DJTeddyBear
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GialmereMrCasinoGames
July 27th, 2021 at 6:12:51 PM permalink
I saw this on Twitter:
Quote:

O'Sheas has a new table game that I have never seen before. If I had to guess, I think it probably plays like sic bo. If anyone out there has played it, please let me know its concept.




https://twitter.com/john_mehaffey/status/1420114269199273991?s=21
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
DJTeddyBear
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MrCasinoGames
July 27th, 2021 at 6:17:30 PM permalink
I have two thoughts:

If it痴 just the repeater bet without the other craps bets, why are there seven 7 spots?

The large layout leaves little room for the players to put their bankroll. And it痴 so large, it痴 ripe for past posting.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Gialmere
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MrCasinoGames
July 27th, 2021 at 6:23:14 PM permalink
I saw this at the Nevada gaming site which shows it on trial at the Linq. I couldn't find anything on it though and assumed it was an updated version of the craps side bet.

But looky here, it's a full blown table game. I wonder how the dice are rolled. A cup?
Have you tried 22 tonight? I said 22.
charliepatrick
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MrCasinoGames
July 28th, 2021 at 8:49:13 AM permalink
Just looking at the 2/12 before 7 (3 for 1) and 3/11 before7 (2 for 1 - i.e. Evens) is terrible House Edge. I can see it as something that would work online, where once the next "game" starts and the rolls can be made automatically, fairly quickly, until the "7 -out"; but the odds need to be better.
mwalz9
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MrCasinoGames
July 28th, 2021 at 8:59:59 AM permalink
Quote: charliepatrick

Just looking at the 2/12 before 7 (3 for 1) and 3/11 before7 (2 for 1 - i.e. Evens) is terrible House Edge. I can see it as something that would work online, where once the next "game" starts and the rolls can be made automatically, fairly quickly, until the "7 -out"; but the odds need to be better.



I've noticed most patrons in the casinos we frequent don't understand house edge, the math behind it or what is a good or bad bet. People will still bet the 2 or 12 as it is the required repeater the fewest times and pays 3:1.

You have no idea how many times I've had people tell me I am stupid for laying max odds on the Don't in craps because I have to lay $2 to win just $1 and that is a bad bet. In reality, people like us know it is the best bet in the casino.

For instance, if there was a craps game where you only had to lay 5/4 on the 4 or 10 in craps, people like us would play that game all day every day, we'd try to get as much action on it as we could. There would still be drunk idiots who have no understanding of math or gambling tell us it's a bad bet!
ChallengedMilly
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MrCasinoGames
July 28th, 2021 at 10:51:11 AM permalink
It is interesting to see this kind of a game design though. I could easily see a "repeater" gimmick using craps fundamentals as the base for such a game. People tend to want to believe things happen in a pattern, and the casino is probably one of the hyper aware places that you see that kind of mentality.
DeMango
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Johnzimbo
July 28th, 2021 at 1:10:17 PM permalink
How much money for 18 yo's?
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
ksdjdj
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July 29th, 2021 at 12:46:24 AM permalink
Quote: DeMango

How much money for 18 yo's?


Probably 54 to 55 billion for one, if they offered a similar bet using the " Sweeter with the Repeater" *** game table.

***: Must hit 18 yo's before a 7
Last edited by: ksdjdj on Jul 29, 2021
Wizard
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MrCasinoGames
August 4th, 2021 at 5:41:10 PM permalink
Sorry for the late arrival to this thread.

So far I have only the images earlier to go off of. I can't see the four bets in red on the left side of the table and none of the surrender values. Hopefully I'll have a chance to visit the Linq soon to get all the rules.

This includes the Repeater bets as already found at some casinos and covered in my page on Repeater Bets.

Here is my analysis of them.

Bet Rolls Pays Probability House edge
2 2 40 0.020408 0.183673
3 3 50 0.015625 0.218750
4 4 65 0.012346 0.197531
5 5 80 0.010240 0.180800
6 6 90 0.008820 0.206209
8 6 90 0.008820 0.206209
9 5 80 0.010240 0.180800
10 4 65 0.012346 0.197531
11 3 50 0.015625 0.218750
12 2 40 0.020408 0.183673


They add some horn type bets that win if any one of a group of bets win. The Even bet wins if any of the even bets (2, 4, 6, 8, 10, or 12) win and pays 12 for 1.

Likewise, the odd bet pays 18 for 1.

A Horn bet pays 12 for 1 and wins if the 2, 3, 11, or 12 bet win.

Much easier mathematically are the following bets:

3 or 11: Wins if a 3 or 11 is thrown before a 7. Pays 2 for 1.
2-3-11-12 (7x): Wins if the shooter throws any horn number 7 times before a total of 7.

You can't just add the probabilities to get a probability of winning, because they are correlated. A correct closed-form answer is pretty trick and worthy of discussion. I plan to bring it up in the Easy Math Puzzles page once the current problem is satisfactorily answered.

The spoiler button below shows my analysis of those.


Bet Pays Probability House edge
Even 12 0.065395 0.215265
Odd 18 0.043176 0.222839
3 or 11 2 0.400000 0.200000
Horn 12 0.062789 0.246528
2-3-11-12 100 0.007813 0.218750



If anyone can save me the trouble and give a full report on the game, please do!
Last edited by: Wizard on Aug 4, 2021
摘xtraordinary claims require extraordinary evidence. -- Carl Sagan
Wizard
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MrCasinoGames
August 6th, 2021 at 8:07:50 PM permalink
Please be the first to see my new page on Repeater Bets Plus.

As always, I welcome all comments, questions, and corrections.
摘xtraordinary claims require extraordinary evidence. -- Carl Sagan
DJTeddyBear
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August 7th, 2021 at 3:52:37 AM permalink
Table 3 is screwed up. You have bet 3 and 4 paying the same and with the same probability and value. The rest of the table is similarly misaligned.

I realized this when I was checking to see which bet had the lowest edge - you uncharacteristically did not include a summary where that would be mentioned.

Otherwise, it was a good write up on a very crappy bet.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
AlanMendelson
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August 7th, 2021 at 4:04:47 AM permalink
Can we see a wide shot of the table? I'm curious to see where the dice are rolled?

Are they rolled on a traditional table and must hit the back Wall or rolled in a cage or what?
DJTeddyBear
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August 7th, 2021 at 6:46:46 AM permalink
The Twitter thread speculated that they are handled by the dealer, and 喪olled sic bo style.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Wizard
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August 7th, 2021 at 2:16:27 PM permalink
Quote: DJTeddyBear

Table 3 is screwed up. You have bet 3 and 4 paying the same and with the same probability and value. The rest of the table is similarly misaligned.



Thank you for that correction. I just fixed it.

Quote: AlanMendelson

Can we see a wide shot of the table? I'm curious to see where the dice are rolled?

Are they rolled on a traditional table and must hit the back Wall or rolled in a cage or what?



There are a couple pictures in the OP. It seems to be played on a blackjack-size table or maybe a little bigger. I'd lay long odds they use small dice and a shaker, like in pai gow.
摘xtraordinary claims require extraordinary evidence. -- Carl Sagan
AlanMendelson
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August 7th, 2021 at 4:24:36 PM permalink
Quote: Wizard

There are a couple pictures in the OP. It seems to be played on a blackjack-size table or maybe a little bigger. I'd lay long odds they use small dice and a shaker, like in pai gow.



Many years ago the Normandie Casino near LA (a poker casino) had card craps with a layout on a blackjack-sized table.

It was not a craps layout, but a layout with boxes to place bets as if you were at a craps table.

The dealer shuffled and chose from two decks of cards.

At the Bicycle Casino near LA when they had a craps table, players would not throw the dice but would shake two dice in a metal cup and then slam the cup on the table. The table was an actual craps table.
Ace2
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August 8th, 2021 at 9:52:20 AM permalink
Assume there's a fire type bet on the repeater table that pays as follows:

Win all ten repeater bets before rolling a 7: 10000 to 1
Win nine repeater bets before rolling a 7: 3000 to 1
Win eight repeater bets before rolling a 7: 1000 to 1
Win less than eight repeater bets before rolling a 7: Loss

What is the house edge?

Like the fire bet, only distinct wins are counted. So, for instance, if you win the 2 three times, that only counts as one win.
It痴 all about making that GTA
ThatDonGuy
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August 9th, 2021 at 6:41:00 AM permalink
Quote: Ace2

Assume there's a fire type bet on the repeater table that pays as follows:

Win all ten repeater bets before rolling a 7: 10000 to 1
Win nine repeater bets before rolling a 7: 3000 to 1
Win eight repeater bets before rolling a 7: 1000 to 1
Win less than eight repeater bets before rolling a 7: Loss

What is the house edge?

Like the fire bet, only distinct wins are counted. So, for instance, if you win the 2 three times, that only counts as one win.



HE = 1 - (53,324,716,837,842,948,521,834,101,764,952,860,337
,479,794,340,838,976,031,734,796,259,597,564,137,929,772,388
,502,720,082,602,106,738,570,389,723,014,763,695,730,257,656
,379,420,034,342,836,497,185,909,061,392,328,048,022,080,590
,390,489,749,983,045,954,828,817,245,768,988,934,637,454,950
,168,370,663,075,325,965,413,107,747,452,805,343,706,609,926
,459,970,444,551,250,523,392,073,018,569,576,096,269,676,229
/
233,473,783,448,792,965,933,524,695,321,952,589,528
,152,298,183,736,457,939,286,839,045,858,409,228,221,650,293
,884,563,811,194,059,089,436,386,958,994,014,107,363,325,671
,964,352,353,886,065,739,979,792,097,652,058,703,112,038,008
,000,165,195,476,192,011,093,871,027,070,114,747,396,048,797
,139,901,142,504,119,068,142,492,631,715,860,366,963,419,710
,442,438,656,000,000,000,000,000,000,000,000,000,000,000,000)
or about 77.1603%

Here is what I think is the Poisson-based method (the above answer is Markov-based):
Define:
P2 = P(2 rolled < 2 times) = e^(-x/36) * (1 + x/36)
P3 = P(3 rolled < 3 times) = e^(-x/18) * (1 + x/18 + (x/18)^2 / 2)
P4 = P(4 rolled < 4 times) = e^(-x/12) * (1 + x/12 + (x/12)^2 / 2 + (x/12)^3 / 6)
P5 = P(5 rolled < 5 times) = e^(-x/9) * (1 + x/9 + (x/9)^2 / 2 + (x/9)^3 / 6 + (x/9)^4 / 24)
P6 = P(6 rolled < 6 times) = e^(-5x/36) * (1 + 5x/36 + (5x/36)^2 / 2 + (5x/36)^3 / 6 + (5x/36)^4 / 24 + (5x/36)^5 / 120)
P7 = P(7 not rolled) = e^(-x/6)

10 wins:
((1 - P2)(1 - P3)(1 - P4)(1 - P5)(1 - P6))^2 * P7

9 wins:
2 P2 (1 - P2) (1 - P3)^2 (1 - P4)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
2 P3 (1 - P3) (1 - P2)^2 (1 - P4)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
2 P4 (1 - P4) (1 - P2)^2 (1 - P3)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
2 P5 (1 - P5) (1 - P2)^2 (1 - P3)^2 (1 - P4)^2 (1 - P6)^2 * P7 / 6
2 P6 (1 - P6) (1 - P2)^2 (1 - P3)^2 (1 - P4)^2 (1 - P5)^2 * P7 / 6

8 wins:
4 P2 (1 - P2) P3 (1 - P3) (1 - P4)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
4 P2 (1 - P2) P4 (1 - P4) (1 - P3)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
4 P2 (1 - P2) P5 (1 - P5) (1 - P3)^2 (1 - P4)^2 (1 - P6)^2 * P7 / 6
4 P2 (1 - P2) P6 (1 - P6) (1 - P3)^2 (1 - P4)^2 (1 - P5)^2 * P7 / 6
4 P3 (1 - P3) P4 (1 - P4) (1 - P2)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
4 P3 (1 - P3) P5 (1 - P5) (1 - P2)^2 (1 - P4)^2 (1 - P6)^2 * P7 / 6
4 P3 (1 - P3) P6 (1 - P6) (1 - P2)^2 (1 - P4)^2 (1 - P5)^2 * P7 / 6
4 P4 (1 - P4) P5 (1 - P5) (1 - P2)^2 (1 - P3)^2 (1 - P6)^2 * P7 / 6
4 P4 (1 - P4) P6 (1 - P6) (1 - P2)^2 (1 - P3)^2 (1 - P5)^2 * P7 / 6
4 P5 (1 - P5) P6 (1 - P6) (1 - P2)^2 (1 - P3)^2 (1 - P4)^2 * P7 / 6
(P2 (1 - P3)(1 - P4)(1 - P5)(1 - P6))^2 * P7 / 6
(P3 (1 - P2)(1 - P4)(1 - P5)(1 - P6))^2 * P7 / 6
(P4 (1 - P2)(1 - P3)(1 - P5)(1 - P6))^2 * P7 / 6
(P5 (1 - P2)(1 - P3)(1 - P4)(1 - P6))^2 * P7 / 6
(P6 (1 - P2)(1 - P3)(1 - P4)(1 - P5))^2 * P7 / 6

Multiply the 10-wins value by 10,001
Sum the 9-wins values, and multiply by 3001
Sum the 8-wins values, and multiply by 1001
Add these three values together to get the expected return; HE = 1 - ER

Ace2
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August 11th, 2021 at 5:08:12 PM permalink
Quote: ThatDonGuy


HE = 1 - (53,324,716,837,842,948,521,834,101,764,952,860,337
,479,794,340,838,976,031,734,796,259,597,564,137,929,772,388
,502,720,082,602,106,738,570,389,723,014,763,695,730,257,656
,379,420,034,342,836,497,185,909,061,392,328,048,022,080,590
,390,489,749,983,045,954,828,817,245,768,988,934,637,454,950
,168,370,663,075,325,965,413,107,747,452,805,343,706,609,926
,459,970,444,551,250,523,392,073,018,569,576,096,269,676,229
/
233,473,783,448,792,965,933,524,695,321,952,589,528
,152,298,183,736,457,939,286,839,045,858,409,228,221,650,293
,884,563,811,194,059,089,436,386,958,994,014,107,363,325,671
,964,352,353,886,065,739,979,792,097,652,058,703,112,038,008
,000,165,195,476,192,011,093,871,027,070,114,747,396,048,797
,139,901,142,504,119,068,142,492,631,715,860,366,963,419,710
,442,438,656,000,000,000,000,000,000,000,000,000,000,000,000)
or about 77.1603%

Here is what I think is the Poisson-based method (the above answer is Markov-based):
Define:
P2 = P(2 rolled < 2 times) = e^(-x/36) * (1 + x/36)
P3 = P(3 rolled < 3 times) = e^(-x/18) * (1 + x/18 + (x/18)^2 / 2)
P4 = P(4 rolled < 4 times) = e^(-x/12) * (1 + x/12 + (x/12)^2 / 2 + (x/12)^3 / 6)
P5 = P(5 rolled < 5 times) = e^(-x/9) * (1 + x/9 + (x/9)^2 / 2 + (x/9)^3 / 6 + (x/9)^4 / 24)
P6 = P(6 rolled < 6 times) = e^(-5x/36) * (1 + 5x/36 + (5x/36)^2 / 2 + (5x/36)^3 / 6 + (5x/36)^4 / 24 + (5x/36)^5 / 120)
P7 = P(7 not rolled) = e^(-x/6)

10 wins:
((1 - P2)(1 - P3)(1 - P4)(1 - P5)(1 - P6))^2 * P7

9 wins:
2 P2 (1 - P2) (1 - P3)^2 (1 - P4)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
2 P3 (1 - P3) (1 - P2)^2 (1 - P4)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
2 P4 (1 - P4) (1 - P2)^2 (1 - P3)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
2 P5 (1 - P5) (1 - P2)^2 (1 - P3)^2 (1 - P4)^2 (1 - P6)^2 * P7 / 6
2 P6 (1 - P6) (1 - P2)^2 (1 - P3)^2 (1 - P4)^2 (1 - P5)^2 * P7 / 6

8 wins:
4 P2 (1 - P2) P3 (1 - P3) (1 - P4)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
4 P2 (1 - P2) P4 (1 - P4) (1 - P3)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
4 P2 (1 - P2) P5 (1 - P5) (1 - P3)^2 (1 - P4)^2 (1 - P6)^2 * P7 / 6
4 P2 (1 - P2) P6 (1 - P6) (1 - P3)^2 (1 - P4)^2 (1 - P5)^2 * P7 / 6
4 P3 (1 - P3) P4 (1 - P4) (1 - P2)^2 (1 - P5)^2 (1 - P6)^2 * P7 / 6
4 P3 (1 - P3) P5 (1 - P5) (1 - P2)^2 (1 - P4)^2 (1 - P6)^2 * P7 / 6
4 P3 (1 - P3) P6 (1 - P6) (1 - P2)^2 (1 - P4)^2 (1 - P5)^2 * P7 / 6
4 P4 (1 - P4) P5 (1 - P5) (1 - P2)^2 (1 - P3)^2 (1 - P6)^2 * P7 / 6
4 P4 (1 - P4) P6 (1 - P6) (1 - P2)^2 (1 - P3)^2 (1 - P5)^2 * P7 / 6
4 P5 (1 - P5) P6 (1 - P6) (1 - P2)^2 (1 - P3)^2 (1 - P4)^2 * P7 / 6
(P2 (1 - P3)(1 - P4)(1 - P5)(1 - P6))^2 * P7 / 6
(P3 (1 - P2)(1 - P4)(1 - P5)(1 - P6))^2 * P7 / 6
(P4 (1 - P2)(1 - P3)(1 - P5)(1 - P6))^2 * P7 / 6
(P5 (1 - P2)(1 - P3)(1 - P4)(1 - P6))^2 * P7 / 6
(P6 (1 - P2)(1 - P3)(1 - P4)(1 - P5))^2 * P7 / 6

Multiply the 10-wins value by 10,001
Sum the 9-wins values, and multiply by 3001
Sum the 8-wins values, and multiply by 1001
Add these three values together to get the expected return; HE = 1 - ER

That is the correct answer !
It痴 all about making that GTA
ThatDonGuy
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August 11th, 2021 at 5:33:10 PM permalink
Quote: Ace2

That is the correct answer !


Well, the answer is right, but I think one of the Poisson values is wrong.

I think the "10 wins" value should be 1 - (((1 - P2)(1 - P3)(1 - P4)(1 - P5)(1 - P6))^2 * P7 / 6).

Why wouldn't the original one be right? Isn't that the probability of making all 10 numbers along with the probability of not making 7 in time x?
Ace2
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August 11th, 2021 at 7:15:15 PM permalink
I don稚 understand the question
It痴 all about making that GTA
ThatDonGuy
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August 12th, 2021 at 4:55:26 PM permalink
Quote: Ace2

I don稚 understand the question


Within a time period of duration x:
The probability of rolling at least two 2s is 1 - (e^(-x/36) * (1 + 1/36))
The probability of rolling at least three 3s is 1 - (e^(-x/18) * (1 + 1/18 + (1/18)^2 / 2))
...
The probability of rolling at least three 11s is 1 - (e^(-x/18) * (1 + 1/18 + (1/18)^2 / 2))
The probability of rolling at least two 12s is 1 - (e^(-x/36) * (1 + 1/36))
The probability of not rolling a 7 is e^(-x/6)
You would think the probability of doing all 11, thus winning the "repeater fire bet", would be the integral over x = 0 to positive infinity of the product of these 11 values, but it's not.
Ace2
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August 13th, 2021 at 7:11:04 PM permalink
For example, the probability of rolling at least three 3s is (1 - ((x/18)^2/2 + (x/18) + 1) * e^(-x/18))

Take the product of such probabilities for all ten points times the probability of zero sevens times 1/6 to get:

((1 - ((5x/36)^5/120 + (5x/36)^4/24 + (5x/36)^3/6 + (5x/36)^2/2 + (5x/36) + 1) * e^(-5x/36)) * (1 - ((x/9)^4/24 + (x/9)^3/6 + (x/9)^2/2 + (x/9) + 1) * e^(-x/9)) * (1 - ((x/12)^3/6 + (x/12)^2/2 + (x/12) + 1) * e^(-x/12)) * (1 - ((x/18)^2/2 + (x/18) + 1) * e^(-x/18)) * (1 - ((x/36) + 1) * e^(-x/36)))^2 * e^(-x/6) * 1/6

Plug that into the integral calculator to get the exact answer shown below. This formula is slightly more efficient than the one I posted on easy math puzzles since it gets you the answer directly (instead of the complement of the answer). You multiply by 1/6 to avoid overcounting.

228296650211142223842235175926562819565310943733017295518075498704612736041415500474174790373335885790742403815919759376329015547908450301690458558920417797769549693735380351101698849553401942470326707347462797442543279964206576929140810851250499537038847980599155971380821671855091570073172919198633489 /

42257698361772482521904922230217663263013990621490761617970468605585232439497131274911233268994399835192106236921993503595172067323864679436391989136613954326164471151500091945731257099763259926492493587351085501547293599982107240564537220360657324137622979804472478269440000000000000000000000000000000000000

=~ 1 in 185,000
It痴 all about making that GTA
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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August 13th, 2021 at 8:07:35 PM permalink
The only thing I don't understand about that is, why multiply by 1/6? What is being "overcounted"?
Ace2
Ace2 
Joined: Oct 2, 2017
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August 13th, 2021 at 9:34:27 PM permalink
This integral is summing the probabilities for all strings that have won all ten repeater bets and have zero sevens for all time values (x). To avoid overcounting you must end each winning sting with a seven (1/6 chance)

If you didn稚 end with a seven, then a winning string at, for instance, x=8 can get counted again at x=9, since some winning strings at x=9 were already in a winning state at x=8 (and maybe at x= 7.2564897567 and at x= 6.2 etc). Ending with a seven ensures a winning string gets counted only once
It痴 all about making that GTA
daveyandersen1
daveyandersen1
Joined: Mar 27, 2015
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November 29th, 2021 at 6:52:00 PM permalink
Quote: DeMango

How much money for 18 yo's? now thats funny
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