The "Vig," derived from the word ?vigorish,? is defined as the percentage edge the house extracts 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 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 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.

You 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 ?vig? is taken out only on a win.

In a random 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 we come to the meat of the matter for controlled shooters. What relationship does the random vig have with the real 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.

In studies of controlled shooting and its impact on the game, such as Dr. Catlin's massive study of the 5-Count, we assume all the other numbers fill in equally for the decreasing appearance of the 7, but that's just an easy way to figure out the math, a short cut, it isn't really all that accurate. In reality, we have something very different going on.

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 ?vig? of the game.

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