I was recently reading an article found here:

http://www.goldentouchcraps.com/Stickman/stick0014.shtml

I had done the math on it before, and agreed with the person who asked the question - that the house edge for a place bet on both the 6 & 8 would be 1.04% (which I considered to be the house edge per bet resolved). This person argued that there were 16 combinations that resulted in a decision, and used this as the basis for his calculations. However, the author of this article had used a different method of calculating the house edge for this combined bet, and stated that "you must do a separate calculation for each bet." He also said that there was an issue with re-booking the bet, and pro-rate this, and pro-rate that. Later in the article he stated that a simulation of 10 million rolls gave an "expected edge" of 1.52% (although no raw data was provided... for all I know, he wrote the program that calculated this).

There was also a calculation in regards to the Iron Cross here:

http://www.goldentouchcraps.com/Stickman/stick0011.shtml

In this article, the author explains in a little more detail about how the calculation works, but I'm not completely sold on it. He offered the same simulated calculation, and even went on to prove the method by calculating the house edge for a single place bet on the 6 (which of course was correct).

This would contradict how I thought it was supposed to be calculated, which lines up with a calculation I saw here about 32-across:

http://krigman.casinocitytimes.com/article/betting-$32-across-in-craps-59413

I'm embarrassed to say that I've wasted my entire day trying to figure this out, and I need some help. Any thoughts?

Of course. Calculate the difference in house edge that is in dispute and multiply that by the amount of your usual bet and the number of bets per hour you would normally make.Quote:kaysirtapI'm embarrassed to say that I've wasted my entire day trying to figure this out, and I need some help. Any thoughts?

Then instead of being embarrassed to have wasted the entire day on the problem, you will be happy to see how much money you saved.

I'm further embarrassed to say... huh? Which method (if any) is correct?Quote:FleaStiffOf course. Calculate the difference in house edge that is in dispute and multiply that by the amount of your usual bet and the number of bets per hour you would normally make.

Then instead of being embarrassed to have wasted the entire day on the problem, you will be happy to see how much money you saved.

If you place both, you should still get an edge of 1.52% per bet resolved... that is, after resolving BOTH bets.

It comes from the idea that after all 16 combinations that cause a decision have been made, the net loss would be $10 ($30 each 6 & 8). The average loss per roll during these 16 decision rolls is $0.625. "The house edge is defined as the ratio of the average loss to the initial bet." - WOO. $0.625/$60=1.04%Quote:dwheatleyThe house edge on Place 6 or 8 is 1.52% per bet resolved, and 0.46% per roll. I'm not sure where your 1.04% figure comes from, I didn't follow your link. My source is the WOO.

If you place both, you should still get an edge of 1.52% per bet resolved... that is, after resolving BOTH bets.

IF the bet on the table were "6 &/or 8", Pays 7 to 12... obviously hits when either the 6 or 8 roll... are we still sticking with the 1.52% edge? Additionally, this 1.52% still does not agree with the article from the first link I posted.

If you read the first entry of this WOO faq, you can see a similar calculation based o the iron cross.

When you try to calculate the house edge of a multiple bet combination, you are looking at the resolved bets (ie, only have 16 combinations that have decisions). However, the proper comparison between that kind of bet, and a single bet, would be to compare their results on a per-roll basis (ie, over all 36 possible combinations).

If you do that with a single place bet (place 6, for example), you will see that the house edge on a PER-ROLL basis is 0.46%. If you look at the place-6 and place-8 combination bet, the house edge on a PER-ROLL basis is also 0.46%.

You, and the OP of the question you linked, are trying to compare the PER-RESOLVED-BET house edges, but this doesn't make any sense. As you yourself indicated, there are different combinations involved. With a single Place-6 bet, there are only 11 results to concern yourself with, while with the combination of both Place-6 and Place-8, there are 16 results to consider. It doesn't make sense to compare these two to each other, since one has 5 more results to think about.

Hope that makes sense. Also, you indicate disapproval of the 1.52% number, but it is correct for a pre-resolved-bet house edge on a place bet. You state that it doesn't agree with the article that you linked, but it is indicated in that article in at least 4 different locations, including the original question asker of that post, so you may want to re-read the article.

Thanks for taking a look...Quote:konceptum

You, and the OP of the question you linked, are trying to compare the PER-RESOLVED-BET house edges, but this doesn't make any sense. As you yourself indicated, there are different combinations involved. With a single Place-6 bet, there are only 11 results to concern yourself with, while with the combination of both Place-6 and Place-8, there are 16 results to consider. It doesn't make sense to compare these two to each other, since one has 5 more results to think about.

Hope that makes sense. Also, you indicate disapproval of the 1.52% number, but it is correct for a pre-resolved-bet house edge on a place bet. You state that it doesn't agree with the article that you linked, but it is indicated in that article in at least 4 different locations, including the original question asker of that post, so you may want to re-read the article.

Indeed, it looks as if the 16/11 decision possibilities are affecting the calculations, since a single bet on the 6 for the same amount as two bets between both the 6 & 8 end with the same loss. However, I don't understand why we can't compare the 6 & 8 together to any other single place bet. Everybody seems to accept the edges of 1.52%, 4.00%, and 6.67%. But a place bet on the 5 only has 10 decision combinations, yet we have no issues with comparing it with the place bet on the 6 which has 11, or even the 4, which has 9 decision combinations. So what's the deal?

I've come to the conclusion that the calculation of 1.04% is actually correct - in that you can say the house keeps 1.04% of the original bet every time the bet is resolved... only the bet is resolved more frequently than either of the numbers alone. So 1.04% x 16 = 1.52% x 11, resulting in the same loss.

I'd be open to criticism of any arguments I have presented.

Also, I was questioning the value of 1.527% - but I have since realized that this is probably just due to rounding.

Yep. I and many others share the same belief.Quote:konceptumI believe that the problem comes from the issue of comparing the house edge of resolved-bet vs house edge per-roll.

If you read the first entry of this WOO faq, you can see a similar calculation based on the iron cross.

All about apples to apples. (per roll vs. per bet resolved)

Same link but the 5th question down.

"Anything but 7" system:

5- place $5

6- place $6

8- place $6

field- $5

total= $22

They claim the house edge is 1.136%.

How is that possible if every individual bet made has a higher house edge?

per roll vs. per decision

What follows is an excellent explanation by the Wizard.

Now,

As to the 32 across article, same thing. not comparing apples to apples. (per every roll vs. per some rolls)

In Krigman's article he actually ignores pushes,

the horn numbers, so he really is not calculating a perfect per roll calculation.

His answer, not counting 2,3,11 and 12s that roll: =((6/30)*9 + (18/30)*7 + (6/30)*-32)/32 = -0.0125

"one-round standard deviation for $32 across is $15.82"

Counting all rolls, because they roll, and it makes session calculations much easier:

=((6/36)*9 + (18/36)*7 + (6/36)*-32 + (6/36)*0)/32 = -0.010416667

variance/stdev = 208.55555897462784 / 14.4414528

anyone can run a sim in Wincraps (after paying a $10 to $15 registration fee)

to see the HA at ~3.73% for most lengths of long term play. (about -12/322)

net/action (per bet resolved)

Here is my table to calculate the 32 place bet across house edge on a 'per bet resolved' basis.

bet | wager | win | ways to win | total | ways to lose | wager | total | net (ev) | action | edge | variance | sd |
---|---|---|---|---|---|---|---|---|---|---|---|---|

6 | $6 | $7 | 5 | $35 | 6 | -$6 | -$36 | ($1) | $66 | -0.015152 | 41.900826 | 6.473085 |

8 | $6 | $7 | 5 | $35 | 6 | -$6 | -$36 | ($1) | $66 | -0.015152 | 41.900826 | 6.473085 |

5 | $5 | $7 | 4 | $28 | 6 | -$5 | -$30 | ($2) | $50 | -0.040000 | 34.560000 | 5.878775 |

9 | $5 | $7 | 4 | $28 | 6 | -$5 | -$30 | ($2) | $50 | -0.040000 | 34.560000 | 5.878775 |

4 | $5 | $9 | 3 | $27 | 6 | -$5 | -$30 | ($3) | $45 | -0.066667 | 43.555556 | 6.599663 |

10 | $5 | $9 | 3 | $27 | 6 | -$5 | -$30 | ($3) | $45 | -0.066667 | 43.555556 | 6.599663 |

total | ($12) | $322 | -0.037267 | 240.032764 | 15.492991 |

the longer one plays it always comes back to edge*action

How did you calculate these variances and is that just how much +\- SD(1,2,3)?

Not a math student, just trying to piece the stuff together.

Thanks