Setting the hard ways There’s nothing more enjoyable than spending an hour or more at a bookstore browsing the stacks. I’m always looking for any craps books I haven’t read, curious to see if I can glean any new information I haven’t read before. Not every book addresses dice control, but when they do, the authors tend to fall into one of two categories:
“…it has been calculated that to overcome the house edge takes only one successful controlled roll out of 43. That is, one roll in which the dice should have sevened-out, but didn’t.”
He goes through the math to show how the House Edge is computed:
House Edge = (House Odds – Correct Odds) x (Probability of Winning)
Jensen continues:
“For a place bet on 8, the house odds are 7/6 (7 to 6), which is the actual casino payoff on the wager. The correct odds are 6/5, which is based on six ways to get a 7 and five ways to get an 8. The probability of winning is 5/11, a figure that represents the ways to win (5) divided by the ways to win plus the ways to lose (5 plus 6). Put these numbers into the formula for the following result:
House Edge = (7/6 – 6/5) x (5/11) = (35/30 – 36/30 x (5/11) = -0.01515
Multiplying -0.01515 by 100 and rounding off gives the commonly stated house edge of -1.52% for a place bet on 6 or 8. The negative result indicates that it is a house edge; if it was positive, it would be a player edge.
Theoretically, if a pair of dice are rolled randomly 216 times, a total of thirty-six 7s should appear and a total of thirty 8s should appear. Assume the shooter, who is just learning to control the dice, is trying to avoid rolling a 7. The shooter, who is not very good yet, gets thirty-five 7s in 216 rolls, missing only one 7.”
Again he goes through the math:
House Edge = (7/6 – 35/30) x (35/65) = (35/30 – 35/30) x (35/65) = 0
“It may be hard to believe that eliminating one 7 out of thirty-six reduces the house edge to zero, but the math proves it. Image how much the house will be losing when the shooter gets better and is able to eliminate three or four 7s in 216 rolls.”
I have to agree. While I have seen random shooters (those who don’t set the dice) have good rolls, I’ve observed that shooters who carefully set the dice will consistently have good rolls. I haven’t practiced my dice control in a couple of years, but when I play I do set the dice and try to have a controlled throw. It has been my observation that I will consistently have longer rolls than the non-dice setters. Thinking back over my playing sessions, the times I’ve won were at tables that had at least one other dice controller playing. Coincidence? I think not. While you may not know how to set the dice, playing at a table that has one or more dice controllers will surely increase your win ratio. After all, the math proves it doesn’t take a lot to decrease the house edge to zero.
- They believe dice control is possible
- Call it a total fraud with no basis in proof
“…it has been calculated that to overcome the house edge takes only one successful controlled roll out of 43. That is, one roll in which the dice should have sevened-out, but didn’t.”
He goes through the math to show how the House Edge is computed:
House Edge = (House Odds – Correct Odds) x (Probability of Winning)
Jensen continues:
“For a place bet on 8, the house odds are 7/6 (7 to 6), which is the actual casino payoff on the wager. The correct odds are 6/5, which is based on six ways to get a 7 and five ways to get an 8. The probability of winning is 5/11, a figure that represents the ways to win (5) divided by the ways to win plus the ways to lose (5 plus 6). Put these numbers into the formula for the following result:
House Edge = (7/6 – 6/5) x (5/11) = (35/30 – 36/30 x (5/11) = -0.01515
Multiplying -0.01515 by 100 and rounding off gives the commonly stated house edge of -1.52% for a place bet on 6 or 8. The negative result indicates that it is a house edge; if it was positive, it would be a player edge.
Theoretically, if a pair of dice are rolled randomly 216 times, a total of thirty-six 7s should appear and a total of thirty 8s should appear. Assume the shooter, who is just learning to control the dice, is trying to avoid rolling a 7. The shooter, who is not very good yet, gets thirty-five 7s in 216 rolls, missing only one 7.”
Again he goes through the math:
House Edge = (7/6 – 35/30) x (35/65) = (35/30 – 35/30) x (35/65) = 0
“It may be hard to believe that eliminating one 7 out of thirty-six reduces the house edge to zero, but the math proves it. Image how much the house will be losing when the shooter gets better and is able to eliminate three or four 7s in 216 rolls.”
I have to agree. While I have seen random shooters (those who don’t set the dice) have good rolls, I’ve observed that shooters who carefully set the dice will consistently have good rolls. I haven’t practiced my dice control in a couple of years, but when I play I do set the dice and try to have a controlled throw. It has been my observation that I will consistently have longer rolls than the non-dice setters. Thinking back over my playing sessions, the times I’ve won were at tables that had at least one other dice controller playing. Coincidence? I think not. While you may not know how to set the dice, playing at a table that has one or more dice controllers will surely increase your win ratio. After all, the math proves it doesn’t take a lot to decrease the house edge to zero.
RSS Feed
