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The Casino Answer Man

Q. Some years ago a fellow worker at my Ford office described to me a craps betting system whereby if you roll a 4, 5, 6, 8, 9, or 10 on the comeout roll, you then bet on the 7 along with backing your line bet with free odds. That way you either make money or break even whether you make the point or not. Is that a good betting strategy?

John R., via e-mail

A. I've seen many players try to "protect" their pass-line wagers by also betting on 7. Winning bets on 7 pay 4-1, so they'll bet $1 on the 7 in hopes of offsetting losses on the pass line. The problem is that the pass line is a multiroll bet, while wagers on 7 are decided on every roll. On rolls that are not the point number and not a 7, the pass bet just stays on the board while the bet on 7 loses.

Let's assume the point is 4. As with the 10, there are only three ways to roll a 4, leaving the pass player at his most vulnerable. In a perfect sequence of 36 rolls in which each possible combination turns up once, the $5 pass bettor will win on 4 three times, collecting $15, but lose six times, dropping $30 for net losses of $15.

That's not an inviting scenario, but it's even worse for a player who also hedges by betting on any 7. He still loses a net of $15 on the pass line. Does he make it up by winning on 7? No. His $1 bet on 7 bet wins six times in the sequence, for $24 in winnings, but loses 30 times for $30 in losses. That's a net loss of $6 on any 7, pushing overall losses up to $21 for the sequence.

Does increasing the bet size on 7 help? No, it makes matters worse. Bet $5 on 7, and losses soar to $30 on 7, to go with the $15 in losses on the line.

Hedging by making a bad bet -- and 7 is one of the worst -- to go along with a good one such as the pass line does not protect the player. It just makes for bigger losses.

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