Bonus Craps for Crapless Craps

I 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.

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?

Is this a progressive jackpot starting at what amount?

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: AlanMendelson

Is 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: Wizard

I 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?
link to original post

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

Quote: SOOPOO

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

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 the math is flawed.

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.

So you roll aces twice in a row you lose???

Quote: Wizard

Quote: SOOPOO

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

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.
link to original post

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

The apparently low progressive seed amount is a problem since if the ten number hits early, the player share can be less than the nine number payout.

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?)

Quote: DeMango

So you roll aces twice in a row you lose???
link to original post

Yep.

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: Wizard

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

Any agreements or disagreements on my probabilities?

link to original post

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: DJTeddyBear

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

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: DeMango

So you roll aces twice in a row you lose???
link to original post

Yep.

Quote: ThatDonGuy

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)
link to original post

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.

You've got a 96.66% chance of winning $10 or $0. Looks like a crapless one-armed bandit to me.

Quote: Wizard

Quote: ThatDonGuy

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)
link to original post

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.
link to original post

I don’t think Poisson/calculus would be an efficient method for this problem. Too many combinations to be summed

Quote: Wizard

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.

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.

My page on the Bonus Craps Progressive is up. I welcome all comments.

Good article. I see you’ve already incorporated some of my comments.


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?

One 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? A $1.00 increase for every $5 bet is going to make the breakeven number a lot different then $0.05 for every $5 bet.

Quote: redneckrounder

One 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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That type of info is rarely, if ever, made public on ANY progressive jackpot.

Easy enough to get a rough estimate though: just watch how much the meter increases after each bet is made. I could see where it may be a little tough to get an exact number, but you could get a rough estimate.