I have been reading the forums and I am still a bit confused on the house edge per roll vs. house edge per bet resolved. I am trying to determine hourly expected loss based on a few different methods of playing. Any help would be greatly appreciated!

I typically play $10 tables and I will play the pass line with single or double odds followed by placing the six or eight.

In calculating the hourly expected loss using this strategy, I am assuming 100 rolls/hour (is this average?)

--Pass Line: ($10 x .42%) x 100 = $4.2 hourly expected loss

--Placing Six: ($12 x .46%) x 100 = $5.52 hourly expected loss

--Placing Eight: ($12 x .46%) x 100 = $5.52 hourly expected loss

Total Hourly Expected Loss = $15.24

Is this the correct calculation?

I am trying to compare this expected loss scenario with playing a three point molly with single odds and double odds. I am a bit confused as to how the house edge is reduced when utilizing odds. My understanding is that odds are a free bet and thus have no house edge but the pass line bet has a house edge of 1.41% no matter what, i.e., if I play a $10 pass line bet with $20 in odds, my expected loss is still 1.41% or .42% per roll.

I am confused when I see that the house edge is .83% when utilizing single odds and .61% when utilizing double odds. Does this percentage apply to the combined wager, i.e., when playing double odds at a $10 table, .61% house edge on the total $30 wagered? If so, how would I calculate the house edge per roll as I am interested in trying to determine expected hourly loss?

Thank you

looks right to meQuote:Craps457<snip>

Is this the correct calculation?

the house edge is NOT reduced. The combined house edge is lower than the line bet but the expected value remains the same. Gambling writers should really stop saying the HE is lower. Cuz one has to know the average bet, and most do not, they just guess.Quote:Craps457I am trying to compare this expected loss scenario with playing a three point molly with single odds and double odds. I am a bit confused as to how the house edge is reduced when utilizing odds.

I agreeQuote:Craps457My understanding is that odds are a free bet and thus have no house edge but the pass line bet has a house edge of 1.41% no matter what, i.e., if I play a $10 pass line bet with $20 in odds, my expected loss is still 1.41% or .42% per roll.

not the combined wager by itself, but the average bet between no odds (1/3 of the time) and with the odds (2/3 of the time.)Quote:Craps457I am confused when I see that the house edge is .83% when utilizing single odds and .61% when utilizing double odds. Does this percentage apply to the combined wager, i.e., when playing double odds at a $10 table, .61% house edge on the total $30 wagered?

so for full 2x odds on 6&8, $10 with $25 and 2x odds on rest, your average bet is $24.72

the math comes out the same, expected loss-wise

for multiple line bets (pass-come) you also need to know the average number of those bets per roll. I think I put that info out before, or maybe someone else did. I do not see the folder where it should be or I would share what I have now.Quote:Craps457If so, how would I calculate the house edge per roll as I am interested in trying to determine expected hourly loss?

Thank you

I would simply simulate this first

so you know about (not exact)

what the ev would be per roll, then calculate.

A sim would also show what the variance is too.

nice question!

Sally

Here is what I found from past simulations.Quote:mustangsallyfor multiple line bets (pass-come) you also need to know the average number of those bets per roll. I think I put that info out before, or maybe someone else did. I do not see the folder where it should be or I would share what I have now.

3 point Molly (pass and max 2 come bets) has an average number of bets at 2.4 per roll (over 100 rolls)

so

2.4*$10*-0.0042*100=-$10.08 per/100 rolls

SN Ethier calculated this exactly (because he can)

after asking the question in one of his books

He used a Markov chain in Mathematica (lucky)

answer

ev per roll = {-(49427/4887675)}

or

-0.0101126 = a

a*$10*100 = -$10.11 per/100 rolls

looks like we agree

hope this helps out!

Sally

you are welcome.Quote:Craps457Thank you so much! This is very helpful!

just lucky I found my data

and I was so happy to see I was right there with a super-genius in Mr. Ethier.

to be fair

in comparing different betting methods, one really should

make sure

the total amount bet, say per 100 rolls (the action), is about equal between the methods of play

or

one just compares an apple to a watermelon

and of course they are not close to being the same...

and won't be.

Sally

The main thing with all that post-graduate math set forth above is to maintain the discipline of making just those bets rather than getting caught up in an emotional wave of exuberance and start pressing bets or listening to the stickman.

The main thing with all that post-graduate math set forth above is to maintain the discipline of making just those bets rather than getting caught up in an emotional wave of exuberance and start pressing bets or listening to the stickman.

100 rolls per hour was used for calculation purposes.Quote:FleaStiffRolls per hour was mentioned somewhere up thread and I'm no longer sure what is common at tables <snip>

one can use almost any value for average rolls as there is not one value that makes all craps players happy.

2 rolls per minute is 120 per hour, hard to sustain with a full table of hardways players.

1 roll per hour may be hard to sustain also

with all betting the hardways and the Bonus craps bets and having hot shooters. Lots of payouts.

3 point Molly with $10 flat bets

rolls | expected loss per X rolls |
---|---|

20 | -$2.02 |

30 | -$3.03 |

40 | -$4.05 |

50 | -$5.06 |

60 | -$6.07 |

70 | -$7.08 |

80 | -$8.09 |

90 | -$9.10 |

100 | -$10.11 |

110 | -$11.12 |

120 | -$12.14 |

130 | -$13.15 |

140 | -$14.16 |

150 | -$15.17 |

160 | -$16.18 |

170 | -$17.19 |

180 | -$18.20 |

190 | -$19.21 |

200 | -$20.23 |

210 | -$21.24 |

220 | -$22.25 |

230 | -$23.26 |

240 | -$24.27 |

250 | -$25.28 |

260 | -$26.29 |

270 | -$27.30 |

280 | -$28.32 |

290 | -$29.33 |

300 | -$30.34 |

3 point Molly with $5 flat bets

rolls | expected loss per X rolls |
---|---|

20 | -$1.01 |

30 | -$1.52 |

40 | -$2.02 |

50 | -$2.53 |

60 | -$3.03 |

70 | -$3.54 |

80 | -$4.05 |

90 | -$4.55 |

100 | -$5.06 |

110 | -$5.56 |

120 | -$6.07 |

130 | -$6.57 |

140 | -$7.08 |

150 | -$7.58 |

160 | -$8.09 |

170 | -$8.60 |

180 | -$9.10 |

190 | -$9.61 |

200 | -$10.11 |

210 | -$10.62 |

220 | -$11.12 |

230 | -$11.63 |

240 | -$12.14 |

250 | -$12.64 |

260 | -$13.15 |

270 | -$13.65 |

280 | -$14.16 |

290 | -$14.66 |

300 | -$15.17 |

example simulation (only 30k in size)

$5 bets 2x odds 3 point Molly

notice the variance and range

also:

the average number of rolls until all bets resolved was over 103 (103.6)

that is a very good reason why the ev in my sim was a little higher as calculated

per 100 rolls

and last too

number of games (sessions of 100 rolls in length) that ended at exactly

-$4: 75

-$5: 82

-$6: 77

-$7: 83

Sally

I think that I am getting most of it but had a few clarification questions.

My ultimate goal is to have a good time while reducing the house edge. I understand that there is an expected loss with playing craps and I want to choose an enjoyable strategy that limits the losses to a minimum.

I would like to have two to three numbers working. I am leaning towards sticking to playing the pass line with single odds and then placing the six and eight. I typically play $10 tables and thus I will have a total of $44 in action ($10 place bet, $10 odds, $12 each on the six and eight). As calculated previously, it appears that my average expected hourly loss will be $15.24.

I like the idea of three point molly but I am concerned on the variance and putting more money on the table. If I play a $10 table, that would be $60 in action assuming that I have the pass line and two come bets with single odds. In the calculations above, I assume that the hourly loss on this scenario would be less at only $10.11. My concern though is that the variance will be much higher given more money on the table, is this correct? $60 in action is a bit much for me.

I would also like to clarify the odds and house edge. Even if I had 5x odds utilizing the three point molly with $10 flat bets, my expected hourly loss is still $10.11, correct? Essentially there is no cost to the odds bet but I still have the .42% edge on the flat bet portion?

My understanding is that the correct "mathematical" method to play assuming that I wanted $40 in action, would be to just play the pass line with 3x odds as I would have $40 in action for only a $4.2 hourly loss (.42% x $10). The issue that I have with this play is that it is not much fun to just play one number and my variance might be high as my action will be on a single number.

Any additional help would be greatly appreciated.

you can't reduce the house edge. the more bets you make, the larger the expected loss. Odds bets have 0 house edge.Quote:Craps457My ultimate goal is to have a good time while reducing the house edge.

your calculations are not correct, imo.Quote:Craps457I like the idea of three point molly but I am concerned on the variance and putting more money on the table. If I play a $10 table, that would be $60 in action assuming that I have the pass line and two come bets with single odds.

What is the variance of method 1 (pass Place 6 & 8) compared to 3 point Molly?

most do not know, they can only guess.

$10 line with $10 odds produces and average bet of $16 2/3

and with 2.4 average bets per roll

2.4 * 16.67 = $40.00

and all bets must resolve as they are contract bets

Place bets can be called off or removed.

This is where simulations come in handy

Method1 vs. Method 2

same rolls

see what happens

and the average bet, variance, and other stats are collected for you.

I suggest you start there.

Sally