition) while a positive game ALWAYS yields a
RoR less than 100%. This is not merely my opinion, it is a mathematical fact.
It is simply impossible for a negative EV game to go below the 100% RoR
"barrier" and it is impossible for a positive EV to have 100% (or higher)
RoR. Of course, if the positive game has a very tiny advantage, the RoR
will be almost 100%, but it will still be below 100% by some small fraction.

To
understand this, we turn to the trusty ol' CDF or PDF and
consider
two fictitious vp games. Game 1 (vp1) has PDF that looks like a
(truncated) Gaussian (otherwise known the "bell curve" or
"normal
distribution") at least for a very large number of hands.
Gaussians
have the property that the mean (EV), mode (peak or most likely
outcome) and median (50% CDF point) all occur at the same value (how
nice!). [ BTW The CDF of this curve is called the error function or
ERF]. Assume for now that the vp1 game has a 0 EV. Hence the most
likely outcome for this game is 0 as is the EV.

If the EV is 0, then half of the curve is negative, so clearly play has been
allowed to continue after players have gone broke. That isn't allowed for
a proper RoR computation.

I don't have time right now to provide a tutorial on the correct method for
computing RoR, but by way of a hint it involves the "characteristic equation"
which is based on the exact probability distribution for a single play. I'm
sure I've described this somewhere in a previous post, but I don't have
a reference. In general, if p(N) is the probability that the game returns
N units to the player from a single unit wagered, and R is the risk of ruin,
then R is given (recursively) by:

R = p(0) + p(1)*R + p(2)*(R^2) + ... p(1000)*(R^1000) + ...

This equation is exact.

For a simple coin flip, where W = p(win) and L = p(lose), the possible
outcomes are zero or two units, and this reduces to:

R = L + W*(R^2)

or

R*(L + W) = L + W*(R^2)

or

W*(R - R^2) = L*(1 - R)

or

W*R*(1 - R) = L*(1 - R)

or

R = L / W

So a coin flip that favors the player by 10% has W=0.55 and L=0.45
to give R = 0.45/0.55 = 9/11. The probability of ruin is thus 81.8181%
and the probability of playing forever is 9.0909%.

The RoR equation for VP games is a bit more complex, but works the
same way.

···

On Monday 22 August 2005 02:12 pm, cdfsrule wrote:

[dipy9111]

I bet this email was not read by 50% of subscribers with a variance
of 40. I was a math major, but I always was a proponent of removing
the statistics from the mathematics major. My grades C, C-, C---.

Give me AA anytime. Math makes sense, but stats don't,
dipy911

···

"Lower Risk"
The term "lower risk" has been thrown around a bit lately,
and most
often it has been used in a misleading way. Each of us probably has

Important note: The PDF (and CDF) for a session reflects the number
of hands played and how the hands are played. That makes sense. BTW,

[Harry_Porter]

cdfsrule wrote:

After months of silence, I feel compelled to put in my $0.02 here.
Frankly, a lot of what I have been reading about EV, variance, best
game, etc, bothers me. Sure, I find much of the vpfree's emails
extremely valuable and interesting, but a lot of the
mathematical "opinions" expressed here are just plain misleading (,
usually because they are incomplete or overly simplified).

Good to have you aboard.

In general, technical posts here will lean toward simplification
rather than explicitness (though not nearly to a great enough degree
for some ;). This is a layman's group and excessive precision will
serve to obscure the big picture entirely. There will necessarily be
some hand waving, some short cuts and some details that are
conveniently sidestepped.

The critical thing is that a substantive argument is presented, in a
fashion that others can readily grasp without a MS or PhD, that leads
the reader to the expressed conclusion, and provides a reasonable
basis on which to carry forward discussion/dissent.

I'll let Steve, if he chooses, explain the appropriateness of his
comments that you cited in your post. Of all people here, I think
you'll find he's by far the last person to be "misleading ...
incomplete or overly simplified".

- Harry

[Jerrod_Ankenman]

<snip>

My statement is true. A negative game ALWAYS yields
100% RoR (based
on the "classical" definition) while a positive game
ALWAYS yields a
RoR less than 100%. This is not merely my opinion,
it is a mathematical fact.
It is simply impossible for a negative EV game to go
below the 100% RoR
"barrier" and it is impossible for a positive EV to
have 100% (or higher)
RoR. Of course, if the positive game has a very
tiny advantage, the RoR
will be almost 100%, but it will still be below 100%
by some small fraction.

Small and unimportant nit -- positive games with
infinite variance can have RoR 100%.
(probable example: optimal backgammon where opp spots
you epsilon points per game)

I don't have time right now to provide a tutorial on
the correct method for
computing RoR, but by way of a hint it involves the
"characteristic equation"
which is based on the exact probability distribution
for a single play.

Here, I'll do it. (quick version - by the way this
is in our upcoming book _The Mathematics of Poker_)

So we have this function R(b) which is the risk of
ruin for a bankroll b.

Eq 1: R(a+b) = R(a)R(b)

This is because losing two bankrolls are independent
events. So after you lose the first one, which happens
with probability R(a), your risk of losing the second
one is just R(b).

ln R(a+b) = ln (R(a)R(b))
ln R(a+b) = ln R(a) + ln R(b)

Now make up a function f(x) such that f(x) = ln R(x).

Eq 2: f(a+b) = f(a) + f(b)

This relation shows that f is linear, ie:
f(1+1) = f(1) + f(1)
f(2) = 2f(1)
f(n) = nf(1)
and so on.

So there's this value q (i usually use alpha, but it's
text now), such that the following is true:

f(x) = -qx

ln R(x) = -qx
R(x) = e^(-qx)

So now we have a general formula for risk of ruin of
any game: it's e raised to (the risk of ruin constant
for that game times the bankroll).

Unfortunately, it's not obvious what the risk of ruin
constant is. We can find it, however, using another
property of risk of ruin.

So we can use the <x> notation to mean "expected value
of x".

Then

R(B) = <R(B+y)> where y is a hand outcome randomly
pulled out of the distribution of hand outcomes for
your game.

So what this says is that the risk of ruin NOW has to
equal the expected value of the risk of ruin after one
play.

Eq 3: R(B) = <R(B+y)>
e^(-qB) = <(e^(-qB))(e^(-qy))>

e^(-qB) is a constant, so we can pull it out of the
expected value expression.

e^(-qB) e^(-qB)<(e^(-qy))>
Eq 4: 1 = <(e^(-qy))>

So now we have an equation with only one unknown
(supposing that we know the distribution of outcomes).
We can solve this equation directly to find q, and
then plug q into the equation R(b) = e^(-qb) to find
risk of ruin for any bankroll. The expected value term
is expanded by simply creating terms for each outcome
which are multiplied by the proability of that outcome
occurring.

This, I'm sure is the "characteristic equation" that
Steve referred to above.

Here's a fast example:

Die Roll:
1-2: +1
3-5: -1
6: +2

From Eq4 we have:

1 = <e^(-qy)>
1 = (1/3)(e^-q) + (1/2)(e^q) + 1/6(e^-2q)

This yields a q of about .234.

So R(b) = e^(-.234b)
R(1) = 79.1%

Also, it turns out that for positive games it can be
shown relatively easily that Eq4 has exactly two roots
(one at q = 0, and one at q > 0).

Jerrod

···

--- Steve Jacobs <jacobs@xmission.com> wrote:

__________________________________________________
Do You Yahoo!?
Tired of spam? Yahoo! Mail has the best spam protection around
http://mail.yahoo.com

[nightoftheiguana2000]

One way to rank games (or mutual funds) is with the Sharpe ratio:
(er-1)/sqrt(variance)
The Sharpe ratio squared is bankroll growth.

[Bill_Cook]

Steve,

Really appreciate your input here, but just want to get picky about
one point. Probability theory tends to break down when it encounters
the infinite. Statements such as the ones you make above only apply
in the infinite. Once we limit ourselves to plausible real-world
scenarios, we get different results.

Take the following two games - each costing $1.

Game A -

999,999/1,000,000 returns $0
1/1,000,000 returns $1,000,001

Game B -

9/10 returns $1.11
1/10 returns $0

Any reasonable definition of lower-risk-game would pick game B as
having lower risk. Real people playing these games for plausible
amounts of time will have lower RoRs for game B than game A. It is
only when we allow the infinite that it switches.