January 2012

Wagering Tricks and Combinations - What Does and Doesn't Work To Gain The Edge
by John Grochowski

When a game offers multiple wagering options, casino players start looking for that winning combination, the right mix of wagers that will pad their bankrolls.

Check out any craps table. How many players do you see sticking just to the pass line, or to a place bet on 6? Pretty much everyone makes multiple wagers.

And what about roulette, the other standard game where it takes more than the fingers on both hands to count the ways to wager? You might see a beginner with a bet just on red or black, but mostly you’ll see other player spreading chips all over the layout, hoping that if one bet loses --- or two or three or six --- then another will cover the losses and then some.

Is there a magic combination, one where the weakness in every wager is covered by the strength of another, where the whole is stronger than the parts, and the player has an edge in the long run? No, there’s not. In any combination, the house edge is a weighted average of the house edges of all the individual wagers in the combo. The house edge on a combination can never be lower that the lowest house edge of one of the component bets.

That doesn’t stop player from trying to catch lightning in a bottle. Let’s taking a look at how some of the combinations floated via readers’ email really work.

ROULETTE

One of the first things roulette players are told is that the house edge comes from the zeroes. Bets are paid at odds that yield no edge to the house if not for the presence of 0 and 00 on an American wheel. So, the combination-seekers reason, if the house edge comes from the zeroes, then why no bet on these house numbers? If you bet on black plus a split bet on 0 and 00, wouldn’t you be co-opting the house edge?

In reality, saying the house edge comes from the zeroes is an imperfect understanding of the game. The house edge come from paying bets at less than true odds, the same way the house creates an edge in any other casino game. Bets are paid at odds that would make them even propositions if there were 36 numbers on the wheel. If the only numbers were 1 through 36 then a 35-1 payoff on a single-number wager would cancel out the 35 losers on all the other numbers. If there were 18 red numbers, 18 black numbers and nothing else, then every red loser would be offset by a black winner, and vice versa.

But that’s not the way it works. There are 38 numbers, not 36, and a 35-1 payoff on a single number doesn’t offset all 37 of the other numbers. Two of the numbers are green, so 18 black winners aren’t enough to offset the total of 20 losers when you combine red and green.

The zeroes themselves are subject to the same odds and house edge as any other number. Win a bet on 0, and you’re paid at 35-1 odds, just as on any other number. And the true odds of 37-1 when all other numbers are considered give the house the same 5.26 percent edge as if you’d bet on 1, 17, 35 or any other number.

With all that as background, what about the proposed combination, one of the simplest among those readers have suggested? A reader’s email suggested betting $10 on black and $5 split between 0 and 00 at a table with $5 minimum bets. That way, if black won, you’d win $10 on black, lose $5 on the zeroes, and take a net profit of $5. If one of the zeroes turned up, you’d win $85 on the 17-1 payoff on the split bet, and lose $10 on black, for a net profit of $75. You’d have 20 winning numbers, compared to only 18 losers on the red numbers. You’d win on the 18 blacks, and win big on the zeroes. What could go wrong?

What goes wrong is that on the 18 red numbers you lose both wagers, meaning $15 in casino coffers. Per 38 spins of the wheel, you’d have 18 winners at $5 each for a total of $90, plus two $75 winners, for another $150. That’s $240 in winnings on your profitable spins. But the 18 losers at $15 a pop cost you a total of $270. Despite more winning spins than losers, and despite a couple of big winners, you still lose $30 more than you win per 38 spins.

To put it on a percentage basis, at $15 a spin you risk $570. The house keeps $30 of that money. Divide $30 in losses by $570 in wagers, then multiply by 100 to convert to percent, and you get a 5.26 percent house edge on this combination. Not so coincidentally, the house edge on betting black by itself, or on just betting the split bet, or on almost any other wager at double-zero roulette is 5.26 percent. The search for combination magic has just spotted the house the same edge it gets on the component bets.

One more, also suggested by a reader” You bet $10 on black, and then you bet $5 on of the third column. The third column has eight red numbers, so now you have 24 numbers covered. The column pays 2-1, so if any of your eight red number hits, your $10 payoff cancels out your $10 loss on black. If any of the four blacks in your column hits you win $10 on the column AND $10 on black. And of any of the other 14 black numbers hit, you lose $5 on the column, but win $10 on black, so you have a profit of $5.

Bottom line: You show a profit on 18 numbers, including four where the profit is $15. And since you have 26 numbers covered, there are only 12 numbers that can beat you.

Problem is that on every one of those 12 losses, you lose both bets. Every loss costs you $15, and in the end the house has its 5.26 percent.

That’s the way it works with roulette combinations. You want to avoid the five-number bet on 0, 00, 1, 2 and 3, where the house edge is 7.89 percent. Any combination that includes the five-number wager will have an overall house edge of higher than 5.26 percent. But any other combination you can think of, with as many or as few component bets as you like, will lead to the house keeping an average of 5.26 percent of your money. There is no magic combo that can reduce that.

CRAPS:

Just as beginning roulette players think of using the zeroes to try to cancel out the house edge, craps combination-builders often start with any 7. They figure that if their pass line bet or place bet loses when a 7 turns up, then maybe betting the 7 can turn the tables on the casino. As it happens, any 7 is a terrible bet, one of the worst in the house on any game. It’s a one-roll proposition that pays 4-1. True odds are 5-1, and that gives the house a 16.67 percent edge.

Still, several readers have suggested that in combination with a bet that loses on 7, it shores up that weakness and helps you win. That’s not really the case.

One reader detailed an easy system. He made $6 place bets on 6 and 8, and he'd hedge with a $3 place bet on any 7. With a 4-1 payoff, he figured he'd win back the $12 he lost on the place bets when a 7 turned up. When either place bet won, he'd win $7 at 7-6 odds, and just lose $3 on his any 7.

Problem is, the any 7 bet also lost when 2 turned up, or 5, or 9, or any other number, while the place bets just stayed in action. He had to bet a new $3 to keep his hedge 20 times per 36 rolls.

The house edge on the combination is 6.67 percent, a weighted average that takes into consideration that he had more money on the place bets at a 1.52 percent house edge than on any 7 at 16.67 percent. Players are far better off to stick with the better bets and not burden themselves with the extra house edge that any 7 brings to the combination .

Let’s look at a more complex combination, one that proponents once claimed had an overall house edge that was lower than the house edge on any of the component wagers. That’s not possible.

The combination is meant to be used at casinos that pay 3-1 on the field when a 12 is rolled. Make a $5 place bet on 5, $6 place bets on 6 and 8 and a $5 wager on the field. , The people who touted the bet claimed a house edge of 1.136 percent.

That would be a a math-defying feat. All the individual pieces of this combination have higher house edges than 1.136 percent. House edges are 4 percent on the place bet on 5, 1.52 percent on the place bets on 6 and 8 and 2.78 percent on the field when rolling a 12 brings a 3-1 payoff.

Problem is, the calculation of a 1.136 percent house edge assumes fresh money on every wager on every roll of the dice, and that’s not how we calculate house edges. We assume each bet that is played to a decision.

Take the place bet on 6. Per 36 rolls, 25 of them make no difference. The only ones that matter in settling the bet are the six ways to roll 7 and the five ways to roll 6. Let’s say we’re betting $6 a pop. For those 11 decisions, we risk $66. On the five we win, we win $35 plus keep our $30 in wagers, leaving us with $65. The house keeps $1. Divide that $1 by the $66 risked, then multiply by 100 to convert to percent, and you have the 1.52 percent house edge on placing the 6.

Now let’s do the calculation a non-standard way, one that assumes we’re betting fresh money on every roll instead of leaving the same money on the table until we win or lose. For example, we bet $6 on 6, the next roll is a 2, we take back our $6, then bet a fresh $6 on 6. Repeat as long as necessary. Sounds silly, doesn’t it? But it’s what we have to do to make house edge calculations match those of the combination proponents.

If we assume we’re betting fresh money on every roll, then we don’t put $66 at risk, and the wager total soars to $216 per 36 rolls. We still lose the same $1, but now we divided it by $216 in wagers, and get a house edge of 0.46 percent.

If we were going to calculate house edges in that way, then our starting points in the suggested combination would be edges of 0.46 percent on 6 or 8 and 1.11 percent on 5. And the house edge on the combination is higher than that on the place components, and lower than that of the field bet. The combination does not reduce the overall house edge, as claimed.

When we loo at the combination with the usual method of calculating house edges, with place bets left on the table until a decision is reached rather than replaced by fresh money on every roll, the overall house edge is 2.5 percent --- lower than the edge on the field or the place bet on 5, but higher than the place bets on 6 or 8. That’s the way it has to be. The house edge on any combination is a weighted average of the house edges on the component bets.

Let’s look at one more, one of the most well-known combinations in craps. It’s called the Dinner Bet, or Darby’s Field. Those who like it tout it as a way to hit and run, stick around just long enough to win dinner money and get out. The getting out part is the bet bet.

The Dinner Bet starts with the field, which wins even money on 3, 4, 9, 10 or 11, wins 2-1 if the roll is 2, and either 2-1 or 3-1, depending on house rules, if the roll is 12. You also make place bets on 5, 6 and 8. That leaves you with every number covered except the 7. How can you lose? One hit, and you’re off to the buffet, with it at least partly paid for from your winnings.

At a $5 minimum table, you’d wager $5 on the field, $5 to place the 5, and $6 each to place 6 and 8. Those $6 bets are important, since place bets on 6 and 8 pay 7-6 odds. If you bet only $5, you win only $5. You have to bet in multiples of $6 to get the odds.

So with all that, you start with $22 in wagers. There are 36 possible combinations of two dice, and only six of them --- the six ways to make 7 --- wipe out all your wagers.

Under the most favorable conditions --- field bets that pay 3-1 if the roll is 12 --- you get the following results per 36 rolls when each number turns up once. (We’ll assume you’re picking up the place bets and leaving on field winners, since the object here is to win once and go): On the field, win $10 on the one 2; $15 on the one 12; and $5 each on the two 3s, three 4s, four 9s, three 10s and two 11s, for a total of $95.On the place bets, $5 wins on the 5s are offset by $5 losses on the field, while the $7 wins on the five 6s and five 8s are partly offset by the $5 field losses, leaving you with a $2 profit for each of the 10 6s or 8s. Your total profit on the non-7s is $115.

But even though there are only six losing rolls, each one costs you $22, for a total of $132 in losses. The old staying is that there is no such thing as a free lunch, and that goes for dinner, too, as far as this combination goes.

Remember, casino games are based on math, not magic, and no amount of combination wizardry can change that. The bet combination to keep the house edge at the minimum is a little knowledge and the willingness to stick to the best bets instead of trying to force combo trickery.

— John Grochowski is the author of The Casino Answer Book, The Slot Machine Answer Book, The Video Poker Answer Book and the Craps Answer Book, available online at: www.casinoanswerman.com.