I just got home from there! I will play it next timeQuote:LuckymanI am excited to announce the launching of the new game, "LUCKY 9", at Harrah's Rincon Casino in San Diego California. Abuan Gaming, LLC (exclusive distributor of Lucky 9) recently signed a 3 year contract with Caesars Entertainment Operating Company, Inc., the game will go LIVE on June 26, 2012 (Tuesday) at 11:00 am. For more information, please check out www.abuangaming.com.

Well, good things are worth waiting for !

CONGRATULATIONS

Good luck, it will be interesting to see if this type of baccarat derivative will broaden the appeal of traditional baccarat.

Does the dealer only draw one more card?

Good luck. Is this an Indian casino? To have a new table game in an Indian casino, CA, you need to have the State approve it first, then the tribe has to approve.Quote:LuckymanI greatly appreciate all your comments and support. Thank You.

Quote:LuckymanI greatly appreciate all your comments and support. Thank You.

Luckyman you may want to check out the site discountgambling.net There are some reviews of your game at rincon.

Quote:HunterhillLuckyman you may want to check out the site discountgambling.net There are some reviews of your game at rincon.

I have not had the opportunity to play the game, given that I am nowhere near Rincon, but having perused the Review at discountgambling.net, I strongly disagree with the review.

My first problem is that he talks about dealer confusion in resolving hands involving Lucky 9. In my opinion, these complaints are a load of garbage. If dealers are capable of figuring out correct payouts in games such as Craps and Roulette as well as the correct payouts for many side bets in other popular casino games, then I cannot see what is so confusing about:

Lucky 9 v. Any dealer two-card 9 = Push

Lucky 9 (No Split) v. Anything else = 3:2

Lucky 9 (After split) v. Dealer 9 = Push

Lucky 9 (After split) v. Anything else = 1:1

All other Player 9's Push against all other Dealer 9's, except a dealer Lucky 9 beats a player three-card 9.

---It's really not that difficult. If I had to choose between learning that and all of the exotic Roulette bets people can make, I'm going to take Lucky 9.

I'm also having trouble with, "There is no value to counting this game." If you look at the result of, "Lucky 9," by itself, then wouldn't knowing the shoe has six tens less than it should be enough to compel someone to bet more for an increased chance of Lucky 9 which pays 3:2 against everything except a dealer two-card nine, thereby resulting in a reduced HE? This must be the case because the only two card values that preclude a player from getting Lucky 9 on a, "First card out," basis are 9's and 10's.

Furthermore, on any hand where a player would take a hit, a ten inherently makes the player's hand neither better nor worse, so an absence of tens, it would seem, would give the player a greater opportunity to improve any two-card hand that would normally be hit, especially in the event that the two cards equal ten (which is zero).

Am I incorrect in assuming that knowing there are less tens in the deck provides an advantage?

It seems to me that the game should play pretty fast when you have a dealer used to dealing it, so the result of that could be some low table minimums (even with the low HE) as well as a game with plenty of excitement. It's tough to imagine that, with a skilled dealer, this game would not be moderately even faster than Blackjack.

All things considered, I would give this game an 8/10, but that's having never played it. I'd really like to play it, though, and get a feel for the speed of the game.

What about 27, 63, or 54 ?

Quote:Hunterhill"This must be the case because the only two card values that preclude a player from getting Lucky 9 on a, "First card out," basis are 9's and 10's. "

What about 27, 63, or 54 ?

That's exactly what I am saying.

A-8-First Card-Can Still Get Lucky 9

9-10-First Card-Can No Longer Get Lucky 9

If you know there are less 10's in the deck than there should be, then you know that Lucky 9 is a more likely result. You also know that you are more likely to improve your hand with a hit, rather than the hit keeping your hand the same.

Quote:Mission146I'm also having trouble with, "There is no value to counting this game."

If Stephen says the EORs are too low, I believe him. I don't care enough about the game to try to come up with an explanation for why the EORs are low. Stick to BJ or Spanish.

Quote:dwheatleyIf Stephen says the EORs are too low, I believe him. I don't care enough about the game to try to come up with an explanation for why the EORs are low. Stick to BJ or Spanish.

He said that there was, "No value to counting this game," that's the part that I disagree with. If you can even lessen the HE, then there's value there, as far as I am concerned. I just misinterpreted his statement. If he had said that, "Counting at this game will not result in an opportunity to bet at an edge for the player," then I may not have contested that. It seems to me that you are saying that is what he meant, so I withdraw my earlier statements and admit to being wrong for misinterpreting him.

You have eight values as, "First Card Out," where Lucky 9 is a possible result.

8 * 2 (Decks) * 4 (Cards per Value/Deck) = 64

52 Cards/Deck * 2 Decks = 104 Cards (Obviously)

If A-8, then Eight possible cards to make, "Lucky 9."

64/104 * 8/103 * $15 = x

.615384 * .07767 * $15 = x

.047797 * $15 = x

$0.72 = x (If dealer does not have, "Natural," 9)

EV of Lucky Nine-Two Decks, Six Less Tens, $10 Bet (Pays $15)

70/104 * 8/103 (This is assuming that there are not more of the second card needed to go with, "First Card Out.")

.6731 * .07767 * $15 = x

.05228 * $15 = x

$0.78 = x (If dealer does not have, "Natural," 9)

***

If the HE on the game is 1.18%, then the EV of a $10 bet is $10 -.118, or $9.88. In the event that the dealer does not have a, "Natural 9," we can see that six less tens in the deck mitigates the Negative EV by $0.06, thereby increasing it to an EV of $9.94 on a $10 bet.

I would actually have to go through each possibility individually to see what other effects a count of six less tens would have on the game. For example, as I have stated, it would decrease the liklihood that you would hit on a total of 0, 1 or 2 and still end up with a hand that automatically loses. It would also decrease the liklihood of hitting on other totals and failing to improve the hand.

It would take an analysis of every possible hand to come up with a true EV for the bet, though. There is one thing that interests me, though:

Odds of Not Improving 0, 1 or 2 to playable hand, Two Decks (Remaining after starting hands), Perfectly Composed

Total: 0

Does Not Play: Ten, Ace, Deuce

48/104 = 46.15%

Total: 1

Does Not Play: Nine, Ten, Ace

48/104 = 46.15%

Total: 2

Does Not Play: Ten, Nine, Eight

48/104 = 46.15%

Odds of Not Improving 0, 1 or 2 to playable hand, Two Decks (Remaining after starting hands), Six Less Tens

---I am going to assume that the composition of A-2's and 8-9's stays perfect for these purposes, but it probably wouldn't.---

Total: 0

Does Not Play: Ten, Ace, Deuce

42/104 = 40.38%

Total: 1

Does Not Play: Nine, Ten, Ace

42/104 = 40.38%

Total: 2

Does Not Play: Ten, Nine, Eight

42/104 = 40.38%

You go from 46/100 automatic losers to 40/100 automatic losing hits. You will automatically lose 6% less of the time on such hits.

In that event, the odds of a first card out possibly leading to that are 100%. We want to find out what the odds of having such a total in the first place are with 105 cards left in the deck, after the first card, because the numbers above are inclusive to two decks remaining. Because the first card out does not preclude a total of 0, 1 or 2, there will be three values that would result in a total of 0, 1 or 2 based upon the first card out.

We're going to assume an equal chance of it being either 24 or 25 cards (with 105 cards left in the shoe) based on the first card out. In other words, if the first card out is a 3, then we don't want a seven, eight or nine. There would be 24 such cards left and possibly one off card (making the 105) so you have three values divided by 13 total values, 3/13 = .23, so we'll say that there are 24.23 cards that would result in such a total.

1 * 24.23/105 = x

1 * .23076 = x

.23076 = x

In other words, 23.076% of hands would result in a total of 0, 1 or 2. We have determined that with a perfect composition of two decks that the odds of a non-playable hand (at this point, taking a hit) are 46.15%, so:

.23076 * .4615 * $10 = x

.1065 * $10 = x

-$1.07 = x

***The EV of a non-playable hand based on a starting count of 0, 1 or 2 with two perfectly composed decks is -$1.07 on $10 bet.

.23076 * .4038 * $10 = x

.0932 * $10 = x

-$0.93 = x

***The EV of a non-playable hand based on a starting count of 0, 1 or 2 with two perfectly composed decks is -$0.93 on $10 bet.

The difference in EV is $0.14 combined with the difference in EV on the, "Lucky 9," result of $0.06 is $0.20. The Expected Loss on the game in a perfectly composed shoe would be $0.12, but based on the difference in these two results, the expected WIN on the game with six less Tens and $10 bet is $0.08.

This would reflect a player advantage of .8%, but admittedly, does not yet consider the other hand possibilities.

Quote:UCivanWhat is EOR?

I couldn't find that, but if it was my post, I probably meant, "ER," for, "Expected Return," and mistyped. My apologies if such is the case.

EDIT: I see what you mean. I just took it as a change in expected value, but I am not sure what that abbreviation means. I believe it may be, "Effect of Removal," for taking cards out of the deck, but that could be wrong.

Quote:UCivanWhat is EOR?

Eor =effects of removal

If EOR is low, then no use to count, just pray. I got it.Quote:HunterhillEor =effects of removal

Quote:Mission146

The difference in EV is $0.14 combined with the difference in EV on the, "Lucky 9," result of $0.06 is $0.20. The Expected Loss on the game in a perfectly composed shoe would be $0.12, but based on the difference in these two results, the expected WIN on the game with six less Tens and $10 bet is $0.08.

This would reflect a player advantage of .8%, but admittedly, does not yet consider the other hand possibilities.

Actually, no it doesn't.

It just occurred to me that some of the hands that would automatically lose might lose anyway. It's certainly going to improve the expected value to have less tens in the game, especially when it comes to having to hit on automatically losing totals, but not quite to the extent that my post indicated because it doesn't account for hitting and losing (to the dealer) as opposed to hitting and losing automatically.

It's really tough to find shortcuts in the world of gambling Math. You can get away with it in Economics (my Bachelor's) because it's all BS anyway, but you can't BS expected value.

It seems that there will be no other choice except to getting around to doing a hand-by-hand analysis on the EoR of six tens. I'm pretty much going to have to account for every possible hand. That's going to take forever, so please excuse me if I do it a little bit at a time (probably starting tomorrow) and post my results as I go.

Two Decks-Perfectly Distributed-Dealer's Hand Probabilities

The dealer would receive (assuming only one player at the table) the 103rd and 101st remaining cards in the deck, however, since both the player and the dealer must have two cards, the dealer's hand can just as well be treated as the 104th and 103rd cards out. It is true that the player's first card will affect the probability in terms of what the dealer can draw (if the player draws a ten, then there is one less ten in the deck) but since both hands come from the same 104-card remaining shoe, this can be ignored for these purposes.

Dealer Hand Probabilities to the sixth decimal place:

A-A, 2-2, 3-3, 4-4, 5-5, 6-6, 7-7, 8-8, 9-9

8/104 * 7/103 = .005228/each---.005228 * 9 = .047052 (All)

Lucky 9: A-8, 2-7, 3-6, 4-5, 8-A, 7-2, 6-3, 5-4

8/104 * 8/103 = .005975/each---.005975 * 8 = .0478 (All)

Dealer 0 (other than 5-5):

10-10, A-9, 2-8, 3-7, 4-6, 9-A, 8-2, 7-3, 6-4

Tens-Any:

32/104 * 31/103 = .092606

All Other Zeroes:

8/104 * 8/103 = .005975/each---.005975 * 8 = .0478 (All)

All Zeroes: .092606 + .0478 = .140406

Dealer Totals One:

A-10, 9-2, 8-3, 7-4, 6-5, 5-6, 4-7, 3-8, 2-9, 10-A

Tens & Aces:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

All Other Ones:

8/104 * 8/103 = .005975/each---.005975 * 8 = .0478 (All)

All Ones: .047797 + .0478 = .095597

Dealer Totals Two (except A-A + 6-6)

10-2, 2-10, 3-9, 9-3, 4-8, 8-4, 5-7, 7-5

Ten-Deuce:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

All Other Twos:

8/104 * 8/103 = .005975/each---.005975 * 6 = .03585

All Twos: .047797 + .03585 + .005228 (Aces) + .005228 (Sixes) = .094103

Dealer Totals Three

10-3, 3-10, A-2, 2-A, 4-9, 9-4, 5-8, 8-5, 7-6, 6-7

Ten-Three:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

All Other Threes:

8/104 * 8/103 = .005975/each---.005975 * 8 = .0478 (All)

All Threes: .047797 + .0478 = .095597

Dealer Totals Four-Except 2-2, 7-7

10-4, 4-10, A-3, 3-A, 5,9, 9-5, 6-8, 8-6

Ten-Four:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

All Other Fours:

8/104 * 8/103 = .005975/each---.005975 * 6 = .03585

All Fours: .047797 + .03585 + .005228 (Twos) + .005228 (Sevens) = .094103

Dealer Totals Five

10-5, 5-10, A-4, 4-A, 2-3, 3-2, 6-9, 9-6, 7-8, 8-7

Ten-Five:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

All Other Fives:

8/104 * 8/103 = .005975/each---.005975 * 8 = .0478 (All)

All Fives: .047797 + .0478 = .095597

Dealer Totals 6-Except 3-3, 8-8

10-6, 6-10, A-5, 5-A, 2-4, 4-2, 7-9, 9-7

Ten-Six:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

All Other Sixes:

8/104 * 8/103 = .005975/each---.005975 * 6 = .03585

All Sixes: .047797 + .03585 + .005228 (Threes) + .005228 (Eights) = .094103

Dealer Totals 7

10-7, 7-10, A-6, 6-A, 2-5, 5-2, 3-4, 4-3, 8-9, 9-8

Ten-Seven:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

All Other Sevens:

8/104 * 8/103 = .005975/each---.005975 * 8 = .0478 (All)

All Sevens: .047797 + .0478 = .095597

Dealer Totals Eight-Except 4-4, 9-9

10-8, 8-10, A-7, 7-A, 2-6, 6-2, 3-5, 5-3

Ten-Eight:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

All Other Eights:

8/104 * 8/103 = .005975/each---.005975 * 6 = .03585

All Eights: .047797 + .03585 + .005228 (Threes) + .005228 (Eights) = .094103

Natural-Not Lucky-Nine

10-9, 9-10

Ten-Nine:

32/104 * 8/103 = .023898/each---.023898 * 2 = .047797

PROOF

Lucky 9 = .0478

Dealer 0 = .140406

Dealer 1 = .095597

Dealer 2 = .094103

Dealer 3 = .095597

Dealer 4 = .094103

Dealer 5 = .095597

Dealer 6 = .094103

Dealer 7 = .095597

Dealer 8 = .094103

Dealer Natural-Not Lucky- 9 = .047797

Sum= .994803 (Deviation from 1.00 due to rounding)

The probabilties will be taken as the same for the player, even though the cards that one draws will affect the other, that cannot be known until the hand is initiated, so will be ignored.

That will be a good enough start for now. I intend to get into the House Rules play tomorrow to determine the probabilties of hand totals with which the dealer will end up. The dealer will stand on a total of five (or better) and will hit on a total of 0-4. I believe I am going to base it on 102 cards remaining because it will be too tedious to have the player's cards affect the dealers cards and vice-versa.

This will be easy (yet just as lengthly) to do because the House Rules are fixed.

The next step will be to determine the player's starting hands (and what the player does based on what the dealer is showing) against the probabilties that the dealer will be showing a certain card. In this case, the probabilities of player starting hands will be considered the same as those of the dealer. (since we are starting with 104 cards remaining and both player and dealer must have a hand). The only thing left to determine, then, is the probability of a dealer showing a certain card, which is easy, because it will either be 32/104 (Tens) or 8/104 (All Others).

Quote:Mission146Start

That will be a good enough start for now. I intend to get into the House Rules play tomorrow to determine the probabilties of hand totals with which the dealer will end up. The dealer will stand on a total of five (or better) and will hit on a total of 0-4. I believe I am going to base it on 102 cards remaining because it will be too tedious to have the player's cards affect the dealers cards and vice-versa.

This will be easy (yet just as lengthly) to do because the House Rules are fixed.

The next step will be to determine the player's starting hands (and what the player does based on what the dealer is showing) against the probabilties that the dealer will be showing a certain card. In this case, the probabilities of player starting hands will be considered the same as those of the dealer. (since we are starting with 104 cards remaining and both player and dealer must have a hand). The only thing left to determine, then, is the probability of a dealer showing a certain card, which is easy, because it will either be 32/104 (Tens) or 8/104 (All Others).

Nowhere have I seen the penetration mentioned? Did I miss it?