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## Poll

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**6 members have voted**

Note the rule that repeating an already rolled total is a loss, unlike Bonus Craps for regular craps games.

A bet costs $5. Here is my return table.

Numbers made | Pays | Probability | Return |
---|---|---|---|

10 | $2,335.00 | 0.000014 | 0.033372 |

9 | $1,500.00 | 0.000221 | 0.331008 |

8 | $250.00 | 0.001610 | 0.402455 |

7 | $50.00 | 0.007397 | 0.369865 |

6 | $25.00 | 0.024101 | 0.602513 |

5 | $10.00 | 0.059357 | 0.593572 |

4 or less | $- | 0.907300 | 0.000000 |

Total | 1.000000 | 2.332786 |

Based on the $2335 jackpot on 11/22/21, the expected win is $2.33. Based on a $5 bet, that is an expected return of 46.66%.

The breakeven meter is $186,619.66.

Any agreements or disagreements on my probabilities?

The question for the poll is would you make this bet?

Ironically, regular Make Em All has been won with the shooter never rolling the same number twice. It's happened to me at least twice and I've seen other shooters do it as well.

One of the "great things" about Bonus Craps is you never need to make a pass to win nor even repeat one number.

Quote:AlanMendelsonIs this a progressive jackpot starting at what amount?

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Yes, it is progressive. I do not know the reseed, but suspect it's $2000. That should not matter to the player's point of view. He should be asking himself what is the value of this bet now.

Quote:WizardI went to the Stratosphere today and noticed Bonus Craps for Crapless Craps. The object is to roll as many individual totals as you can before rolling a seven or repeating a number. Here is a picture of the pay table.

Note the rule that repeating an already rolled total is a loss, unlike Bonus Craps for regular craps games.

A bet costs $5. Here is my return table.

Points made Pays Probability Return 10 $2,335.00 0.000014 0.033372 9 $1,500.00 0.000221 0.331008 8 $250.00 0.001610 0.402455 7 $50.00 0.007397 0.369865 6 $25.00 0.024101 0.602513 5 $10.00 0.059357 0.593572 4 or less $- 0.907300 0.000000 Total 1.000000 2.332786

Based on the $2335 jackpot on 11/22/21, the expected win is $2.33. Based on a $5 bet, that is an expected return of 46.66%.

The breakeven meter is $186,619.66.

Any agreements or disagreements on my probabilities?

The question for the poll is would you make this bet?

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I thought in Vegas you need an RTP of at least 75%? Does that rule not apply to progressives?

Quote:SOOPOOI thought in Vegas you need an RTP of at least 75%? Does that rule not apply to progressives?

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I think the way that works is the overall RTP must be 75% over an infinite period of time. However, for variable-state games, like progressives, they can go below 75% temporarily.

I think you're basing it on a payout of 100% of the progressive. Read the second line of fine print on the meter display: "The top 100% award will be shared ..."

Put another way, your math is correct if there's only one participant when it hits.

I clearly remember having this same problem when developing my Poker For Roulette side bet. I *thought* I discussed it and my unique solution with you.

Because all players in the Bonus Craps Progressive are betting on the same result, and all win or lose at the same time, it has the exact same problem I had.

At the time I came up with what I thought was a logical solution. Let the meter advance the amount taken from a single participant anytime there is at least one participant. The number of participants is irrelevant. When it hits, all participants get the meter value. In other words, there is a progressive bank that is advancing faster, based on the number of participants, while the meter on the display advances as if there was never more than one participant.

In my head, the logic behind this was simple. While there are times when it hits with more participants than at other times, there are also times when there is more money going into the bank than others. That means there are times where the payout is more than what's in the bank, but also times when it's less. Over time, it will all average out.

While it was sometimes difficult to get people to understand my logic, once they did, they agreed with the concept but argued that it would be impossible to do math on it. Simulations might help, but casinos are not in business to gamble on that sort of thing.

The alternative is to advance the meter correctly based on the number of participants each time. Then when it hits, share the jackpot. The advantage is that the meter advances more quickly, but correctly shows the amount in the bank to be divided. And if it should hit when there are few participants? Good for them!

Bottom line, this progressive might create the same type of situation as the lottery. More participants when the meter is higher, making it harder for this to ever be a mathematically positive expectation. Of course, since there is a limited number of participants at any time, it doesn't have the exact same problem as the lottery.

Quote:WizardQuote:SOOPOOI thought in Vegas you need an RTP of at least 75%? Does that rule not apply to progressives?

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I think the way that works is the overall RTP must be 75% over an infinite period of time. However, for variable-state games, like progressives, they can go below 75% temporarily.

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I did a very quick perusal of

NGC regulations and was only able to find the 75% theoretical RTP mentioned in conjunction with gaming devices.

https://gaming.nv.gov/index.aspx?page=51

This bet *might* be more popular if it wasn't progressive, and players could bet any amount. (There are plenty of people who bet more than the minimum on the regular Bonus Craps, so why not here too?)

Yep.Quote:DeMangoSo you roll aces twice in a row you lose???

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And that's probably one of the reasons this bet will fail.

Another is that dealers will hate it since they have to mark, or unmark after every roll. And if they are unmarking, then they are also either paying winners or getting replacement bets from losers.

And since paying or getting replacement bets can happen many times mid-roll, it will be unpopular because of the interruption.

At least with the regular bet, that all happens before a come out roll.

----

On a different note, there's nothing about this bet that restricts it to Crapless Craps.

Similarly, there's nothing about regular Bonus Craps that restricts it to regular craps.

And why the name Bonus Craps Progressive? Other than the artwork, there's little similarity to Bonus Craps - which everyone calls All Tall Small.

Why not just call it Make 'Em All?

Quote:WizardA bet costs $5. Here is my return table.

Numbers made Pays Probability Return 10 $2,335.00 0.000014 0.033372 9 $1,500.00 0.000221 0.331008 8 $250.00 0.001610 0.402455 7 $50.00 0.007397 0.369865 6 $25.00 0.024101 0.602513 5 $10.00 0.059357 0.593572 4 or less $- 0.907300 0.000000 Total 1.000000 2.332786

Any agreements or disagreements on my probabilities?

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I get the same numbers you do. To be more specific:

Prob(10) = 4,375 / 306,110,016 (0.00001429224713770883)

Prob(9) = 33,775 / 153,055,008 (0.00022067229580622393)

Prob(8) = 985,565 / 612,220,032 (0.0016098215486029684)

Prob(7) = 9,057,545 / 1,224,440,064 (0.007397295520052497)

Prob(6) = 546,475 / 22,674,816 (0.0241005263284165)

Prob(5) = 897,275 / 15,116,544 (0.05935715200511452)

Prob(Lose) = 42,331 / 46,656 (0.9073002400548698)

Quote:DJTeddyBearPut another way, your math is correct if there's only one participant when it hits.

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I agree with that. I was aware of the jackpot sharing rule. What I posted was not the finished product, just the one-player case.

Quote:DeMangoSo you roll aces twice in a row you lose???

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Yep.

Quote:ThatDonGuyI get the same numbers you do. To be more specific:

Prob(10) = 4,375 / 306,110,016 (0.00001429224713770883)

Prob(9) = 33,775 / 153,055,008 (0.00022067229580622393)

Prob(8) = 985,565 / 612,220,032 (0.0016098215486029684)

Prob(7) = 9,057,545 / 1,224,440,064 (0.007397295520052497)

Prob(6) = 546,475 / 22,674,816 (0.0241005263284165)

Prob(5) = 897,275 / 15,116,544 (0.05935715200511452)

Prob(Lose) = 42,331 / 46,656 (0.9073002400548698)

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Thank you for the confirmation! Did you do it with calculus? I did a Markov chain and verified it only for hitting all 10 numbers with your usual method.

I didn't want to fuss with creating and solving 252 equations for all the ways to hit exactly five numbers, but maybe you know something I don't.

I don’t think Poisson/calculus would be an efficient method for this problem. Too many combinations to be summedQuote:WizardQuote:ThatDonGuyI get the same numbers you do. To be more specific:

Prob(10) = 4,375 / 306,110,016 (0.00001429224713770883)

Prob(9) = 33,775 / 153,055,008 (0.00022067229580622393)

Prob(8) = 985,565 / 612,220,032 (0.0016098215486029684)

Prob(7) = 9,057,545 / 1,224,440,064 (0.007397295520052497)

Prob(6) = 546,475 / 22,674,816 (0.0241005263284165)

Prob(5) = 897,275 / 15,116,544 (0.05935715200511452)

Prob(Lose) = 42,331 / 46,656 (0.9073002400548698)

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Thank you for the confirmation! Did you do it with calculus? I did a Markov chain and verified it only for hitting all 10 numbers with your usual method.

I didn't want to fuss with creating and solving 252 equations for all the ways to hit exactly five numbers, but maybe you know something I don't.

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Quote:WizardThank you for the confirmation! Did you do it with calculus? I did a Markov chain and verified it only for hitting all 10 numbers with your usual method.

No - I used brute force computing. Well, I suppose it's identical to a Markov chain.

However, the probability for 10 is simple; there are 10! permutations, each of which has a probability of 1 x 2 x 3 x 4 x 5 x 5 x 4 x 3 x 2 x 1 / 36^10 of occurring, so the result = 3,628,800 x 14,400 / 6^20.

First, I generated all 10! = 3,628,800 permutations of {2, 3, 4, 5, 6, 8, 9, 10, 11, 12}:

1. Start with {2, 3} and {3,2}

2. For each number from 4 to 12 (except 7), replace each previous permutation with multiple permutations where the new number is in each possible position (e.g. replace {2, 3} with {4, 2, 3}, {2, 4, 3}, and {2, 3, 4}; when all of the size-3 permutations are done, replace {4, 2, 3} with {5, 4, 2, 3}, {4, 5, 2, 3}, {4, 2, 5, 3}, and {4, 2, 3, 5}).

Then, for each 10-permutation:

1. Calculate the probability of the first 5 numbers coming up then one of those numbers or 7 (i.e. a 5-roll win).

If the first five numbers are {2, 8, 6, 10, 3}, this is 1/36 x 5/36 x 5/36 x 3/36 x 2/36 x (1 + 5 + 5 + 3 + 2 + 6)/36.

2. Do the same for the 6-roll wins, the 7-roll wins, and so on through the 10-roll wins.

3. Sum the 5-roll win probabilities, then divide by 5! as each permutation of 5 numbers appears in 5! of the 10-permutations; this is the overall probability of a 5-roll win.

4. Sum the 6-roll win probabilities, then divide by 4! as each permutation of 6 numbers appears in 4! of the 10-permutations; this is the overall probability of a 5-roll win.

5. Do the same for the 7-roll through 10-roll wins.

Intro:

It’s not Stratosphere anymore. It’s Strat. Or actually: STRAT Hotel, Casino & SkyPod

Rules paragraph 1:

The All bet should be the All/Tall/Small bet.

Maybe I am underestimating the intelligence of your readers, but I’d like to see an explanation of what the expected return per $1,000 means.

I’d also like to see a comment, or even a chart, that unless the progressive is at least $1,500 per player, it pays less than a nine hit pay out - or is there a special rule for that?

That type of info is rarely, if ever, made public on ANY progressive jackpot.Quote:redneckrounderOne thing I didn’t see mentioned is what is the meter rate, i.e. how much of your five dollar bet goes into the progressive?

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