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...

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

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 |

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 |

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.

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.

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)

So the shooter first establishes his point and at that moment you make your 6 place bets. Let’s say the point is 4 and on the next roll the shooter hits his 4 so you win your place bet on that number.

Now....do you take down all $14 ($5 stake plus $9 winnings) from the 4 or just the $9 in winnings, leaving the original $5 wager ?

Pardon my ignorance if I’m missing something obvious or customary. I never make any bets besides pass/don’t/come/don’t/odds. The Wizard doesn’t approve the others

by default,Quote:Ace2So the shooter first establishes his point and at that moment you make your 6 place bets. Let’s say the point is 4 and on the next roll the shooter hits his 4 so you win your place bet on that number.

Now....do you take down all $14 ($5 stake plus $9 winnings) from the 4 or just the $9 in winnings, leaving the original $5 wager ?

if one does NOT say anything to the dealer as what to do with a winning place bet, the winnings are handed off to the player and the existing bet is left alone,

set down on the table in front of the player. Seen this many many times.

The dealer may ask "same bet?" and do the same thing, but not to wait and slow the table down as the player thinks long and hard about what to do next.

on the come out roll, by default, all 'place bets to win' are OFF, meaning they can not win or lose on the come out roll.

I think the OP is after just that.

of course a small -ev is dominated by a large variance per roll

many do not get that part either

Sally

Jeremíah, please confirm you only take down winnings and you leave bets off on come out rolls.

But to answer your question, yes, I would take down winnings as they occur; leave the place/buy bets off on the come-out rolls; then look to begin prudently pressing the hot numbers should someone take off on a run of luck. Experimental? Certainly. Laughable? Yeah, sure. But what the hell, I'm a dice player. Thanks again for your help.

This table shows the amount bet, what a win would pay, the number of winning and losing combinations, and the expected value per roll.

Point | Bet | Pays | Winning combinations | Losing combinations | Exp value |
---|---|---|---|---|---|

4 | 5 | 9 | 3 | 6 | -0.083333 |

5 | 5 | 7 | 4 | 6 | -0.055556 |

6 | 6 | 7 | 5 | 6 | -0.027778 |

8 | 6 | 7 | 5 | 6 | -0.027778 |

9 | 5 | 7 | 4 | 6 | -0.055556 |

10 | 5 | 9 | 3 | 6 | -0.083333 |

Total | 32 | -0.333333 |

The lower right cell shows the player can expect to lose 33 1/3 cents per roll. Compare this to the $32 bet, and you have a house edge of (1/3)/32 = 1/96 = 1.04%.

You might ask, "Wiz, how can this be, each individual bet has a higher house edge than that. It is because the house edge on place bets is traditionaly defined as the expected loss per bet resolved. If the house edge on place bets counted ties, then it would be much less.

it appears to meQuote:WizardHere is my answer. Please note that I'm counting a pushes. So, this would answer the question at hand on a "per roll" basis.

OP does not have any place bets working on the come out roll.

the -$1/3 is where place bets work on the come out roll and that is a rare player indeed that wood due that.

The OP needs to understand why his method would result in a lower ev and what the probability of a point roll would be instead of all rolls.

good point showing 'per roll' and 'bet resolved'

Sally

1.04 % doesn’t seem reasonable to me.Quote:WizardThe lower right cell shows the player can expect to lose 33 1/3 cents per roll. Compare this to the $32 bet, and you have a house edge of (1/3)/32 = 1/96 = 1.04%.

I assume we can all agree that the house edge is 3.90 % on all money wagered, which is just the summation of all place bet edges at the 5/5/6 ratio.

So for the per-roll edge to be 1.04 %, that implies that only 1 in 3.75 (390/104) of all bets are resolved.

I believe that 30 of 36 bets are resolved, which would put the per-roll edge at 3.25%. The only time that bets aren’t resolved is when a 2,3,11,or 12 is rolled (6 out of 36 combinations). Come out rolls are insignificant since bets either haven’t been placed (new shooter) or place bets are off (same shooter just hit point). So come out rolls are not ties, they simply don’t exist since nothing is ever wagered unless a point has been established (original poster confirmed). Every other number - 4 through 10, 30 of 36 combinations, is a resolved bet for this scenario.

Or I am missing something fundamental here..

or my or others simulation resultsQuote:Ace21.04 % doesn’t seem reasonable to me.

when one has as total action bets resolved, not $$$ risked. IF $$$ risked (bets on the table) mattered why not include all money in the rail too?

I think many others have show that to be 3.73% but what is more important really,Quote:Ace2I assume we can all agree that the house edge is 3.90 % on all money wagered, which is just the summation of all place bet edges at the 5/5/6 ratio.

HE or EV?

the Wizard used $32 at risk, for what reason, still remains unknown. for 10 million rolls, that would mean that $32 million was resolved bets. do knot C that happening.Quote:Ace2So for the per-roll edge to be 1.04 %, that implies that only 1 in 3.75 (390/104) of all bets are resolved.

I think not. IF the math for this combinations of bets was so easy to do, everyone would do it. and of course they all do.Quote:Ace2Or I am missing something fundamental here..

now what are the results and what do they mean and are there any simulations that match any calculated results.

it is also a rare place bettor that keeps the wagers at the same unit all the time. Press to win more $$$.

Press to lose more $$$.

Sally

I would suggest getting your plan together and simulate it. that would give you results from many trying your system to see what kind of range of wins and losses to expect.Quote:jeremiah22then look to begin prudently pressing the hot numbers should someone take off on a run of luck.

MOST ALL place bettors I have seen think they win most all the time and only lose because of

1) poor money management,

2) bet management (not pressing their bets when the shooter does get hot, or taking down their bets B4 the shooter gets cold - locking up a profit)

3) betting on poor shooters (should be on the don't side for those)

well, have fun no matter what you get done here

Sally

had to add this

flat-betting across place bets

this was calculated (Markov chain)

rolls(where wagers are working)/chance of net loss (flat betting)

10/51.49%

100/57.94%

1000/76.56%

2000/84.89%

5000/94.93%

10000/98.98%

the more bets you make, the GREATER the chance U have of showing a net loss over X # of equal wagers.

not good I say

I now agree with that number for this scenario. The 3.90 % would be correct if place bets were taken down individually as they won. But as I learned that is not common practice nor what the original poster does.Quote:mustangsallyI think many others have show that to be 3.73% but what is more important really,

So the place bets start out at a $5/5/6 ratio but the subsequent wins and additional bets will come at a 3/4/5 ratio, slightly skewing the edge down since a higher % (5/12ths not 6/16th) of new wagers will go to lower edge bets 6&8 and a lower % (3/12ths not 5/16th) of new wagers will go higher edge bets 4&10.

It’s already been shown that on average a shooter’s turn lasts 8.53 rolls, and exactly 6 of those rolls, on average, are made with an established point. The residual 2.53 rolls are come out rolls which have no action in this scenario. Incidentally you can get to that number by knowing that on avg it takes 1.5 rolls to establish a point. We also know that on average a shooter hits 0.68 points, so he has (1 + .68) * 1.5 = 2.53 comeout rolls. Therefore we know that our bettor averages 6 active rolls per turn, one of which is the final seven out and one of which (6/36 * 6) is a 2,3,11 or 12 which is a tie. The remaining 4 rolls are wins and, for example, 3/24 * 4 * $5 = $2.50 of new wager gets added on to the intial $5 placed on the 4.

Crunch the other numbers and the combined bet ratio (initial plus additions) is $7.50 / 8.33 / 11.00 = $ 26.83 per side. Apply the individual place bet house edges (7.50 * 1/15 + 8.33 * 1/25 + 11.00 * 1/66) / 26.83 = 3.73% combined.

So the per roll number would be 3.73% * 5/6 = 3.11%.