November 6th, 2018 at 1:44:34 PM
permalink

I've tried to calculate what the house edge would be for betting all 6 place numbers at craps. I'm not a math guy but used the following calculations and would like you good people to please point out where I might be wrong.

My first calculation is that on each roll of the dice there would be 24 ways to win (total possibility on all 6 place numbers) vs 6 ways to lose (on 7 appearing. That would provide a ratio of 24:6, which reduces to 4:1. That would, in turn, mean a player could expect, on average, 4 winning hits to every wipe-out loss of 6 units.

I've used numbers that would be put in play at a $5 minimum table. Making the minimum bet on all 6 numbers across would mean you were risking $32 on each attempt. I further calculated that with the 4&10 paying $9 per hit; and the 5,6,8 & 9 paying $7 per hit, the average profit per hit would be $7.5.

That would mean that, on average, a player could expect to win $30 (4 hits at $7.5 each) for every wipe-out loss of $32. Or, that a player could expect, on average, a loss of $2 for every $32 attempt. That's a ratio of 2:32 or 1:16. 1/16th comes out to a percentage of 6.25%, which I would take to mean as the house edge of betting all 6 numbers across.

Have I figured this out correctly? If not, please show me where I went wrong. Thanks...

My first calculation is that on each roll of the dice there would be 24 ways to win (total possibility on all 6 place numbers) vs 6 ways to lose (on 7 appearing. That would provide a ratio of 24:6, which reduces to 4:1. That would, in turn, mean a player could expect, on average, 4 winning hits to every wipe-out loss of 6 units.

I've used numbers that would be put in play at a $5 minimum table. Making the minimum bet on all 6 numbers across would mean you were risking $32 on each attempt. I further calculated that with the 4&10 paying $9 per hit; and the 5,6,8 & 9 paying $7 per hit, the average profit per hit would be $7.5.

That would mean that, on average, a player could expect to win $30 (4 hits at $7.5 each) for every wipe-out loss of $32. Or, that a player could expect, on average, a loss of $2 for every $32 attempt. That's a ratio of 2:32 or 1:16. 1/16th comes out to a percentage of 6.25%, which I would take to mean as the house edge of betting all 6 numbers across.

Have I figured this out correctly? If not, please show me where I went wrong. Thanks...

November 6th, 2018 at 8:55:16 PM
permalink

Should be 5/16 * 3 /12 * 1/15 + 5/16 * 4/12 * 1/25 + 6/16 * 5/12 * 1/66.

Summation of Bet size * bet frequency * edge for placing 4, 5 and 6.

Summation of Bet size * bet frequency * edge for placing 4, 5 and 6.

It’s all about making that GTA

November 7th, 2018 at 2:02:38 AM
permalink

The usual way to work out House Edge for Craps bets is consider active rolls, for instance in the UK £3 on 5 would pay £4.20. So the House Edge would be (4*£4.20-6*£3)/(4+6)/£3 = £1.20/10/£3 = 4%. You can see this at https://wizardofodds.com/games/craps/appendix/2/

However you can also look at all rolls and get a lower House Edge (NB I've assumed UK odds which pay 9.5 to 5 on 4 or 10 and a typical minimum bet of £3 on each number). With multiple bets running it might be easier to look at the cost per roll given your total bet is £18.

24

However you can also look at all rolls and get a lower House Edge (NB I've assumed UK odds which pay 9.5 to 5 on 4 or 10 and a typical minimum bet of £3 on each number). With multiple bets running it might be easier to look at the cost per roll given your total bet is £18.

24

Roll | Perms | Payoff | Total |
---|---|---|---|

2 | 1 | £0.00 | 0.00 |

3 | 2 | £0.00 | 0.00 |

4 | 3 | £5.70 | 17.10 |

5 | 4 | £4.20 | 16.80 |

6 | 5 | £3.50 | 17.50 |

7 | 6 | -£18.00 | -108.00 |

8 | 5 | £3.50 | 17.50 |

9 | 4 | £4.20 | 16.80 |

10 | 3 | £5.70 | 17.10 |

11 | 2 | £0.00 | 0.00 |

12 | 1 | £0.00 | 0.00 |

Total | 36 | -5.20 | |

Average cost/roll | 36 | -0.802% | -£0.14 |

Averagecost /active roll | -1.204% | -£0.22 |

November 7th, 2018 at 7:29:13 AM
permalink

An alternative way to look at it is how much you have to wager to win $100 if any number (4 5 6 8 9 or 10) is thrown given the fair price would be $400. It is equivalent to making a total wager of $100 and only receiving $23.83.513c instead of a fair value of $25.

Again I've used UK place odds.

Again I've used UK place odds.

Roll | Odds | Bet Needed |
---|---|---|

4 | 19/10 | $52.63.158c |

5 | 7/5 | $71.42.857c |

6 | 7/6 | $85.71.429c |

8 | 7/6 | $85.71.429c |

9 | 7/5 | $71.42.857c |

10 | 19/10 | $52.63.158c |

Total bet | $419.54.887c |

November 7th, 2018 at 4:31:11 PM
permalink

I used US place odds since I see no indication to assume otherwise. To my knowledge there are like 5 Craps tables in all of England...last year I played at 2 of them in the bottom level of the Hippodrome casino in London. Anyway it looks like the only difference is that the 4/10 pays 19 to 10 instead of 9 to 5.

My calculation is for the house edge on all money wagered. To get a per-roll edge I’d multiply that edge by 7.02 * 30/36 / 8.52.

8.52 being the average roll length until seven out. Minus 1.5 avg rolls to establish a point is 7.02 rolls that aren’t coming out after a seven out. 30/36 of those rolls have a resolved bet since only 2,3,11 and 12 have no resolution when a point had been established.

My calculation is for the house edge on all money wagered. To get a per-roll edge I’d multiply that edge by 7.02 * 30/36 / 8.52.

8.52 being the average roll length until seven out. Minus 1.5 avg rolls to establish a point is 7.02 rolls that aren’t coming out after a seven out. 30/36 of those rolls have a resolved bet since only 2,3,11 and 12 have no resolution when a point had been established.

Last edited by: Ace2 on Nov 7, 2018

It’s all about making that GTA

November 7th, 2018 at 5:47:22 PM
permalink

Ace2 and charliepatrick, I want to thank you both for taking the time and effort to reply to my post. Please know it is very much appreciated. I may not have been clear in my question. Or, more likely, you've both already given me the answer I was looking for, but I don't have adequate left-brain skills to recognize it. Here's the situation I was trying to address...

A shooter has established his point. I now place a single unit wager on each of the 6 point numbers, including the point to be made. What I was trying to work out in the meatball-math I laid out in my original post is what the house edge -- expressed as a simple single percentage -- would be in that situation on each individual roll of the dice. This would be similar to craps books telling the novice player that the house edge on placing the 6 or 8 is 1.5%; or that the house edge on placing the 5 or 9 is 4%.

Again, in my naivete, I may not realize it's not possible to give a simple, single house-edge percentage on what is essentially an accumulation of 6 wagers being made at once. Whatever the case may be, I thank you both once more for the time and effort you've taken in this regard. Best of luck to you at the tables.

A shooter has established his point. I now place a single unit wager on each of the 6 point numbers, including the point to be made. What I was trying to work out in the meatball-math I laid out in my original post is what the house edge -- expressed as a simple single percentage -- would be in that situation on each individual roll of the dice. This would be similar to craps books telling the novice player that the house edge on placing the 6 or 8 is 1.5%; or that the house edge on placing the 5 or 9 is 4%.

Again, in my naivete, I may not realize it's not possible to give a simple, single house-edge percentage on what is essentially an accumulation of 6 wagers being made at once. Whatever the case may be, I thank you both once more for the time and effort you've taken in this regard. Best of luck to you at the tables.

November 7th, 2018 at 6:07:38 PM
permalink

It will be 1/15th of what you have on the 4&10 plus 1/25th of what you have on the 5&9 plus 1/66th of what you have on the 6&8. You said the wagers will be $6 on 6&8 otherwise $5. That’s 3.90% house edge on your total wager. On a per roll basis it’s that * 30/36 = 3.25 % since 6 of 36 rolls (2,3,11,12) have no resolution.

It’s all about making that GTA

November 7th, 2018 at 6:18:49 PM
permalink

Thanks so much, Ace2. That's what I was looking for. And thank you, too, charliepatrick. Grateful to you both.

November 8th, 2018 at 10:49:42 AM
permalink

I agree that per roll is a better calculationQuote:jeremiah22What I was trying to work out in the meatball-math I laid out in my original post is what the house edge -- expressed as a simple single percentage -- would be in that situation on each individual roll of the dice.

expected value = avgResolvedwager * house edge

so once we know the ev we can divide by the avg Resolved wager and get the HE per roll.

alo remember that over 100 rolls there are, on average, only 70 non-comeout rolls that you are after.

that messes things up

ok,

this can and has already been calculated B4 here, I do not have those links, so I have some actual dice rolls and rng computer dice rolls (10 million) in Wincraps

you should be able to get a real close answer as the expected he of 3.73% was very close in this simulation.

REMEMBER, NOT ALL PLACE BETS RESOLVE ON EVERY ROLL.

from wincraps classic

I get, if the wind has not clouded my brain today

from

https://wizardofodds.com/games/craps/appendix/2/

ev per roll = about -0.332976 cents... but WAIT HERE!

IS NOT THAT EVERY ROLL? COUNTING ALL COME OUT ROLLS??

different from the simulation results

-0.2361136 cents per roll

or the shown expected value

-0.2346643

REMEMBER, NOT ALL PLACE BETS RESOLVE ON EVERY ROLL.

this is not fun at all, so I say have fun with this.

you seem to have basic math skills, just remember the difference between 'per roll' and 'per bet resolved'

maybe I can find those links somewhere.

Sally

added: forgot to mention

wincraps wagers it tracks are 'resolved' wagers. meaning that when $32 is working and a 2,3,11 or 12 rolls

there is NOT any bets resolved being ADDED to the tally.

that complicates things more (for calculations) but makes some things simpler at the same time

apples and oranges

last addition:

I like this description about the 32 across method (place all numbers)

the game now is a coin flip (lol)

50% chance of winning $7 on one roll (5,6,8,9)

50% of something else on one roll split equally into 3 results

1/6 chance of winning $9 (4,10)

1/6 chance of winning $0 (2,3,11,12)

1/6 chance of LOSING $32 (7 out)

that ev = -$1/3 (per roll)

probability that any roll will be a point roll (NOT a come out roll) = 392/557

ev of $32 across = -1/3 * 392/557

about -0.234590066 cents per roll (not counting come out rolls)

Last edited by: mustangsally on Nov 8, 2018

I Heart Vi Hart

November 8th, 2018 at 5:42:14 PM
permalink

Thank you for your detailed response, Sally. Very much appreciate it. I also take it by your (lol) parenthetical to the line "the game is now a coin flip" that when you say you "like this description" you are more amused by it than impressed by any validity it might have.