Re: Math Vs. Skill

On Fri, 11 Nov 2005 13:04:44 -0500, "ACDOC" <ACDOC@xxxxxxxxxx> wrote:

>Batter up!
>
>"OK, I'm game. Please show your derivation of Bankroll Volatility (BV)
>for 100 decisions on a $1 bet on 12, where you either lose $1 or win
>$30 (assume a 30-to-1, not 30-for-1 casino).
>Step up to the blackboard, ACDOC!
>I'm ready to be instructed.
>Cheers,
>
>Alan Shank"
>
>First, this method is a more accurate for low EVs and more even payouts.

Why would that be? The actual standard deviation of a bet is true for
all bets, no matter the ev or the payout. For Christ's sake, ACDOC,
that's what you're trying to measure!!! The variance, volatility,
standard deviation, i.e. the magnitude of the swings you can expect.

>Also, it's more accurate if we arrive at the number of decisions based on
>an estimated amount of time (hours) at the table. But, let's give it a
>spin (100decisions =100rolls).

You can't really use time, because the number of decisions is what
matters. Obviously, you just have to make an assumption as to the
number of rolls per hour, then divide by the average number of
rolls/decision, which in this case is 1.
>
>First, for 1 decision! Yes, you will have to match it Shank (sucker).
>
>Win Range=
>(square root of 1 X 30) + (1 X 1 X -0.1389)=
>$30-$.1389= + $29.86 .
>
>Lose Range=
>(square root of 1 X -1) + (1 X 1 X -0.1389)=
>-$1-$0.1389= -$1.14 .
>
>So the volatilty range for 1 decision (1 stddev) is +$29.86 to -$1.14 .
>
>That resembles reality alot better than your method does.

Well, no, because I've never seen a casino pay $29.86 on a 12 winner,
or demand an extra 14 cents on a loser. The volatility range for a
single decision is $31, since you will always either win $30 or lose
$1. Your method is totally artificial, as if they just subtracted the
vig from the winnings. My method (which isn't MINE at all, of course,
just basic probability) takes into account the actually probabilities
of winning the bet, which yours doesn't.

So: one $1 bet on the 12:

win $30, probability .0278
lose $1, probability .9722
range $31, -$1 to +$30
SD:

Well what is the standard deviation of a bet? It's the square root of
the variance. What is the variance? It's the sum of the WEIGHTED
squared differences from the mean expectation of each possible result.
Weighted by what? By the probability of each result. This is where
DiMaurCOC's method is totally flawed, as it doesn't even address the
relative probabilities of the outcomes.

First, we have to figure out the mean expectation. That's easy.

35 * -1 = -35
1 * +30 = +30
---
- 5 / 36 = -.1389

This is why ACDOC gets the figure he (artificially) adds to the
outcomes. This figure means nothing for one bet, or a very few, but it
is "woven into the fabric of the Universe," so to speak. It is the
RESULT of the two factors: 1) probability of winning or losing the bet
AND 2) the payoff for winning the bet.

So, the variance:

30 - -(.1389) = 30.1389 squared = 908.3533 * 1 = 908.3533
-1 - -(.1389) = -.8611 squared = .7415 * 35 = 25.9523
--------
934.3056
to "unweight", we divide by 36 = 25.9529

So, the variance of a $1 bet on 12 is $25.9529. Why do we square the
differences? Because we're interested in the amount of difference, in
either direction. Squaring results in all positive numbers. The
weighting accounts for the fact that the -1 outcome is much more
likely to occur than the +30, which DiMaurCOC's method completely
ignores.

Now, we squared all the differences; to get the number back to the
magnitude we started with, we take the square root. This is the
standard deviation:

25.9529^.5 = $5.0944

If you look in WinCraps, in Configure/Advantage, you will see a figure
of 509.44 for the Standard Deviation of the 12 bet. That is 509.44% of
the bet amount, which for $1 is $5.0944. I know that ACDOC does not
accept Steen as an authority, but I just want to demonstrate that
Steen and I come up with the same figure.

Now, the standard deviation of one bet is not particularly useful in
analyzing the outcome of one bet, because there are only two possible
outcomes. So, we just do what I did, above and state the outcomes and
their probabilities.
>
>Match that first Shank!

Done.
>
>Now for 100 decisions:(square root of 100=10)
>
>Win Range=
>(10X30) + (100 X 1 X -0.1389)= $300 - $13.89 = $286.11 .
>
>Lose Range=
>(10X-1) + (100 X 1 X -0.1389)= -$10 + -$13.89=
>-23.89 .
>
>So the volatility range (1 stddev) for 100 decisions, on the 12, at 31 for
>1, is +$286.11 to -$23.89 .
>
>In the short term, we see why some crapsters find the prop bets
>attractive. You will be somewhere within that range for 1 standard
>deviation. Have I seen prop betters hit it big at the tables? Of course I
>have. Don't assume that it is a symetrical distribution, it isn't.

So, are you saying that you're equally likely to win $286 as to lose
$24 in 100 $1 bets on 12? Would that that were true! To win $286,
you'd have to win 13 of 100 bets ($303 net), the probability of which
is less than .000004. To win $272, you'd have to win 12, the
probability of which is less than .00002. Out to four decimal places,
the probability of winning 11 or fewer out of 100 is 1.0000.

The standard deviation is defined in terms of the mean; it measures
dispersal around the mean. Your applying this idea to the range has no
basis at all.

Let's examine the range. How does it increase with the number of bets?
Well, the range for one bet is $31. If you make two bets, you can lose
a maximum of $2 and win a maximum of $60, so the range is $62. Is that
more than $31 by the square root of two? No, it's more by 2. For 3,
the range is $93, etc. As it is easy to see, the range increases
directly with the number of bets, not with the square root of the
number of bets.

It seems clear that the DMaurCOC method simply makes no sense. He
starts with a range, then treats it as though it were a standard
deviation. Nowhere does the probability of winning the bet appear.

I ran a simulation of 65,000 sessions of 100 $1 12 bets each. The mean
outcome was -$14, with a standard deviation of $51. Only one session
made at least $286, with $302; the next best was $240.

sessions:

>= +100 1464, or 2.3%
>= + 50 9658, or 14.9%
> 0 19,932, or 30.7%
<= -50 14,647,or 22.5%
<= -100 3821, or .5.9%

BTW, the theoretical probability of losing all 100 bets is 35/36^100,
or .0598.

Between +1 SD and -1 SD there were 62.6% of the values.

The theoretical SD for 100 $1 12 bets is simply 5.0944 * 10 = $50.94,
so the simulation SD was very close to the theoretical.

For very short sessions, you just have to grind out the binomial.
For two $1 12 bets:

p (+60) = .0278^2 = .00077
p (+29) = .0278.* .9722 * 2 = .05415
p (-2) = .9722^2 = .94517

This becomes tedious fast. You can imagine, though, that as you get
more and more decisions, you will start getting a curve. So, when can
you switch to doing it by "normal approximation to the binomial
distribution?" The general rule is that when np > 5 AND nq > 5. This
just means that the number of trials times the probability of winning
AND the number of trials times the probability of losing both have to
exceed 5. So:

100 * .9722 = 97.22
100 * .0278 = 2.78

So, with highly skewed bets like the 12, you should simulate until you
get to around 200 bets. 200 * .0278 = 5.56.

It's interesting to see why the DiMaurCOC method seems to work OK for
the pass line. Since the standard deviation of a pass line bet is
almost exactly the bet amount (WinCraps gives 99.99%), then you can
just increase the +- outcomes by the square root of the number of
bets. The closer the SD of the bet is to 1.00 times the bet, the less
of an impact their reasoning error has. Look at ADCOC's figures for
the place 6:

[QUOTE]Say you wanted to Place the 6 for 100 decisions. How do we find
the
Bankroll Volatiltiy win/loss range (for 1 standard deviation or 68% of
the
sessions)?
Win= Square root of 100 X 7 - (100 X .01515)
Loss=Square root of 100 X 6 + (100 X .01515)
That's a volatility range of +68.485 to - 61.515 ."[END QUOTE]

The standard deviation of a place 6 bet for $6 is actually $6.47, a
result of the win probability of .4545 and the 7-to-6 payoff. The ev
for 100 decisions is -$9.09, so the 1 SD range for 100 decisions is
-$73.79 to +$55.64. The reason his figures are not too far off is that
the SD is only 1.078 times the bet amount and the win probability is
not too far from .5.

Even here, though, you can see that his figures do not make sense,
because the midpoint is above zero, which implies a player advantage.

When he got into comparing the odds on the front vs. the back line,
the method broke down entirely, showing a large player advantage for
the front-line and a large house advantage for the backline odds. Here
again, you can see that the method simply ignores the probability of
winning/losing the bet.

It's really hilarious! MoneyLA and ACDOC are like Jack Spratt and his
wife: MoneyLA thinks the probability of winning is the only factor,
and ACDOC thinks the payoff is the only factor. Between the two of
them, they "lick the platter clean." >:-)
Cheers,
Alan Shank
.

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