Combining bets can't reduce the house edge

It's one of the mathematical truisms of craps that combination bets can't reduce the overall house edge below that of the best bet in the combination.

If you bet the field, a one-roll bet with a house edge of 2.78% provided the house pays 3-1 on 12, and bet an equal amount on any 7, with a house edge of 16.67%, then the house edge on the combination is going to fall midway between 2.78 and 16.67%. If you bet the pass line, with a house edge of 1.41%, and follow up with place bets on 6 and 8, with house edges of 1.52%, the overall edge is going to fall between 1.41 and 1.52.

With a three-bet combination such as placing the 6 along with the field plus any 7, the overall edge on the combination will be something higher than the 1.52% on placing the 6 and lower than the 16.67% on any 7. It'll be a weighted average of all the bets.

That doesn't stop players from trying to come up with that magic combination that will somehow yield a lower house edge, something where the interaction of the wagers turns the math toward the players.

I received an e-mail from a player who tried to convince me that a simple combination, a place bet on 6 and a place bet on 8 reduces the house edge from the 1.52 on each bet to 1.04% overall.

It's an argument others have made. Here's the logic:

In evaluating craps bets, we throw out rolls that push as irrelevant. Rolls such as 2, 5 or 11 do not yield a decision on 6 or 8, so we disregard them in calculating a house edge. That leaves only 16 rolls that matter: the 6 ways to make 7, the 5 ways to make 6 and the 5 ways to make 8.

If you wager $6 each on 6 and 8, you risk $192 on the 16 rolls. We get $7 in winnings on each of the five 6s and five 8s, for a total of $70. We lose $12 on each of the six 7s for a total of $72. That's a net $2 loss. Divide that by the $192 risked, then multiply by 100 to convert to percent, and you get a house edge of only 1.04%.

A mathematical miracle? No, just a flawed assumption.

That calculation treats pushes on 6 and 8 as money at risk, unlike any other pushes. In reality, a shooter's roll of 8 is no different than a 2, 5 or 11 when it comes to deciding a place bet on 6. When we calculate the house edge on the combination, we have to disregard money bet on 6 when the roll is an 8, just as we disregard rolls of 4, 9, 10 or any other push.

Now the risk becomes $72 for the 12 7s, $30 for the five 6s and $30 for the five 8s, for a total of $132.

You still win $70 on the 10 wins, and lose $72 on the 12 sessions. The loss is still $2 overall, but it's compared to a total risk of $132, not $192. Divide $2 by $132, and you get 0.0151515. Multiply by 100 to convert to percent, round up and you get 1.52%.

The house edge when you combine two 1.52% bets is 1.52%, just as it has to be. The house edge on any craps combination is a weighted average of all its components.

ONE-ROLL EDGE: The e-mail asking about the above system opened with a calculation of the house edge on placing 6 and 8 as 0.46% if you always took the bet down after one roll. That's true enough, but I've always had a strong preference for assuming the bet is played to a decision, leaving a house edge of 1.52%.

Let's say you and I went to a craps table together. I placed the 6 and left it on the table until the shooter rolled a winner 6 or a loser 7. You placed the 6 with the same size wager, but always took our bet down after each roll, and always replaced it with fresh money on the next one.

You and I would have the same amount of money on the table on each roll. We'd win on the same rolls and lose on the same rolls. At the end of the day, our balance would be exactly the same. But the house edge calculations would show you as facing a 0.46% house edge, while I faced a 1.52% house edge.

I think the 1.52 is more realistic in that the bets are usually played to their conclusion, and also more useful in comparing to bets where the wagers must remain on the table until a decision, such as pass or come.