Jan

17

2012

# A Craps System that Wins More Often Than it Loses?

In games that offer multiple wagering choice, it’s easy to design a system that will win more often than it loses. That doesn’t mean it’s to your advantage to play the system.

Case in point: A craps player told me he had a simple system. He always made place bets on both 6 and 8, hoping the shooter would roll one of those numbers before rolling a 7.

Since there are five two-dice combinations that total 6, five that total eight, and six that total that loser 7, that leaves 10 ways to win, and only 6 ways to lose. He knew there had to be a flaw in the system, and asked me what it might be.

That simple system, placing both 6 and 8, will win more often than it loses, but it will lose more MONEY than it wins. Problem is, each winning roll wins only one bet, but that 7 makes losers of them both at one time.

There are 36 possible rolls of two dice, but once you’ve placed the 6 and 8, the only rolls that matter are 6, 8, and 7. On any other number, your money either just stays on the table, or you can take the bets down.

Let’s say you bet \$6 each on 6 and 8. In an average 36 rolls of the dice, you get 7-6 payoffs on five 6s and five 8s, for \$70 in winnings. However, on the six 7s, you lose \$12, for \$72 in losses — you lose more money than you win.

The house edge is 1.52 percent on 6, 1.52 percent on 8, and 1.52 percent when you bet both at once. Combining the bets can’t change the house edge.