Re: You Can Buy Your Way to a Lower Edge, but Beware the Cost

On Sun, 06 Nov 2005 20:21:43 GMT, "Mason"
<mrzer0_remove_@xxxxxxxxxxxxx> wrote:

>ACDOC uses the term "volatility" to describe the difference between wining every
>bet and losing every bet?
>
>It is hard to argue with the assertion that when you risk more and lose, you
>lose more. It seems intrinsic in the idea of a quantified risk at the crap
>table.
>
>I suppose that this might be some sort of useful insight. I just can't imagine
>to whom.

OK, Mason, you forced me to go back and dig through the myriad posts
about this. Here is my analysis of the "DiMauCOC" theory of Bankroll
Volatility:

"Let's examine these two very different concepts.

Range

If we bet one unit on the passline and take two units odds, then, if
the
point is 4 or 10, it is possible to win 5 units (1 on the flat, 4 on
the
odds, since 4/10 pays 2:1), but only possible to lose 3 units.

If we bet one unit on DP and lay double odds (so we can win two
units),
if the point is 4 or 10, we would have to lay 4 units, so we could
lose
5 units, but only win 3.

So, the ranges of outcomes are:

pass -3 to +5
DP -5 to +3

The standard deviations, OTOH, which are a measure of volatility
taking
into account all the factors relevant to the bet, as I have shown in
painstaking detail, are:

pass 2.857 units
DP 2.894 units

Although the ranges are quite different, the standard deviations are
almost the same. The reason for this is that the DP odds bets have a
much higher probability of being won. This fact seems to escape Mr. Di
Mauro and his "acolyte," ADCOC.

The ev's are simply equal to the HAs -.01414 and -.01403, since we're
talking about units here.

Let's talk about a session of about an hour, 120 rolls, about 36
resolutions of a line bet.

For the expected value, we just multiply the unit ev's by 36, so they
are:

pass: -.509 units
DP: -.505 units

For the standard deviations, we just multiply the SDs per bet by the
square root of the number of bets, or 6, so they are:

pass: 17.14 units
DP: 17.36 units

To find the figures that are likely to enclose roughly 68% of the
outcomes, we just take the ev and add/subtract one standard deviation,
getting:

pass: +16.63 -17.65
DP: +16.86 -17.87

So, a pass and a DP player, both taking/laying double odds, are both
going to fall between those very similar figures about 68% of the
time.
About 95% of the time, they will fall between +- 2 SD, or:

pass: +33.77 -34.79
DP: +34.22 -35.23

What about the ranges? Well, it's theoretically possible for a bettor
to
either win or lose all of 60 bets, but even then the maximum wouldn't
be
achieved unless every bet went to a point of 4. The probability of
winning 36 pass bets is 8.71 * 10^-12, the odds against which are over
100 billion to 1, and that includes the 33% chance to win on the
comeout, where the ranges are exactly the same for pass and DP. So, is
information about the ranges really of any use or even interest here?
Of
course not.

Let's examine Di Mauro's logic with this in mind:
I quote from ADCOC's posting of DiMauro's "volatility" calculations:

"He determined the ?range (his words)of Odds?, to be:
Square root of (327 x (24/36)) = (+-) 14.765 Bets Behind the Don?t
Pass
Line,
Square root of (324 x (24/36)) = (+-) 14.697 Bets Behind the Pass
Line."

24/36 is the ratio of decisions based on a point being rolled to total
decisions. OK. Then he takes the square root of that number of bets (a
figure whose derivation is also completely confused, but...). Since
DiMauro and ADCOC don't know that a standard deviation is not always
one, they just multiply times the square root of the number of bets,
and
think they are getting a standard deviation, which they, of course,
are
not. (The average SD on odds taken is about 1.22/unit, that on laid
odds
about .83/unit.) So, these figures of around 15 dont' represent
anything
that is relevant to this problem.

He continues:

"With Double Odds, he determined that the maximum amount lost behind
the
Pass Line = 2 x 14.697 = 29.394 units Pass Odds Lost.

With Double Odds, he determined that the maximum amount won behind the
Don?t
Pass Line = 2 x 14.765 = 29.530 units Don?t Pass Odds Won.

With Double Odds, he determined that the maximum amount lost behind
the
Don?t Pass Line = ((4 x (6/24)) + (3 x (8/24)) + (2.4 x (10/24))) x
14.765
= 44.295 units Don?t Pass Odds Lost.

With Double Odds, he determined that the maximum amount won behind the
Pass Line = ((4 x (6/24)) + (3 x (8/24)) + (2.4 x (10/24))) x 14.697 =
44.091 units Pass Odds Won."

Now, he is introducing the concept of range into an already hopelessly
confused calculation of volatility. He multiplies the maximum units
that
can be lost on the pass odds times a figure that he thinks might be
the
standard deviation for 216 pass odds bets (it's not). What does that
yield? A figure that has no utility whatsoever. Ignoring, for the
moment, that this "standard deviation" is totally incorrect, why would
you multiply it times the range of a single bet? What sort of logic is
at work here? I think we should call this figure the "coefficient of
confusion." >:-)

The rest of the calculations just involve adding these meaningless
figures to the actual +- 1 SD figures for the flat bets, which are
correctly only because the SD for the flat part of the bets is very
close to 1.0. The result is neither "fish nor fowl." It's not a range,
it's not a standard deviation, it's not a variance. It's utterly
without
meaning or utility in examining craps.

This man is profoundly confused. He has a few mathematical tools, but
he
doesn't know how to use them, so he just applies them more-or-less
randomly, producing utter garbage.

It's over."
Cheers,
Alan Shank
.

Relevant Pages

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