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# Does craps have the opposite of a Place bet, one you win if a seven hits before the number?

2 September 2013

Question: In craps, you can bet on four, five, six, eight, nine, or 10, winning if the number appears and losing if the throw is a seven. The trouble is, sevens roll more than any other result so the odds are always against you. Is it possible to bet with the house, so you win on a seven and lose on the selected number?

Answer: Yes. These are called "lay" bets. As you note, you're favored to win. In fact your odds of winning are 2-to-1 on four or 10, 3-to-2 on five or nine, and 6-to-5 on six or eight. But, before you quit your day job figuring the casino bosses have made a mistake, consider the whole picture.

First, winning lay bets pay less than what you stand to lose, the proportion being the inverse of the odds. So, if you bet \$60, you'd win \$30 on four/10, \$40 on five/nine, and \$50 on six/eight.

Second, the minimum bets are too high for many players. You must lay at least enough for the payoff to be \$20. This means \$40 on four/10, \$30 on five/nine, and \$24 on six/eight.

Third, while you're favored to win, the house still has an edge. This is collected as a "vigorish" or "vig," which you must give the dealer when you make the bet. The vig is 5 percent of what you hope to win, rounded down to the nearest dollar. For the minimum bets cited above, 5 percent of the \$20 payoff is \$1. You therefore have to give the dealer \$41 to lay \$40 on the four or 10, \$31 to lay \$30 on the five or nine, and \$25 to lay \$24 on the six or eight. The house keeps the vig whether you win or lose.