mhernandez116
mhernandez116
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August 13th, 2011 at 10:20:43 PM permalink
Apologies if this has been covered before. I've seen similar threads, but not quite what I'm looking for.

What is the average total bet for a pass line bet and x number of come bets, at 3-4-5x odds? I generally like to have 3 points working at any given time, but on the last few trips I've started to be concerned that I'm overbetting my bankroll.

Assuming this style: PL bet, full odds, followed by come bets with full odds until 3 pts are established. Then I stop betting. I only add another bet as one is resolved (or 7 out), never more than 3 pts. On average, how much money do I have out there?
TheNightfly
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August 13th, 2011 at 10:31:53 PM permalink
You haven't said what your bankroll is. You could bet any portion of your bankroll that you want depending on what your goal is and your level of risk aversion.
Happiness is underrated
vert1276
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August 14th, 2011 at 3:45:51 PM permalink
5.16 units is the average bet......assuming you bet 1 unit P/L with max 3-4-5x's odds.
mhernandez116
mhernandez116
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August 15th, 2011 at 5:00:38 PM permalink
Thank you for the response.

Would you mind giving a quick run through of the math you used to calculate that? Again, I like to make one pass line bet and two come bets, with max 3-4-5x odds.

So if the avg bet is 5.16 units, would avg bet with 3 points be 5.16 x 3?
soulhunt79
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August 15th, 2011 at 6:07:40 PM permalink
I don't think it would be just 3 times the value. You are unable to just put the 2 come bets out at the start so you have a few rolls at the start of each shooter where you don't have all 3 bets out there. Also you wouldn't be putting a come bet out when you had 3, so anytime one of the 2 hit, the next roll would only have your pass + 1 come bet.
vert1276
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August 16th, 2011 at 12:10:15 AM permalink
Quote: mhernandez116

Thank you for the response.

Would you mind giving a quick run through of the math you used to calculate that? Again, I like to make one pass line bet and two come bets, with max 3-4-5x odds.

So if the avg bet is 5.16 units, would avg bet with 3 points be 5.16 x 3?



36 rolls by the numbers

12 rolls will be front line winners or losers........8 winners 4 losers

24 rolls will points will be established
4 and 10 est. 6 times......1 unit P/L plus 3 units odds... 4 units total 4X6=24 units
5 and 9 est 8 times......1 unit P/L plus 4 units odds... 5 units total 5X8=40 units
6 and8 est. 10 times.....1 unit P/L plus 5 units odds...6 units total 6X10= 60 units

124 total units 24 rolls........average bet per 5.16

SO if you had 1 P/L with full odds and 2 comes bets working with full odds you will have about 15.5 units in play
odiousgambit
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August 16th, 2011 at 1:51:42 AM permalink
Quote: mhernandez116

PL bet, full odds, followed by come bets with full odds until 3 pts are established. Then I stop betting. I only add another bet as one is resolved (or 7 out), never more than 3 pts. On average, how much money do I have out there?



That it was something close to 5 units was a gimme, and that 3 numbers working with odds was about 15 units. So I wonder if that really was the question you intended.

Quote: vert1276

36 rolls by the numbers

12 rolls will be front line winners or losers........8 winners 4 losers

24 rolls will points will be established
4 and 10 est. 6 times......1 unit P/L plus 3 units odds... 4 units total 4X6=24 units
5 and 9 est 8 times......1 unit P/L plus 4 units odds... 5 units total 5X8=40 units
6 and8 est. 10 times.....1 unit P/L plus 5 units odds...6 units total 6X10= 60 units

124 total units 24 rolls........average bet per 5.16

SO if you had 1 P/L with full odds and 2 comes bets working with full odds you will have about 15.5 units in play



For a variant question that I often ask myself, "as you play, betting full 3x4x5x odds but sometimes of course just getting immediate front line resolution, what is your average bet?" This when working perhaps with an estimate of come-out rolls per hour and want to know your average bet based on that. Now it's 136 units per 36 rolls, approx. 3.78 units per come-out roll. Unless I missed something.

When you do come bets, basically you are increasing the percentage of come-out rolls per actual rolls. I've always had to guess at how many come-out-equivalent bets I was making. For one thing I am not consistent. But I figure about 40 per hundred rolls when adding 'some' come bets, instead of the average of about 30 per 100 when never betting come, is right for me. If you were consistent about your style, never failing to make a come bet when your strategy called for it, the numbers could be crunched on that too.
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mhernandez116
mhernandez116
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August 16th, 2011 at 3:47:36 PM permalink
Again, thanks for the follow up.

I was confusing myself trying to account for the front line decisions as well as the times the shooter will 7-out before I get all three points.

Very helpful, much appreciated.
Harmony
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August 22nd, 2011 at 10:58:33 AM permalink
I'm fairly new to craps and usually play $5 tables. Lately most of the tables have $10 minimums. What are the minimum bets for Come line and placing the 6 and 8? Obviously the Come bet minimum is $10 but what about the 6 and 8. Can you still wager $6 or do you have to go up to $12??

Thanks
jc2286
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August 22nd, 2011 at 11:00:29 AM permalink
Quote: Harmony

I'm fairly new to craps and usually play $5 tables. Lately most of the tables have $10 minimums. What are the minimum bets for Come line and placing the 6 and 8? Obviously the Come bet minimum is $10 but what about the 6 and 8. Can you still wager $6 or do you have to go up to $12??

Thanks



It would have to be $12 (likewise, $18 for a $15 min, $30 for a $25 min).
Harmony
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August 22nd, 2011 at 11:40:34 AM permalink
Thanks,

Would take a pretty good size bankroll.
FleaStiff
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August 22nd, 2011 at 2:08:36 PM permalink
You can find craps tables as low as one dollar but all these one, three and five dollar tables are often either a trek or problematical in that by whim they are often raised when things get busy.

As a matter of fact given the importance of dealer competency, you might actually be better off at a ten dollar table with a well trained and experienced crew than a five dollar table in a break in joint where the dealers are a bit inexperienced.

You might as well do the ten dollar line bets. And a place bet on the six or eight will be 12 dollars. Frankly the "extra" money for making those bets is probably cheaper than limiting yourself to remote casinos or early morning gambling only. Often the cabfare to get to a cheap casino is just too absurd since you are often much better off playing wherever you happen to be.

Giving a good portion of your "action" to the casino where you are staying might even earn you a room comp, treking to another casino won't.
rudeboyoi
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August 22nd, 2011 at 3:52:54 PM permalink
if you really really want to know the average instead of just an estimate, heres the formula. youre gonna have to put the probabilities in though. assuming a $5passline/comeline bet backed up with 3x4x5x odds.

on the comeout roll, you can roll a nonpoint number or establish a point.
once one point is established you can either roll a nonpoint number or establish a 2nd point.
once two points are established you can either roll a nonpoint number or establish a 3rd point.

$5(P(nonpointnumber"NP"))+$30(P(6or8))+$25(P(5or9))+$20(P(4orT))+$35(P(6or8thenNP))+$30(P(5or9thenNP))+ $25(P(4orTthenNP))+$60(P(6or8then6))+$55(P(6or8then5or9))+$50(P(6or8then4orT))+$65(P(6or8then6thenNP))+ $60(P(6or8then5or9thenNP))+$55(P(6or8then4orTthenNP))+$55(P(5or9then6or8))+$50(P(5or9then5))+$45(P(5or9then4orT))+ $60(P(5or9then6or8thenNP))+$55(P(5or9then5thenNP))+$50(P(5or9then4orTthenNP))+$50(P(4orTthen6or8))+ $45(P(4orTthen5or9))+$40(P(4orTthen4))+$55(P(4orTthen6or8thenNP))+$50(P(4orTthen5or9thenNP))+ $45(P(4orTthen4thenNP))+$85(P(6or8then6then5or9))+$80(P(6or8then6then4orT))+$85(P(6or8then5or9then6))+ $80(P(6or8then5or9then5))+$75(P(6or8then5or9then4orT))+$80(P(6or8then4orTthen6))+$75(P(6or8then4orTthen5or9))+ $70(P(6or8then4orTthen4))+$80(P(5or9then5then6or8))+$70(P(5or9then5then4orT))+$85(P(5or9then6or8then6))+ $80(P(5or9then6or8then5))+$75(P(5or9then6or8then4orT))+$75(P(5or9then4orTthen6or8))+$70(P(5or9then4orTthen5))+ $65(P(5or9then4orTthen4))+$70(P(4orTthen4then6or8))+$65(P(4orTthen4then5or9))+$80(P(4orTthen6or8then6))+ $75(P(4orTthen6or8then5or9))+$70(P(4orTthen6or8then4))+$75(P(4orTthen5or9then6or8))+$70(P(4orTthen5or9then5))+ $65(P(4orTthen5or9then4))

P(NP)=1/3
P(6or8)=10/36
P(6)=5/36
P(5or9)=8/36
P(5)=4/36
P(4orT)=6/36
P(4)=3/36

plug those values into the formula and multiply them together and you will get your answer.
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