by Frank Scoblete

*Craps Vig* from the
word “vigorish,” is defined as the percentage edge the house takes for
every
dollar gambled. The vig is often misleading when it comes to how much a
casino
actually makes and how much a player actually loses in a random craps
game and
it is seriously misleading when it comes to an advantage player at
craps who
changes the odds by his controlled throwing.

For example, if Place betting is your style and if you Place the 6 and
8 in
multiples of $6, the vig is considered 1.52 percent. You should get
paid $7.20
for a winning $6 bet on the 6 or 8, but you only get paid $7 when you
win.
You'll have five winners ($5 X $7 = $35) and six losers when the 7
rears its
ugly head (6 X $6 = $36).

In those 11 decisions, you'll be down a dollar because the
casino kept that dollar as its share. You've wagered $66 dollars on
those 11
decisions, lost one dollar (1 divided by 66 is 0.01515). There's the
craps vig
for the normal, random placing of the 6 and 8.

So you bring $100 to the casino and you figure you're going to bet $6
on the 6
and 8, which is $12 total, thinking that you only stand to lose about
1.52
percent of your money, a dollar fifty to make it rounded. So you think
you're
going to go home with about $98.50 in the long run using that same
$100. But
you won't.

In the long run the 1.52 percent house edge will wipe your
$100 away and safely tuck it into the casino coffers. Why? Because in
the long
run, or even over one or a few sessions, you will bet far, far more
than that
$100.

Your money will be going back and forth, back and forth, and
with each back and forth, the house edge is subtly chop, chop, chopping
away at
your cash.

The placement of the 6 and 8 will see it acted upon approximately 44
times per
hour, if we assume 100 rolls of the dice in that hour. So assuming in
100 rolls
the numbers we’re concerned with, the 6, 8 or 7, will appear (on
average) about
44 times.

The 6 and 8 will appear approximately 28 times (winning you
$196, while the 7 will pop up 17 times (losing you $204). I'm rounding
up the
fractions here so that's why we have 44 percent but 45 appearances.
Darn math!

In 100 rolls, you can expect to be down $8. One hundred rolls of the
dice is
about one hour’s worth of play, sometimes less in a fast game. Now that
$100
has been whittled away to $92. In the second hour, you'll lose another
$8 and
be down to $84 and on down it goes over time.

Of course in the real world of casino craps, the losing of your $100
will not
be smooth. You might win a whole bunch of rolls right from the get-go
and be
substantially ahead. Conversely, you might lose a whole bunch of rolls
and be
so down, emotionally and economically, that the only thing you want to
do is
slink out of the casino and return to your room to suck your thumb.

And what about the casino itself? Chances are with all those craps
tables
seeing sustained action, the casino achieves the long run in short
order and
that means whatever swings, up or down, any given table is experiencing
at any
given moment will all smooth out according to the math as the other
tables
contribute to the casino's bottom line.

Anyway, in a strictly random contest, the wise player just goes with
the best
mathematical bets if he wants to see his bankroll last as long as
possible.
These would be Pass with odds, Come with odds, the placing of the 6 or
8, and
the buying of the other numbers if the craps vig is taken out only on a
win.

In a random craps game, a player has no chance to be a
long-term winner if he actually plays a lot, the math will grind him to
dust
and that is the reality of the situation.

Here math and reality are joined like Siamese twins that
can't be separated. So my advice when paying against any random shooter
or in
any random game is to follow the math. The casinos do and they do quite
nicely,
economically speaking, thank you.

Now to
the meat of the matter for controlled craps shooters.

What relationship does the random craps vig have with the
real craps vig when someone is changing the odds of the game by
reducing the
appearance of the 7? Do all the numbers fill in equally if someone's
SRR [seven
to rolls ratio] is 1:8?

Or do some numbers fill in more than others based on the set
and the skill of that particular setter? Keep in mind that in a
strictly random
game of craps the SRR is 1:6.
It's the latter, unquestionably, as the numbers *do not*
fill in equally.

A skilled controlled shooter, staying reasonably on axis (meaning his
dice stay
pretty much as he originally set them without flopping to this or that
side),
will be avoiding not just the 7 but other numbers as well. It would
take much
too much time to go into which and why those numbers would be for every
set,
but suffice it to say that there's a whole new set of mathematical
principles when
it comes to controlled shooting.

To conclude this craps vig lesson,
wise and skillful dice controllers will develop an understanding of
which
numbers they tend to hit more than other numbers as they are reducing
the 7.

These "signature numbers" will be like having the
casinos name on one of those big, fat oversized checks, only this will
be a
cashable check, not a cardboard one.

It will say: Pay to the Order of This Controlled Shooter.

Such "signatures" will be money in the bank
despite the craps vig of the game.

Craps
Vig is followed by Craps Players and the roll of the dice

OR

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to *Learn**
to Play Craps *Program

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Craps Bets: Don’t Pass with odds and Don’t Come with odds

Always put the least allowed on the don’t pass or the don’t come and the most in odds behind it. The house will have a 1.403 percent edge on the don’t pass and don’t come bets but no edge on the odds bets.