Other Casino Games - FAQ

Thanks for pointing out this variation to me. They use different semantics to explain casino war at the Casino Niagara. What they fail to mention is that the original wager always loses if in the event of a tie.

What is really going on is they pay two times the Ante for a win after a tie, and three times the ante for a tie after a tie. The usual rules still pay two times the ante for a tie after a tie. This rule change decreases the house edge from 2.88% to 2.33%.

I could talk about this all day. Part of my income is derived from analyzing games such as these. The gaming authorities require such analysis before a game can be licensed to play. Usually, these games are invented by an individual. Here in Nevada, after the game owner receives a license, then it must go through a 30-day trial period. If the trial period went well, then the owner can then apply for a permanent license. The entire process is very slow and it is difficult to get a casino to be the guinea pig for the trial period. Casinos are actually quite risk averse in their business decisions. Yes, the game owner will usually seek a copyright to protect others from stealing the idea.

You can find much more information about the business of marketing casino table games in my Game Inventors Corner.

Ask and ye shall receive. Please see my page on Catch a Wave.

I have never seen solitaire played for money in Vegas. I understand in the early days of Vegas people wagered on the standard Klondike variation of solitaire but I don't anything else about it.

I saw this at the World Gaming Expo, but have never studied it. After I move to Las Vegas in February I will be better at covering new games such as this.

Let's let d be the number of decks. The probability of a tie on the first round is (4*d-1)/(52*d-1)= 0.073955. The probability of a tie in the second round is 12*4*d/(52*d-2)*(4*d-1)/(52*d-3)+(4*d-2)/(52*d-2)*(4*d-3)/(52*d-3) = 0.073974. Lets call p₁ the probability of a tie in the first round and p₂ the probability of a tie in the second round. Then the player return is p₁*(2*p₂ +(1-p₂)/2*(1-2))= -0.023301. Multiply by -1 and you have the house edge of 2.33%. I hope I didn't go over this too quickly.

I’ve seen it at the Regent, New York New York, and Palace Station. I hear it is also at the Sunset Station and Santa Fe Station.

Thanks for the compliment. I just saw the game at the California casino here in Las Vegas, but it wasn’t open yet. I got the rule card and will work on when I have the chance. At this time I have no information about it at all.

I’ve seen it there too, and fortunately took some notes. It is similar to Three Way Action as used to be found at the Las Vegas Club. In Triple Shot the player may make any combination of three bets. The first is a regular blackjack wager. The second is a poker hand. The third is a war bet. I don’t remember if the poker bet is based on the player’s or dealer’s hand but the best five out of six cards are used. If the blackjack hand doesn’t contain six cards then more are added to make six. The following odds table for the poker bet shows the house edge is 3.20%.

Triple Shot

Hand	Combinations	Probability	Pays	Return
Royal flush	376	0.000018	100 to 1	0.001847
Straight flush	1468	0.000072	30 to 1	0.002163
Four of a kind	14664	0.000720	15 to 1	0.010804
Full house	165984	0.008153	7 to 1	0.057071
Flush	205792	0.010108	5 to 1	0.050542
Straight	361620	0.017763	4 to 1	0.071050
Three of a kind	732160	0.035963	3 to 1	0.107890
Two pair	2532816	0.124411	2 to 1	0.248821
Pair	2252472	0.110640	1 to 1	0.110640
Nothing	14091168	0.692151	-1 to 1	-0.692151
Total	20358520	1		-0.031321

Finally there is a war game, the player’s first card against the dealer’s up card, highest card win. In the event of a tie the player loses half. The house edge in the war game is 2.94%. I don’t know of any other casinos that have the game. It is probably in trial period and only at the Treasure Island.

I don’t know. If you find out please tell me. I’m particularly interested because Cryptologic casinos just introduced two versions of Klondike solitaire.

Fortunately I am a big James Bond fan and have all the Bond movies on DVD. I checked Dr. No and it seems he is playing Chemin De Fer. The scene was spoken in French, which doesn’t help me. There is a similar scene in For Your Eyes Only. In that movie it looks like Bond is playing baccarat, acting as the banker, but after the player acts he pauses and another character tells Bond, "The odds favor standing pat". This would imply that Bond had free will in whether to take a third card, an option you don’t have in baccarat. As I understand my gambling history, the American version of baccarat is a simplified version of Chemin De Fer, in which the drawing rules are predetermined. Incidentally, according to www.casino-info.com American baccarat originated at the Capri Casino in Havana, Cuba.

Please see my faro page for the answer to that question.

From my Sic Bo appendix we see the probability of a total of 5 or 16 is 6/216, a 6 or 15 is 10/216, a 7 or 14 is 15/216, and a 3 is 1/216. So on any one throw the probability of "big" winning is 31/216, "small" is 31/216, and "aces" is 1/216. The number of ways any of these could win is 2*31+1=63. So given that one of these events did occur, the probability that it was big is 31/63, small is 31/63, and aces is 1/63. The house edge on all three bets is 1.59%.

Following is the median high hand according to the number of players. This is based on an assumption of independence between hands, which is not the case, but the table should still be a very close estimate.

The standard deviation in Pick ‘em Poker is 3.87. The standard deviation in conventional video poker tends to run from about 4.4 to 6.4. I don’t have any risk of ruin tables for Pick ‘em Poker. So the best advice I can offer is to use the jacks or better table in my video poker appendix 1. Jacks or Better has the lowest standard deviation in that appendix at 4.42, so you can be a little more aggressive than that table calls for.

Generally speaking, you want to put it on a long shot. This is because you don’t get to keep the coupon on a win, which lowers the value by the probability of winning. The less the probability of winning the less the value is reduced. Following are three tables for the three games listed. You’ll see the best bet is a tie between the 12, 30, 60, triple, and any triple in sic bo.

Baccarat

Free Bet Coupon Value in Baccarat

Bet	Pays	Probability	Return
Banker wins	0.95	0.458597	0.481484
Player wins	1	0.446247	0.493175
Tie	8	0.095156	0.761248

Big Six

Free Bet Coupon Value in Big Six

Bet	Pays	Probability	Return
1	1	0.444444	0.444444
2	2	0.277778	0.555556
5	5	0.12963	0.648148
10	10	0.074074	0.740741
20	20	0.037037	0.740741
Joker	40	0.018519	0.740741
Logo	40	0.018519	0.740741

Sic Bo

Free Bet Coupon Value in Sic Bo

Bet	Pays	Probability	Return
Small, Big	1	0.486111	0.486111
4, 17	60	0.013889	0.833333
5, 16	30	0.027778	0.833333
6, 15	17	0.046296	0.787037
7, 14	12	0.069444	0.833333
8, 13	8	0.097222	0.777778
9, 12	6	0.115741	0.694444
10, 11	6	0.125	0.75
Triple	180	0.00463	0.833333
Any triple	30	0.027778	0.833333
Double	10	0.074074	0.740741

I must confess that my 0.17% figure was an error. I discovered the flaw in my analysis when I recently updated it for the Microgaming rules. To all those who played it because of my 0.17% figure, I apologize.

The probability that any given player will have a dragon is 4¹³/combin(52,13) = 0.000106. The probability that exactly one player is dealt a Dragon could be closely approximated as 4*0.000106*(1-0.000106)³ = 0.000424, or 1 in 2,359.

Usually these free bet coupons are limited to even money bets, so this is an interesting case. My advice is to use the free bet on a long-shot, to minimize the effect of the rule that you lose the free bet, even if you win. The biggest long-shot in Big Six is the joker/logo, with a probability of winning of 1/54. I’m not sure whether the Mohegan Sun pays 40 or 45 on the joker, but assuming 45 the value of the free bet is (1/54)×45 = 83.33% of face value. In Sic Bo the biggest long-shots are on the six triples. I’m also not sure what they pay for a specific triple, but I would guess 180. In that case the value of any one of the six triple bets would be (1/216)×180 = 83.33% of expected value. So, we have a tie in terms of expected value. In that case I would go for the bet with the greater probability of winning, the joker/logo in Big Six, but that is up to you.

You're right, you can lower the element of risk by deviating from my strategy, and raising on hands with expected values of slightly less than -1. In your example, 5/2 has an expected value of -1.019987, under the Las Vegas rules. That means if you raise on that bet on average, then you can expected to lose about 1.02 times your original bet by the time the hand is over. After the initial raise, and possible additional raises after the flop and turn, the average wager on that hand will be 3.627374 units. The way I would look at it, making the raise bet is worth -0.0109987 units to the player, over 2.627374 additional units bet. The ratio of marginal additional win to marginal additional bets, by raising, is -0.0109987/2.627374 = -0.00761. That is less than the overall expected value of the game of -2.04%. So, if your goal is to minimize money lost to total money bet, including raises, then, yes, you should deviate from my basic strategy and raise on that hand. Other examples could be made in lots of games that involve raising.

To summarize, if you are trying to minimize money lost per hand, then you should follow the house edge minimizing strategies on this site. If you are trying to minimize money lost per total amount wagered, then you should opt to bet more on very borderline plays.

The expected number of times any number will appear exactly n times in 12 games is combin (12,n)×(6/45)ⁿ×(39/45)^n-12. The following table shows the expected number of occurrences from 0 to 12.

So, to answer your question, you will see the same number exactly six times about 0.099 times per set of cards, or about once every 10.1 times. The same number appearing exactly seven times will happen 0.0131 times per set of cards, or once every 76.6 times.

I hope you're happy; I watched this scene over and over for at least an hour, trying to make sense of the rules. I’ve played guts lots of times, over many years and locations, and have never seen it played as was done in that movie. Let’s call the first player to act Player 1, and the second player to act (the dealer) Player 2. Here is my understanding of how they played.

1. Both players ante (or re-ante).
2. Each player gets two cards.
3. Player 1 must declare “in” or “check.” If he checks, go to rule 4. If he goes in, go to rule 7.
4. Player 2 must declare in or check. If he checks, go to rule 5. If he goes in, go to rule 6.
5. Although two checks never happened in the movie, I assume both players would start over from step 1.
6. The action goes back to player 1, who must declare in or fold. If he goes in go to rule 8. If he folds, go to rule 9.
7. Player 2 must declare “in” or “fold.” If he goes in go to rule 8. If he folds, go to rule 9.
8. The two hands are compared; and the higher hand wins. The winner collects the pot, and the loser must match it, creating a new pot. This is equivalent to the loser just paying the winner the amount of the pot. Although there was never a tie in the movie, I assume no money would move. Next, go to rule 10.
9. When a player folds, the other player collects the pot. Then repeat with a new hand from step 1.
10. An additional card is given to each player, to add to his existing 2-card hand, making a 3-card hand. The third card is dealt face down, on top of the face-up two card hand. I do not know whether straights or flushes counted at the 3-card stage. I prefer to play where they do count (but not at the 2-card stage).
11. Steps 3 to 9 repeat. If both playes go "in," then go to rule 12.
12. An additional two cards are given to each player, to add to his existing 3-card hand, making a 5-card hand. The fourth and fifth cards are dealt face down, on top of the face-up three card hand.
13. Steps 3 to 9 repeat. Then start over at step 1.

If you watch the movie carefully, Huck should have lost $11,000 in total, when he had $10,000 to begin with. I watched the scene lots of times to try to find this missing $1,000. My best guess is that when he went in on the last two-card hand, he should have matched the $4,000 pot, but had only $3,000 left. I assume that, much as in regular poker, he could only stand to win what he was risking. In the last hand, Huck folded. I’m not sure if this was because his three-card hand couldn’t beat his father’s two-card hand on the table, or if he was forced to fold, because he didn't have the money to match the pot if he lost.

If my understanding of the rules or analysis of the scene is in error, I welcome correction.

The house edge under that pay table is a comparatively low 1.85%. Kudos to the Borgata, assuming your information is correct.

Based on viewer feedback, the Borgata lowered the win on a three of a kind to 30 to 1 sometime during 2008.

My Super Pan 9 page shows the probability of a tie is 11.3314%. So if a tie paid 8 to 1, the expected return would be 9×0.113314 − 1 = 0.019826. Although a 1.98% player advantage is less than your figure, it is still a great bet. Where can I play it?

I'm told that game had to be pulled out of the U.S. casinos, because the game of patent infringement. According to the Fourth Quarter 2008 Statistical Report of the Nevada Gaming Control Board, the following are the table game counts in Clark County.

Unfortunately, they don't say what the 243 "other" games are, so this isn't of much help to answer your question, but it is still worth mentioning.

Not that you asked, but you have a 43.4% advantage if your first card is a jack. It is the dealer’s fault for flashing the card. Contrary to what some members of the casino staff, especially in security, incorrectly believe, you are legally allowed to make use of whatever information made available to you under normal playing conditions.

Morally, you should follow your own conscience. You have to live your own life. That said, I think most players, including me, would be okay with increasing the bet in that situation. For one thing, game security is not the player’s job. For another, the casinos take advantage of, if not rely on, player mistakes. For example, consider the big 6/8 bet in craps. The casinos have no compunction about accepting a bet on that, when the place bet on 6 or 8 pays on exactly the same thing, but has better odds. See if you are offered forgiveness if you foul your hand in pai gow poker, even if the correct setting is totally obvious.

If it happens again, don’t get too greedy, and act nonchalant. If you suddenly go from a $10 to a $500 bet, it will set off all kinds of red flags. A good dealer would realize why, and ultimately the bet would not be accepted, or a card would be burned.

I’m getting asked about this more and more frequently. Unfortunately, the number of combinations in this game would be an unholy gigantic figure. A brute-force looping program might take thousands of years to finish. However, a good programmer can find short cuts. Weighing the costs and benefits, I don’t find this project to be a good use of my time. If I lived in Russia or the Baltics, I would likely feel differently.

For the benefit of other readers, Royal Poker is like Caribbean Stud Poker, with the following added options. It is my understanding that the player may invoke any or all of these options, except he may not invoke options 2 and 3 both.

1. If the player can make two paying hands, which both beat the dealer, and neither hand is entirely within the other, then both are paid. I am not sure whether the ante and/or raise are paid twice. For example, if the player had six cards and could make a straight and a flush, then the player could be paid for both hands.
2. The player may switch one to five cards for the price of the Ante.
3. The player may buy a sixth card for the price of the Ante.
4. The player may buy "insurance" before the dealer turns over his four face-down cards. The insurance bet pays even money if the dealer does not qualify.
5. The player may force the dealer to switch his highest card for the next one in the deck for the price of the Ante wager.
6. There is an "AA Bonus" side bet, which pays 7 to 1 if the player’s first five cards are a pair of aces or higher.

I can say that the AA Bonus side bet and Insurance Option should never be taken, and thus are not worth anything. The house edge on the AA Bonus is 12.99%. The following table shows the house edge on insurance to range from 8.57% to 33.57%, depending on the dealer’s up card.

Insurance in Russian Poker

Dealer’s Up Card	Combinations	Probability	Exp. Value
A	132804	0.335714	-0.328571
K	132804	0.335714	-0.328571
Q	108528	0.457143	-0.085714
J	108732	0.456122	-0.087755
10	108936	0.455102	-0.089796
9	109140	0.454082	-0.091837
8	109140	0.454082	-0.091837
7	109140	0.454082	-0.091837
6	109140	0.454082	-0.091837
5	109140	0.454082	-0.091837
4	108936	0.455102	-0.089796
3	108732	0.456122	-0.087755
2	108528	0.457143	-0.085714

I also know from my page on Oasis Poker that just the option to switch cards lowers the house edge from 5.22% to 1.04%. I tend to think the rule about being double-paid, getting to keep the sixth card (as opposed to switching), and forcing the dealer to switch a card will get the game to a worthwhile player advantage, if you knew the proper strategy. Sorry to pull a Fermat on you, but that is that best I can do at this time.

P.S. I have heard some casinos add a rule that if the player wins, then the ante bet only pushes. This would work significantly in the casino’s favor, I think wiping out any advantage.

For those unfamiliar with the rules, in Pick ’Em Poker the player is dealt two cards, plus the choice of one of two more. The game then gives the player two more cards to complete a five-card poker hand. The question at hand is what is the probability of having at least a pair of nines on the deal. Let’s call the players initial two cards that he must keep the "pocket," and the other two cards the "field." This could be accomplished the following ways:

• Four of a kind: 13 combinations
• High (9-A) three of a kind: 1,152 combinations
• Low (2-8) three of a kind with the singleton in the field: 672 combinations
• Two high pairs: 540 combinations
• One high pair, one low pair, with at least one high card in the pocket: 1,260 combinations
• High pair, with at least one in the pocket: 31,680 combinations

The sum of those combinations is 35,317. The total number of ways to pick 4 cards out of 52 is combin(52,4)=270,725. So the probability of getting a pair of nines or better on the deal is 31,680/270,725 = 13.05%. The probability of not getting a pair of nines or better is 100%-13.05%=86.95%. The probability of going 66 hands without a pair of nines or better is (1-(31,680)/270,725))⁶⁶ = 0.00009848, or 1 in 10,155. That could have just been ordinary bad luck, and doesn’t rise to the level to make a convincing case of foul play. A bigger sample size is warranted to make a better case.

Assuming eight decks, the probability of winning is 52*combin(8,2)/combin(52*8,2) = 1,456/86,320 = 1.69%. The house edge is 13.98%.

That rule increases the house edge from 2.19% to 7.97% and the Element of Risk from 0.53% to 1.90%. This is because the dealer won't qualify more often and it will be harder to win on the Blind bet, which requires a straight or better.

For more details of my analysis, please see my new page on the Auckland variant of Ultimate Texas Hold 'Em.

For discussion about this question, please see the thread ULTIMATE IN NEW ZEALAND in my forum at Wizard of Vegas.

The red and black bets seem the most vulnerable. I would do a simple red/black count, as follows:

• Let C = count (where red cards are +1 and black cards are -1)
• J = count of jokers left in the shoe
• if J-C < 0 then bet on black
• If J+C < 0 then bet on red

For example, if after some play cards played are:

red = 100
black = 75
jokers = 10

The remaining cards would be:

red = 108
black =133
jokers = 14

The count would be +25. Jokers remaining - C = 14-25 = -9. Because that is less than zero, bet on black, because there are more good cards left (133) than bad cards (122) on the black bet.

The probability of a winning royal flush is 1 in 32,487. Each time this happens you are shortchanged $20,000, or 400 Ante bets. That is a cost of 400/32,487 = 1.23% of all money bet on the Ante. That increases the house edge (as measured relative to the Ante bet) from 2.185% to 3.416%.

This question is asked in discussed in my forum at Wizard of Vegas.

Yes, indeed Bond was playing Chemin de fer, not baccarat. As a reminder, the main differences are:

• The turn to bank rotates around the table. The banking player automatically assumes the Banker* hand.
• Both Player* and Banker have free will to draw a third card, as long as neither has a two-card natural 8 or 9.

Note:
As usual, when writing about baccarat or chemin de fer, I capitalize the names of the bets, to avoid confusion with the players playing the game.

That said, the first such scene is in the Dr. No clip below, where not only does the bank rotate, player's have free will in the third card, but Bond actually refers to the game verbally as "chemin de fer" in the 2:11 point in the video:

The next video shows three chemin de fer scenes from three different movies.

Here are the starting points of each scene:

• 2:09 — Thunderball
• 4:19 — On Her Majesty's Secret Service
• 7:30 — For Your Eyes Only

In all three we see the players banking and free will in the third card decision, especially in For Your Eyes Only, where the odds are discussed.

Finally, below is the scene from Goldeneye.

Here we again see the turn to bank going back and forth and free will in drawing a third card. However, the female character does refer to the game as "baccarat" at the 0:45 point. I would file this under "character error."

In closing, there can be no doubt Bond was playing chemin de fer in every movie. Technically, I think the judges were incorrect to accept "baccarat" as a correct answer. However, I don't blame them. Even by Jeopardy standards, expecting the average player to know baccarat from chemin de fer would be too much. As shown by accepting an answer of "The Mooch," in lieu of Anthony Scaramucci, shows they are getting more lenient.

This question is asked and discussed in my forum at Wizard of Vegas.

Assuming correct 4x raising strategy, there is the probability of each outcome, given a 4x raise:

• Win: 58.82%
• Loss: 38.47%
• Tie: 2.72%

If we factor out the ties, the probability of winning is 60.46%. In 96 hands resolved, the expected number of 4x wins is 58.04. So, 66 wins is ahead of expectations, but not significantly.

This question is asked and discussed in my forum at Wizard of Vegas.

This is a good question because some dealers have been known to not enforce the Blind bet rule. The Blind bet has a huge house advantage, so not having to make it would be very beneficial to the player.

Assuming the player follows optimal strategy based on the correct rules (required a Blind bet), then the player advantage would be 29.28%. It would be even higher following a strategy based on no Blind bet being required.

Good question. Mississippi Stud players definitely should be aware that maximum payouts may reduce what they win on a royal flush, which serves to increase the house edge.

The top pay in Mississippi Stud is 500 to 1 for a royal flush and it applies to all bets. The player has two chances to raise up to 3x his Ante, so his final bet can be up to 7x his Ante. If the player has hope for a Royal, he should be maxing the maximum raises.

The most the player should bet on the Ante in Mississippi Stud, without being affected by a payout cap, is the maximum payout/3500. For example, if the maximum payout is $80,000, then the most I advise betting on the Ante is $22.86. I would round that down to $20.

The following table shows the house edge at various bets and common maximum wins. The table assumes optimal player strategy with no cap on wins. Note how the house edge increases as bet size increases and the cap decreases.

This question is asked and discussed in my forum at Wizard of Vegas.

The exact answer is 450528/407170400 =~ 0.001106485 =~ 1/904.

The following table shows the number of ways to make each type of initial hand and the number of combinations that tie in the second hand. With straights and trash hands it matters whether there are two or three suits represented in the first hand.

The bottom right cell shows there are 450,528 ways to tie in Three Card Poker. The total number of combinations for two hands is combin(52,3)*combin(49,3) = 407170400. Thus, the answer is 450528/407170400.