November 21st, 2011 at 5:00:59 PM
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Quote:AlanMendelson

By the way, the reason why the crew does not question your "late bet" on the passline with odds, is that the casino is more likely to win that "late bet" on the passline wih odds, which is why "put bets" are also welcome. But just try to place a "dont bet" after the point is established because no casino will allow it because the casino is more likely to lose a don't bet on the point after the point is established.

Its funny you say that b/c the guy playing next to me was watching how I was betting. After about 5 or 6 rolls, suddenly after the come out roll he throws his bet down on the don't pass. The dealer immediately told him he could not play that bet. The guy looked at me and then when back to hopping the 9.

November 22nd, 2011 at 8:50:56 PM
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At $3, it won't be much more than $0.25 each come out roll you are losing.

It might have been mentioned above, but this only works with a decent amount of odds. I think you are turning your come bet into about a 11% house advantage by placing it after the come out roll. You need a lot of odds to bring that down below the house edge of the place bet on the 6/8. So in the case of 3/4/5 craps, I think you would be better off just doing the place bet and forgetting about the line.

It might have been mentioned above, but this only works with a decent amount of odds. I think you are turning your come bet into about a 11% house advantage by placing it after the come out roll. You need a lot of odds to bring that down below the house edge of the place bet on the 6/8. So in the case of 3/4/5 craps, I think you would be better off just doing the place bet and forgetting about the line.

November 23rd, 2011 at 12:26:06 PM
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I assume we could just weigh the house edge of Put bets vs. Pass Line bets (flats only) - per bet resolved.

33.33% on the 4 and 10

20.00% on the 5 and 9

9.09% on the 6 and 8

(personally, I calculate a weighted average house edge of 18.79%)

vs.

1.41% on the Pass Line

Multiply your $3 flat times the number of come outs multiplied by the difference in house edges, and you should have your theoretical answer.

However, since you only wanted to play the 6 and 8, I assume boymimbo's calculations are more relevant to you.

33.33% on the 4 and 10

20.00% on the 5 and 9

9.09% on the 6 and 8

(personally, I calculate a weighted average house edge of 18.79%)

vs.

1.41% on the Pass Line

Multiply your $3 flat times the number of come outs multiplied by the difference in house edges, and you should have your theoretical answer.

However, since you only wanted to play the 6 and 8, I assume boymimbo's calculations are more relevant to you.

November 27th, 2011 at 2:10:53 PM
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What you are "sacrificing" is the win that would occur when a seven or an eleven is rolled on the come out roll. You would also avoid losing when rolling a 2, 3 or 12. The 11 would result in a win at the same frequency as a 3 so they would cancel each other out. A seven would occur 6/36 or 1/6 of the time. Either a 2 or a 12 would occur about 2/36 or 1/18 of the time (1/36 for each one). After doing the math (assuming a line bet of $3) you would end up with a net gain (that you would be sacrificing) of $0.33 per come out roll (regardless of the number rolled).

However, since you were doing this to avoid having to play any points other than 6 or 8, the $0.33 gain can be offset by the avoidance of the loss that would occur from losing the other points (4, 5, 9, 10) more often than winning them. Mathematically, if the point were 4 or 10, you could expect to lose 2/3 (66.67%) of the time. If the point were 5 or 9 you could expect to lose 60% of the time. If we assume the line bet to be $3 (the odds bet is irrelevant because it has no house edge), the math works out to show that you would have a net loss of $2 per occurance on the 4 and 10 points, and a net loss of $1.80 per occurance on the 5 and 9 points.

You could expect a come out roll to be a 4 or a 10 6/36 of the time and a 5 or a 9 8/36 of the time. So, on a come out roll of a 4 or a 10 the "sacrificed" $0.33 gain would be offset by a $0.33 loss (6/36 x $2), leaving you even. On a come out roll of a 5 or a 9 the "sacrificed" gain would be offset by a $0.40 loss (8/36 x $1.80) leaving you up $0.07 each time.

In summary, by not betting the passline initially, you would sacrifice $0.33 per every come out roll except those that would establish a point of 4, 5, 9, or 10. On those rolls you would be even on the 4 and 10 and up $0.07 on the 5 and 9. That nets out to be about $0.22 for every come out roll including the 4, 5, 9 and 10.

However, since you were doing this to avoid having to play any points other than 6 or 8, the $0.33 gain can be offset by the avoidance of the loss that would occur from losing the other points (4, 5, 9, 10) more often than winning them. Mathematically, if the point were 4 or 10, you could expect to lose 2/3 (66.67%) of the time. If the point were 5 or 9 you could expect to lose 60% of the time. If we assume the line bet to be $3 (the odds bet is irrelevant because it has no house edge), the math works out to show that you would have a net loss of $2 per occurance on the 4 and 10 points, and a net loss of $1.80 per occurance on the 5 and 9 points.

You could expect a come out roll to be a 4 or a 10 6/36 of the time and a 5 or a 9 8/36 of the time. So, on a come out roll of a 4 or a 10 the "sacrificed" $0.33 gain would be offset by a $0.33 loss (6/36 x $2), leaving you even. On a come out roll of a 5 or a 9 the "sacrificed" gain would be offset by a $0.40 loss (8/36 x $1.80) leaving you up $0.07 each time.

In summary, by not betting the passline initially, you would sacrifice $0.33 per every come out roll except those that would establish a point of 4, 5, 9, or 10. On those rolls you would be even on the 4 and 10 and up $0.07 on the 5 and 9. That nets out to be about $0.22 for every come out roll including the 4, 5, 9 and 10.

November 27th, 2011 at 3:35:10 PM
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Many people can't really understand what is going on and are simply following a fellow player or bets they see put down earlier without quite knowing what is proper. I'm sure this is part of the "house edge" for many games. When some bored dealers standing at an empty table called out to passersby I once replied that I had "never figured out where the tiny little goal posts were on the craps layout". They seemed to enjoy the laugh and the sports analogy. Most people know Blackjack is related to the sum 21. Perhaps people know Baccarat is related to the number 9. Its not just alcohol, but utter unfamiliarity with the games that helps the house margin.

February 22nd, 2012 at 8:29:50 PM
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Here's what I tried doing once... I placed $6 6 and 8. If the shooter hit one, I would take that number down. If he hit the other, I would take that one down also, leaving no more action, and not putting up any more action until the next shooter. This would always ensure 2 decisions per shooter. This would yield an average loss of $2 per 11 shooters. Multiply that by 5, placing for $30, yields $10 of loss per 11 shooters. If you put the 6 or 8 for $3 with $30 odds, you lower the house edge from 1.51% for place bets to 0.8%. So go ahead and do that.

What I don't recommend is placing all the numbers for $3. So you have the option of taking $27 across, or $15 across. While the monetary risk is less, the house edge is higher, so on average over the course of time, your loss is pretty much the same either way, but you have a chance to win more doing $27 across. I'd rather do that.

What I don't recommend is placing all the numbers for $3. So you have the option of taking $27 across, or $15 across. While the monetary risk is less, the house edge is higher, so on average over the course of time, your loss is pretty much the same either way, but you have a chance to win more doing $27 across. I'd rather do that.