They have the following "Bonus Craps" bets listed (20x Odds).

^{ALL SMALL OR ALL TALLAll the Small (2, 3, 4, 5, 6) numbers or all the Tall (8, 9, 10, 11, 12)numbers will be rolled before a 7.ALL OR NOTHING AT ALLAll the numbers BUT seven will be rolled before a 7. Pays 175-to-1.PETE & REPEATAny random number will be rolled twice before a 7. Pays Even Money.TWICE IS NICEAny random hardway... 2 (1-1), 4 (2-2), 6 (3-3), 8 (4-4), 10 (5-5) or 12 (6-6)... will be rolled twice before a 7. Pays 6-to-1.DIFFERENT DOUBLESAt least three different doubles will be rolled before a 7.Three Different Doubes pays 4-to-1Four Different Doubles pays 8-to-1Five Different Doubles pays 15-to-1Six Different Doubes pays 100-to-1}

I think I have learned enough from the folks here to calculate some of the more obvious bets, but some of these are beyond my ability. Out of all of them, the Pete and Repeat is the most interesting I think. If I understand it right, if ANY 2 numbers are rolled twice, before a 7 shows up, I win even money. I wonder when you make the bet? Do you have to make it on the comeout roll, and of course, you lose on the comeout 7?

I think it would be interesting just to see how they keep track of this bet.

I assume that the hardway bets are the ones you can calculate yourself, but I'll share my answers for reference.

DIFFERENT DOUBLES

The house edge is 27.9%, as shown.

Different doubles | Pays | Exact | Probability | Return |
---|---|---|---|---|

0 | -1 | 1/2 | 0.5 | -0.5 |

1 | -1 | 3/11 | 0.27272727 | -0.27272727 |

2 | -1 | 3/22 | 0.13636364 | -0.13636364 |

3 | 4 | 2/33 | 0.06060606 | 0.24242424 |

4 | 8 | 1/44 | 0.02272727 | 0.18181818 |

5 | 15 | 1/154 | 0.00649351 | 0.09740260 |

6 | 100 | 1/924 | 0.00108225 | 0.10822511 |

Total | 1 | 1 | -0.27922078 |

TWICE IS NICE

The probability of winning is 0.10086323, or 4183/41472 exactly, for a house edge of 29.4% when it pays 6-to-1.

The first three bets would be inconvenient to analyze without some type of programmable computer. They would require considering at least 3^5 = 243 possible states, although with a computer it is most convenient to analyze 2^10 possible states (treating the numbers in each of the five pairs that have the same probability separately rather than together). It is a nice exercise in dynamic programming.

ALL SMALL OR ALL TALL

The probability of winning is 0.04745011, or 8586422437/180956833200 exactly. The house edge depends on the payout.

ALL OR NOTHING AT ALL

The probability of winning is 0.00525770, or 126538525259/24067258815600 exactly, for a house edge of 7.46% when it pays 175-to-1.

PETE & REPEAT

The probability of winning is 0.47106649, or 1730378051/3673320192 exactly, for a house edge of 5.79% when it pays even money.

It is also highly worth looking at the Wizard's craps side bets page, which lists the All or Nothing at All bet as the All bet offered at Sam's Town (note that that bet pays 175-for-1, or 174-to-1, rather than 175-to-1, which could just be an error on some casino's web page, but this has no bearing on the probability of winning, only the return).

His page also lists the Small bet and the Tall bet, and together with the All bet we can calculate the All Small or All Tall bet from the Wizard's numbers. Let S, T, and A be the events of winning these bets.

P(S or T) = P(S) + P(T) - P(S and T)

However, winning both the Small and Tall bets (both placed at the same time) has the same probability as winning the All bet, so:

P(S or T) = P(S) + P(T) - P(A) = 2 * 0.026354 - 0.005258 = 0.04745

As for the Pete and Repeat bet, you'll just have to trust that I know what I'm doing. I'm willing to explain further, but I fear it is of limited interest except to computer programmers.

Sometimes I cover the hardways at $5 apiece because when I win three times before the 7-out it's a thrill. I figure you could get the same kind of thrill by throwing $1 at this long bet, at a lower house edge than hardways.

Pete and Repeat doesn't seem like that much fun, as there are a lot of ways to get even money or nearly even money bets on the craps table. Why pay a premium house edge for such a low return. Just put more money on your odds.

Generally, I think it is OK to pay a higher house edge to get more of a thrill out of a big return. But in general just remember No side bet was ever invented to help the player!.

MS has pretty favorable craps rules, so I am sure it is most likely 175 to 1.

Yeah, giving up 5.7% for an even money bet is not very interesting. As you noted, there are lots of other ways to achieve even money and not give up that much HA.

Quote:pacomartinI think the all or nothing is a fun bet. The way I figure it, an 8% house advantage is not a big price to pay for that kind of return on a dollar bet.

Sometimes I cover the hardways at $5 apiece because when I win three times before the 7-out it's a thrill. I figure you could get the same kind of thrill by throwing $1 at this long bet, at a lower house edge than hardways.

Pete and Repeat doesn't seem like that much fun, as there are a lot of ways to get even money or nearly even money bets on the craps table. Why pay a premium house edge for such a low return. Just put more money on your odds.

Generally, I think it is OK to pay a higher house edge to get more of a thrill out of a big return. But in general just remember No side bet was ever invented to help the player!.

Yeah paco, I agree that the All bets could be interesting to throw a few bucks at it. I rarely play the hardways, so I may throw those few bets towards this bet and see what happens. You are right, hitting that bet would be fun times.

It seems like the rule is no side bet can be created unless it produces a double digit HA.

This is another bet I just heard of so I had to calculate it.Quote:ZPP

As for the Pete and Repeat bet, you'll just have to trust that I know what I'm doing. I'm willing to explain further, but I fear it is of limited interest except to computer programmers.

I'm not a programmer so I used Excel. It's a matter of listing all possible combinations (1,024) that you can roll without a resolution, and then win. From there it's just multiplying out and summing all the possibilities. This is similar to the way I calculated the Fire Bet, though for that one the order of points made does matter for calculation purposes.

One question I have is how you came up with the fraction of 1730378051/3673320192.

My number in Excel is 0.471066490410645, which shows no sign of repeating at max digits allowed.

You forgot the word 'financially.'Quote:pacomartinGenerally, I think it is OK to pay a higher house edge to get more of a thrill out of a big return. But in general just remember No side bet was ever invented to help the player!.

Every side bet, and game for that matter, is designed to seperate a player from his bankroll.

The successful ones do it by helping the player enjoy the process.

((x/36+1) * (x/18+1) * (x/12+1) * (x/9+1) * (5x/36+1))^2 / (6 * e^x)

Then taking the complement gives 1730378051/3673320192 or about 0.471

Quote:Ace2The Pete and Repeat bet can also be calculated by integrating the function:

((x/36+1) * (x/18+1) * (x/12+1) * (x/9+1) * (5x/36+1))^2 / (6 * e^x)

Then taking the complement gives 1730378051/3673320192 or about 0.471

Can you please expand a bit on why this works?

I must have reduced that formula. Here is a formula in expanded form:Quote:WizardCan you please expand a bit on why this works?

(((x/36)/e^(x/36)+1/e^(x/36)) * ((x/18)/e^(x/18)+1/e^(x/18)) * ((x/12)/e^(x/12)+1/e^(x/12)) * ((x/9)/e^(x/9)+1/e^(x/9)) * ((x*5/36)/e^(x*5/36)+1/e^(x*5/36)))^2 * 1/e^(x/6) * 1/6 dx

In this case it's much easier to calculate the chance of losing the bet since there is only one way (7) to lose and ten ways (pair of 2,3,4,5,6,8,9,10,11,12) to win.

We lose whenever a seven is thrown before a pair is rolled. For example, a 3 will be rolled every 36/2 = 18 rolls. Therefore the chance of it being rolled once is (x/18)/e^(x/18) and the chance of it being rolled zero times is 1/e^(x/18). The chance of it being rolled less than twice (no pair) is the sum of those: ((x/18)/e^(x/18)+1/e^(x/18)). Do that for all numbers 2-6, take the product, then square it since 8-12 have the same average waiting times. The term 1/e^(x/6) is the probability of 7 being rolled zero times.

The integral sums the probability over all time of being in the following state: numbers 2,3,4,5,6,8,9,10,11,12 have been rolled less than twice, a 7 has not been rolled, and then a 7 is rolled (multiply by 1/6 probability) to lose the bet.

The chance of losing is 1942942141 / 3673320192 =~ 0.529

(exp(-x/36)*(1+x/36))^2*(exp(-x/18)*(1+x/18))^2*(exp(-x/12)*(1+x/12))^2*(exp(-x/9)*(1+x/9))^2*(exp(-5x/36)*(1+5x/36))^2*exp(-x/6)/6

Integrating from 0 to infinity, the probability of no winner is 1942942141 / 3673320192 = 0.5289335095893541.

This is the actual integral, before inserting the bounds of integration: (25*(-x^10-10*x^9-90*x^8-720*x^7-5040*x^6-30240*x^5-151200*x^4-604800*x^3-1814400*x^2-3628800*x-3628800)*e^(-x)+4110*(-x^9-9*x^8-72*x^7-504*x^6-3024*x^5-15120*x^4-60480*x^3-181440*x^2-362880*x-362880)*e^(-x)+290421*(-x^8-8*x^7-56*x^6-336*x^5-1680*x^4-6720*x^3-20160*x^2-40320*x-40320)*e^(-x)+11639700*(-x^7-7*x^6-42*x^5-210*x^4-840*x^3-2520*x^2-5040*x-5040)*e^(-x)+293947380*(-x^6-6*x^5-30*x^4-120*x^3-360*x^2-720*x-720)*e^(-x)+4903428960*(-x^5-5*x^4-20*x^3-60*x^2-120*x-120)*e^(-x)+54884392128*(-x^4-4*x^3-12*x^2-24*x-24)*e^(-x)+408146688000*(-x^3-3*x^2-6*x-6)*e^(-x)+1934615301120*(-x^2-2*x-2)*e^(-x)+5289581076480*(-x-1)*e^(-x)-6347497291776*e^(-x))/38084983750656.

Thanks again Ace!