# Baccarat - FAQ

I was playing baccarat online and out of 75 hands the banker won 52 and the player 23. This is a difference of 29, what is the probability of that happening?

First, I'm going to assume that you are not counting ties. In other words, you mean 75 bets resolved. It would be very unlikely to go 75 hands without a tie. The expected number of banker wins out of 75 bets resolved is 38.00913745. The standard deviation is the square root of the product of 75, the probability of a banker win, and the probability of a player win. The probability of a banker win, given that there wasn't a tie, is 0.506788499 and the probability of a player win is 0.493211501 . The standard deviation is thus 4.329727904 . Then you'll have to make a half point correction for a binomial distribution and look up the Z statistic in a standard normal table (this step is left to the reader). The final answer is that the probability of the banker getting 52 or more wins is .0009. Your question also allowed for the possibility of the banker winning 23 or fewer times (also a difference of 29 more more) which has a probability of .0004 . So the final answer is that the probability of a difference of 29 or more is .0013, or 1 in 769.

First of all, thanks for providing reliable gambling info. You are one of only about four or five sites on the net doing so. In your opinion, is it possible that a mathematically sound method (card counting etc.) could ever be devised to give a positive expectation in baccarat? There has recently been some speculation (and wild claims) on the bj21 and other gambling forums.

Thanks for the compliment. I address the vulnerability to card counting in my baccarat appendix 2. To make a long story short, no, baccarat is not countable unless you use a computer.

Recently, at a casino I occasionally visit, they had a "teaser" for Baccarat that reduced the commission to 4% during certain periods of the day. I know this would have some effect on the 1.06% house edge on a banker bet, but is it enough to create a +EV for the banker bet? How much would it drop the house edge with a 4% commission on banker bets?

The probability that the banker will win is 45.843%, player 44.615%, and tie 9.543%. So the return of the banker bet with a 4% commission is .45845*.96 - .44615 = -.00606 . So the house still has a thin 0.6% edge.

I have a thick, and I mean thick, friend who is intoxicated with having won a fair amount betting Player only in baccarat. He plays $10 units and does the dumb play that 9 straight Banker wins won't happen. He escalates to a risk of $3,980 by going 10 10 30 60 120 250 500 1000 2000. How can I find solid mathematical evidence to try to convince him to stop?

This is a close variation of the Martingale betting system, in which the player doubles after every loss. Usually, the Martingale player will win but occasionally he will have more consecutive losses than he can handle and suffer a major loss. Assuming your friend is betting on the player, the probability that any given bet will begin a streak of nine losses in a row is (2153464/(2153464+2212744))^{9} =~ .001727, or 1 in 579, assuming ties are ignored. There is more information available about the folly of the Martingale in my section on betting systems. However, the more ridiculous a belief is the more tenaciously it tends to be held. It usually takes a big loss to possibly convince a believer in any particular betting systems to stop.

Dear Mike, in baccarat the odds for Banker, Player and Tie are 45.843%, 44.615% and 9.543% as mentioned. Can you please clarify on how you arrive at 1.064% and 1.228%, please?

The expected player return per unit wagered on the banker is .45843*.95 + .44615*-1 = -.01064. The player's loss is the casinos gain. Thus the house edge is -1*-.01064 = .01064 = 1.064%. Likewise, the expected return on the player bet is .45843*-1 + .44615*1 = -.01228. Thus, the house edge on the player bet is 1.228%.

Mike, On my last trip to Vegas, a dealer I've come to know said he was "toying with the idea" of standing on a 16 against a dealer's 7 because only 5 of the 8 cards give the dealer an automatic win. How does this strategy play out?

This would be a bad play. For example, my blackjack appendix 9B shows the return both ways by playing 10 and 6 cards against a dealer 7. Hitting has an expected loss of 39.6% of the bet. However, standing has an expected loss of 47.89%. There is no easy explanation I can give why hitting is better. You have to consider everything that can happen, weight it by its probability, and take the sum. Overall hitting is better of two bad plays.

Dear Mr. Wizard, thank you for your very informative website. My question is about baccarat. What are the good cards for the Player and what are the good cards for the Banker. Kindly give them in their order of strength. Thank you, and more power to you.

I have a whole page on the topic of card counting in baccarat. Briefly, the best card for the Player is 4, and for the Banker is 6.

I was playing baccarat at OasisCasino.com, where they claim that six decks are shuffled for each game. However, there is no indication of the end of the shoe, hence you can play endlessly without knowing where the shoe starts and where it ends. My question is. How many hands are in a six-deck shoe? Also, how many hands should I play until I start recording the plays (hands) as a new game?

First it is very possible that they are shuffling after every hand. Even if they do penetrate into the shoe I doubt they play through the entire thing. The average number of cards per hand is 4.94. Assuming 15 burn cards, a six-deck baccarat shoe would have about 60 hands. However, mathematically speaking, it doesn't make any difference when they shuffle.

My question concerns baccarat, is a 1,2,3,5,8,13,21,etc. progression a profitable method of play? I add the previous two bets and bet the sum on a loss and subtract one level of the progression on a win. Any two wins in a row and I go back to 1 unit. i always bet on banker. I've tried this online in several casinos (as well as your site's baccarat game, and it has worked great. Is it flawed?

All betting systems are flawed. Progressive systems like yours usually do when but with occasional large losses. Over the long run, you will do no better nor worse than the flat bettor or user of any other system.

Enjoy your site more than any other gambling site. I am curious about how realistic the randomness of your Java baccarat game is. I've played it for many hours and have been using a strategy that appears to win every time now. I am fearful to try my strategy at a casino though, because I'm unsure of how random your game is. The strategy is to begin with a $5 bet on Banker and with each loss bet one more unit and with every win subtract one more unit. I've only been down about $300 at the most, but usually will get to $1,100 or $1,500 within about 200 hands. What are your thoughts?

My Java games are based on the random number generator that comes with Visual J++. For personal play, it should be quite fair. I speculate that any bias would only show up over millions of hands. Your results are not the result of a biased random number generator but of both luck and a progressive betting system.

I am thinking of taking the following strategies to play mini baccarat. I only bet after either the Banker or Player has appeared four times in a roll. I double my money if I don't win the first time. However, if the second time I don't win, I stop betting for the time being until the next four continuous appearance coming again. Once I win, I also stop betting until the next 4 continuous appearance coming. Please evaluate my strategy. Thank you!

Waiting for streaks of four in a row is not going to help. The cards do not have a memory. Doubling after a loss is also not going to help. I would recommend betting on the banker every time. Skipping hands is fine, in fact not playing at all is the best possible strategy.

I agree with you that there are no system that can beat a negative expectation game. Anyway, I take a look at the cancellation system and keep wondering ... what about using it in a bet like the banker in baccarat, where you have a POSITIVE expectation outcome? In which extension would the commission payed to the house erode your gains in the long run? I apologize for my shady English.

The Banker is baccarat is not a positive expectation bet. You're confusing the probability of winning the bet with having a positive expectation. Even without a betting system, you will probably win any banker bet but you will win less than what you bet, because of the 5% commission. This makes the banker bet a negative expectation bet.

In baccarat, you address the question of counting to zero the house edge. But what about the (true) count (for ’player’) necessary to make ’player’ the better bet than ’dealer’? According to your table in baccarat appendix 2, removing cards 5 through 9 increases the relative probability of a ’player’ win. In an 8-deck shoe, is there a point at which the house edge on ’player’ falls below that on ’dealer,’ even though the edge on both may remain positive? Where is that point?

Here are the values to assign each rank for counting the Player bet, from my blackjack appendix 2. The true count is the running count divided by the number of decks remaining.

Player Bet Count | |
---|---|

Removed | Player |

0 | -178 |

1 | -448 |

2 | -543 |

3 | -672 |

4 | -1195 |

5 | 841 |

6 | 1128 |

7 | 817 |

8 | 533 |

9 | 249 |

I show that if the true count exceeds 17,720 then the Player bet house edge is reduced to 1.06%, and becomes as good of a bet as the Banker. At true counts greater than 17,720, the Player is the better bet.

I can't help but say that you can just walk over to a blackjack table and have a much lower house edge with basic strategy.

What is the house edge for baccarat with 10% commission on a winning banker?

The probability of the banker winning is 45.86% and the player winning is 44.62%. So the house edge would be 44.62%-.9*45.86%=3.346%.

I was just reading Peter Griffin’s Theory of Blackjack and found something in the back of the book that caught my attention. In his analysis of a baccarat count system in order to get true count he divided the running count by the number of cards remaining rather then the number of decks remaining. Is that correct? Thanks for your attention.

It is more accurate to divide by the exact number of cards remaining. He was trying to show that for all practical purposes baccarat was not countable, even for a computer perfect counter. So there was no need to devise a more practical count. If baccarat isn’t worth playing for a perfect counter then it certainly isn’t worth playing for a mere mortal.

What is the best game to use a match play coupon on?

First let’s define a match play coupon for those who don’t know. This is something often found in casino fun books. If the player accompanies a match play coupon with a real even money wager then the match play will be converted to a like amount of cash if the player wins. For example if the player has a $5 match play and uses it along with a $5 bet on red in roulette then if the player wins his $5 will win $5 and his match play will be converted to $5. Whether the player wins or loses he will lose the match play coupon. In the event of a push, the player gets to keep the match play coupon.

If used in blackjack, the Match Play will usually only pay even money. This decreases the value of the Match Play itself by 2.3%, which is way too much. Of the true even money bets, the best game to use a match play on in the Player bet in baccarat. That has a probability of winning of 49.32% of bets resolved. For the don’t pass in craps, that probability is 49.30%.The value of a Match Play on the Player bet is 47.95% of face value, assuming you wouldn’t have bet otherwise.

What table game has the best odds for winning and is user friendly for a novice gambler? Thanks in advance.

For the person with no casino gambling experience who puts an emphasis on something easy to play I would start with baccarat. Just bet on the banker every time.

In an 8 deck baccarat, what is the probability of getting an Ace and an 8 of Diamonds for both the player and the banker in a same deal?

(8^{2}/combin(416,2))* (7^{2}/combin(414,2)) = 0.00000043, or 1 in 2308093

Is there a progressive wagering system for baccarat? Is there a specific site for this?

There are lots of them, and they are all worthless.

I read your topic in Roulette on the Martingale method. I have tried this method a few times on the computer and I have been up $500. Then I went to the casino and lost over $1000. Because black came up 8 times in a row. But I’m just starting to learn baccarat. I was trying it on the computer and again I have been up $500, by betting on the banker. Starting at $20 then going to $40 then $80 and so on. I was up $500 even with paying the 5%on each hand. Do you think this method would work in a casino? I thought I would ask before I go and lose another $1000. Like I said black came up 8 times in a row. But do you think that the player hand would win 8 times in a row? Plus this game is good because a tie is a push, where in roulette 0, or 00, is a loss.

The Martingale is dangerous on every game and in the long run will never win. However it is better to use in baccarat than roulette, just because of the lower house edge. The probability of the player winning 8 times in a row is 0.493163^8 = 1 in 286. Also keep in mind you could win a hand late in the series and still come out behind because of the commission. For example if you started with a bet of $1 and you won on the 7th hand you would win $60.80 ($64*95%), which would not cover the $63 in previous loses.

I have two friends that have a bet on which game (craps or baccarat) have the best odds for the player. Could you help me settle this. They are both casino workers and are sure they are right.

It depends on how the games are played. If optimal strategy is compared to optimal strategy then craps is better. By betting only the line bets and taking maximum odds the combined house edge in craps is well under 1%. The best you can do is baccarat is bet on the banker at a house edge of 1.06%. However it wouldn’t surprise me if the actual house edge in craps is higher, due to all the sucker bets players make.

In 500 hands of baccarat betting only on banker, what percent of the time will player win over 46% of the decisions. Thanks

First I’m going to assume you want me to ignore ties. From my baccarat section we see the probability of a player win is 49.32%, given that there wasn’t a tie. We’ll use the normal approximation to the binomial distribution for this problem. The expected number of player wins is 500*0.4932 = 246.58. 46% of decisions is 230. The standard deviation is (500*(0.4932)*(1-0.4932))^{1/2} = 11.18. So...

pr(player wins > 230) =

pr(player wins-246.58 > 230-246.58) =

1-pr(player wins-246.58 <= 230-246.58) =

1-pr(player wins-246.58+0.5 <= 230-246.58+0.5) =

1-pr((player wins-246.58+0.5)/11.18) <= (230-246.58+0.5)/11.18) =

1-Z(-1.44) =

1-0.075145503 =

0.924854497

So the answer is 92.49%.

We have a casino here offer zero commission for the baccarat game. But pay 1/2 if banker win on 8. Is this a favorable edge for the house ,compare to the 5 % commission?

The house edge on the banker bet is 4.07%.

Are the on line casino baccarat games like a slot machine with the payout set at 98.8% or do they use a random chip? How would you be able to check that out? Wouldn’t it make a difference? Is there a casino you are sure uses a random chip? Thanks.

From what I know of the business the major software companies deal the cards in a fair and random way. I personally have examined the log files of Odds On, Infinite Casino, and IQ Ludorum and found them to be fair. The laws of mathematics state that the more hands are dealt the more the actual return will approach the theoretical return. If you want to prove otherwise I would suggest keeping track of the cards and putting the results through statistical tests. See my blacklist for more about that.

Mabuhay!! Great site!!! Ive learned a lot from you! Had I not learned the math behind the casino games, I probably would be a compulsive gambler by now. I used to gamble to win, but after learning that one cannot beat the house, I learned how to play for fun. I’m not sure if you are familiar with Super 6. Its a commissionless baccarat that pays 1:2 on a winning 6. What is the house edge (for banker and player) on this? Also, there’s this side bet which pays 12-1 on a winning 6, would this be a sucker’s bet? Thanks.

Thanks for the nice words. I already address commission free baccarat in my baccarat section. Yes, the winning 6 is a sucker bet. The probability of the banker having a winning 6 is 5.39% and the probability of the player having a winning 6 is 6.26%. The house edge on the banker is 30.00% and on the player is 18.68%.

Dear sir, I’ve read your FAQs with great interest. I’ve a question of my own. In the game of baccarat how many times on average can you expect a B/P winning streak of nine times in a row. Can you show the mathematics of it. Thank you.

The probability of a single banker win is 0.50682483 and single player win is 0.49317517, ignoring ties. So the probability that the next 8 hands will be banker win, skipping ties, is 0.50682483^{8} = 0.004353746 . The probability of the same thing on the player is 0.49317517^{8} = 0.003499529.

Your site is amazing. Here’s my question. Does match play change basic strategy at all? My non-math-based instincts tell my that surrender becomes a bad idea, that is if you have to surrender your coupon.

Thanks for the compliment. You are right that you shouldn’t surrender if they take the match play away. There are some other strategy changes but I never worked out a list. Generally the casinos don’t allow doubling the match play chip, in which case you should be less inclined to double. ’Basic Blackjack’ by Stanford Wong indicates when to double if doubling the match play is allowed. My advice is to use the match play on the Player bet in baccarat.

Can you recommend a free baccarat game for the Mac?

My webmaster Michael Bluejay is a loyal Mac user and has a helpful page about Macintosh casino games.

Hello, wiz. Really great site. Thanks for all the valuable information that saves us readers countless money on sucker bets. The society needs more people like you to educate us common folks. I come from southern California, and instead of charging 5% on winning, the local casinos here charge commission by each hand you play ($1 for every $100 bet). My question is, what’s the house edge for both banker and player in this case?

Thanks for the compliment. As I state in my pai gow poker section the probability of a banker win is 29.98%, a player win is 28.55%, and a tie is 41.47%. So if you are charged 1% the expected return as banker in a head to head game would be .2998-.2885-0.01 = 0.0043, or a player advantage of 0.43%. As player the expected return is .2855-.2998-0.01 = -0.0243, or a house edge of 2.43%.

Recently I played at one MG casino (Viper version) High Limit Baccarat and by betting on Banker only, I get an awful result as follows:

Player 44 (64.7%)

banker 19 (27.9%)

Tie 5 (7.4%)

Total 68

What’s the chance of this happening? I appreciate your reply if you can, and hopefully with the formula so that I can calculate it myself next time.

It is bad practice to look back at past play and ask about the odds. Rather, I prefer to state a hypothesis and then gather data to prove or disprove it. However, if we must, I would phrase your question this way: "I played the banker bet 68 times and lost 25.95 units (44-0.95*19). What is the probability of losing this much or more?"

To answer this question we must first find the variance of a single bet on the banker. Here are the possible outcomes and their probabilities, as found in my baccarat section, based on the Microgaming single-deck rules.

Win: 45.96%

Loss: 44.68%

Push: 9.36%

So the variance on a single wager is .4596*(.95)^{2} + .4468*(-1)^{2} +.0936*0^{2} - (-0.010117)^{2}= 0.861468877.

The variance on 68 of these bets is simply 68 times the variance of one bet, or 68*0.861468877= 58.57988361. The standard deviation of the 68 bets is simply the square root of the variance, or 58.57988361^{1/2} = 7.653749644.

The house edge on the banker bet in a single deck game is 1.01%. So over 68 bets you could expect to lose .67 units. You lost 25.95 units, which is 25.28 more than expectations. So your results were 25.28/7.653749644 = 3.30 standard deviations below expectations. You then use a normal distribution table to find the probability of this. Excel has a feature to do this calculation, simply put: =normsdist(-3.30) in any cell and the result is 0.000483424, or 1 in 2069. So this is the probability of losing as much as you did or more. I appreciate that you didn’t make any accusations about foul play. However, if you had, I don’t think this rises to the level to prove anything. It could easily be explained as simple bad luck.

Hi, wiz. Love your site, please keep it up. I have 2 questions would like to ask.

1) Does card counting only work with blackjack? Is it useless or simply not as effective for other card games like baccarat?

2) In your blackjack card counting section, you mentioned that the Ken Uston’s Plus/Minus strategy counts 3-7 as small cards. Doesn’t it seem more reasonable to count 2-6 as small, and 7-9 as natural?

Thanks for the kind words. To answer question one, baccarat is not countable for all practical purposes. I have wondered about your second question myself. I used to use Uston’s Plus/Minus but switched to Wong’s Hi/Low. Looking back I don’t think Wong’s hi/low is much more powerful, but there is much more information about it. My blackjack appendix 7 shows that removing a 2 from each deck adds 0.39% to the player’s return and removing a 7 only adds 0.29%. So if you must track only one the 2 is better. The Knockout Count tracks both the 2 and 7. My opinion is if you haven’t taken up counting yet then the 2-6 Hi/Low is the marginally better way to go, however if you already use something else you should probably stick with that.

There is a story today about a British man who will bet his life savings on one roulette roll. My friend and I have been debating about what the best casino bet is for this type of wager. If you can only place one bet, and you wish to maximize your odds, what is the best game to play and what is the best bet?

First, let me say this guy was a fool. He bet $138,000 on a normal American roulette wheel which has two zeros and a house edge of 5.26%. This amounted to an expected loss of $7,263. However had he taken a 10 minute ride to the Bellagio, Mirage, or Aladdin he could have made the bet on a single zero wheel which follows the European rule of giving half an even money bet back if the ball lands in zero. He planned to make an even money bet anyway. So, at these wheels with full European rules his house edge would have been only 1.35%, for an expected loss of only $1865.

To answer your question, if forced to make just one even money type bet I would have chosen the banker bet in baccarat with a house edge of 1.06%.

Recently, I was watching an episode of the new a "high-roller" playing, I believe, blackjack. Apparently started to lose more and more, he would tear up the cards! I would have thought this a severe breach of etiquette, if not some actual gaming commission regulation, but when asked to stop, he was insulted that they would ask him! Is this sort of thing generally tolerated and I've just never seen it, or is this guy just used to being allowed to get away with that sort of thing because he's losing tons of money, or something else entirely?

Baccarat (at the big tables) is the only casino game in which players are allowed to damage the cards. An explanation I heard is that Asian players bend the cards anyway as they slowly peak at them that they only use each card once. Therefore as long as the dealer is replacing the cards after one usage the casino may as well let the players do anything with them. Being able to identify cards is of little value to baccarat players anyway because the dealer doesn't take a hole card (as the dealer does in blackjack) and the player has no choice as to whether to hit or stand. However, there are also gaming regulations that stipulate that the tapes must show all the cards in case of a dispute, which isn’t possible if the player tears them up first. In the show you mention the player didn’t know this and I think both parties handled it badly, which led to the hard feelings that the show captured. Had I been the casino manager I would have explained what I just said and then asked the player to lay the card face-up on the table before ripping it into tiny pieces.

On a related note yours truly will be on __The Casino__ sometime this season. The story is some college students try to parlay $1000 into $5000 as quickly as possible. They seek my advice on how to do achieve this goal.

Update: That episode never aired. Probably because of me.

In Holland there is a version of baccarat in which the banker bet pays even money, except a winning 5 pays 1 to 2. What is the house edge on this variation?

The house edge is 0.93%. More details can be found in my baccarat appendix 6.

Can you please, please, please put me out of my misery and answer a question that’s been plaguing me for months and I just can’t seem to find an answer to. I play baccarat mostly for leisure, and have created my own decision rules for when to bet Banker or Player, betting only 1 unit per hand (no betting systems for me). Out of curiosity I tried my decision rules on the both Zumma books (a total of 1600 shoes) and returned a tidy profit (betting an average of 60 hands per shoe). Zumma states that when betting this many hands, his shoes are enough to validate a strategy on a conceptual basis. Yet I’ve have read that 1600 shoes are not deemed a significant sample size given the large number of possible B/P combinations in a baccarat shoe. I thought about the effect of large populations when selecting sample size (where at a certain level increasing the population does not materially increase the required sample size), and after using different on-line sample size calculators, get around 2,400 shoes as being a significant enough sample for a population as large as the baccarat BP combinations (4,998,398,275,503,360 according to your calculations). So is it 2,400 or 1,000,000+? P.S. Learning so much from your site, definitely the best I’ve found.

There is no magic number of at which you enter the long run or to determine when a sample size is big enough to prove a hypothesis. It is always a matter of degree. However we can say the standard deviation of the sample mean is inversely proportional to the square root of the sample size. Your question is rather vague so let me rephrase it: what is the sample size required so that the sample mean will be within 1% of the actual mean with 95% probability? From my house edge section we see the standard deviation of the banker bet is 0.93 and of the player bet is 0.95. Since you go back and forth we’ll use the average of 0.94. Now I’ll wave my hands and get an answer of 33,944 hands. At 60 hands per shoe that comes to 566 shoes.

You say "No betting systems for me", but decision rules as to when to bet banker or player is definitely a betting system. But I’m still skeptical that you return a tidy profit over 1600 shoes.

On baccarat, are the odds perpetual (as in dice and roulette) or do the odds change as cards are dealt out of the shoe (as in blackjack)? I know that it is not at all probable, but is it mathematically possible for the Banker to win every single hand in the baccarat shoe?

In an effort to debunk betting systems I used to say that the past does not matter in gambling. However once in a while somebody would rebuke me by saying that the past does matter for card counters, which is true. So now I say that in games of independent trials, like roulette and craps, the past does not matter. As I show in my baccarat appendix 2 a shoe rich in small cards favors the player and a shoe rich in large cards favors the banker. Thus, in baccarat, there is an extremely slight disposition that the next outcome will be the opposite of the last. So, yes, the odds do change in baccarat as the cards play out, but only to a very small extent. For all practicable purposes the game is not countable. I do not know if the banker could win every hand but I speculate that the answer is yes.

Dear Mr. Wizard, If you had $5,000 to bet and wanted to win only $200 what game would you play? Please assume European rules and choose only among roulette, black jack, or baccarat.

I would bet $200 on the player bet in baccarat. If it wins, walk, if it loses then bet $400 (or whatever you lost). Then just go into a Martingale until you win your $200 or lose your entire $5,000.

I am just learning how to play Baccarat and since every player can bet on either player and banker and are not really playing each other, I was wondering what game is played in the James Bond movies? For example, In Dr. No it seems as if Bond is against a woman and he is winning her money? Is there something I am missing or is it a different game? Thank you for your time.

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.

I just read your answer about baccarat as played in 007 movies, and I would like to let you know that in South America with a point of 5 the player can choose taking a hit or not. As this option should be made prior the bank showing his card, only a fool would take a hit, since in this situation there are 4 cards that favors the player and 5 that hurts him. Best regards from your loyal fan

Thanks for your comments. I just watched the scene in question from For Your Eyes Only several more times and am still not sure what is going on. It doesn’t help that the dealer giving the running commentary is doing so in French. It also doesn’t help that the table is mostly plain, like a poker table, unlike an American table where you can tell a bet by its location.

We see Bond dealing the cards but an unseen dealer is paying players. Bond is apparently betting the opposite of what the only other bettor at the table is doing. In the first hand the other character turns over a 2-card natural 8, Bond turns over a 2-card 5, and Bond wins the hand. This would imply that the other player bet on the banker hand, and thus Bond on the player hand. In the second hand the other bettor increases his bet from half a million to on million, at the goading of his wife. After receiving his first two cards he requests a third. Bond turns over his two cards, revealing a face card and a 5, and gives the other bettor a third card. The other bettor’s cards are not turned over yet but he seems pleased with his hand. Then a third character, who just walked up, comments to Bond, "The odds favor standing pat." However Bond takes a card anyway, which is a 4, for a total of 9. The other player storms off without turning over his cards.

This is consistent with what you said, except Bond is acting last, or as the banker. I tend to think the American makers of the movie didn’t understand European baccarat rules and incorrectly gave the banker the free will take card a card, as opposed to the player. It certainly wouldn’t be the first time a gambling scene was depicted incorrectly in the movies. I have seen numerous card counting scenes in the movies and television, and yet to find anything close to being realistic.

I agree that if given the choice the odds favor standing on 5 as the player. Assuming the banker rules are the same either way then if the player stands on a 5 the following is the house edge per bet, based on an 8-deck game.

### Player Hits 5

Bet |
House Edge |

Banker |
0.79% |

Player |
1.52% |

Tie |
17.27%. |

So if the player consistently hits on 5 the house edge goes up by 0.29% on the player bet. The player will get a 5, while the dealer does not have a natural, 9.86% of the time, for a cost per 5 of 2.94%.

I looked in your baccarat appendices and did not find odds for a "Super 6" that paid 12-to-1 as found the Casino Filipino Online. In sum they state: In general, winning hands are paid even money. If the final count of the Banker is 6 while Player has less than 6, bets on the Banker and Super 6 win and the game is a Super 6. In this case, Banker bets are paid 0.5-to-1, while Super 6 bets are paid 12-to-1. The Super 6 bet is based on the proposition that a particular deal would result in a Super 6. Bets on the Super 6 lose when the deal results in a Draw, or any other outcome that is not a Super 6. I was wondering how bad this game could be. Thanks.

The same game without the side bet was once played in Atlantic City and is analyzed in my baccarat appendix 6. There it show that the probability of a Super 6 is 5.3864%. So the house edge at 12 to 1 would be 29.977% (ouch!).

I was playing baccarat online at USACasino which uses a live dealer and Playtech software. Apparently Playtech has instituted a new rule that the dealer burns a card after each hand is dealt. This is not the way it is played in casinos. What bearing, if any, does this have on the odds of the game. I can’t believe the casino would institute a new rule that wasn’t in their favor.

Burning cards has no effect on the basic strategy player. They probably are doing this to discourage card counters. However, they may as well just shuffle earlier. For card counting purposes what is important is the number of cards seen, it doesn’t matter whether the unseen cards are burned or behind the cut card.

Love the site. Regarding Baccarat, what is the house edge if one bet’s less than $5.00 per hand on the Banker? Some mini-baccarat tables used to have a $3.00 minimum. The commission would be calculated as 5% of the nearest $5.00 (so, $0.25 on a $3.00 wager, rather than $0.15.) Thanks for your consideration.

Thanks. Paying a 25-cent commission on a $3 bet amounts to an 8.33% commission. Assuming you only play as the player, you will win both bets 28.61% of the time. So the normal cost of the 5% commission rule is 0.2861×0.05=1.43%. The losing on copies rule costs the player 1.30%, for a total house edge of 1.43%+1.30% = 2.73% normally. In the case of this game, the cost of the commission is 0.2861×0.0833=2.38%. So the total house edge is 2.38%+1.30%=3.68%.

In baccarat, how much edge do a player get if the first card showing (accident by the dealer) is 9. I saw it happens twice in the casino and everyone bet a huge amount (included me) on player hand and lose both time. My basic math tell me that 50% of the time the player hand will stand on 6 points or higher and will have a about 70 percent chance of winning. Is it a good time to max out my bet? Is it just bad luck or the advantage is not that great for an all in bet?

I hope you’re happy, I just made an entire section to answer this question on flashing dealers in baccarat.

I was recently presented with numbers for the house advantage for Baccarat (Banker=1.17%; Player=1.36%). The problem is the calculations were done without taking the tie wager into consideration. In the past I had seen these numbers in a book though I don’t remember where. My question is, why would someone do the calculations like that? Is there some reason I am missing? It seems to me that this is a flawed method for presenting the house advantage. What is the best way to present to them how/why this is flawed (if it indeed is flawed).

The reason some sources differ on the house edge in baccarat has mostly to do with how the house edge is defined. I prefer to define the house edge as the ratio of the expected casino win to the initial wager. Other gambling writers define it as the ratio of the expected casino win to bet resolved. The difference is in whether or not ties are considered as a possible outcome. In an eight-deck game the following are the probabilities in baccarat:

- Banker wins: 45.8597%
- Player wins: 44.6274%
- Tie wins: 9.5156%

Here is how I calculate the expected return on each bet by counting ties.

- Banker: 0.458597*0.95 + 0.446274*-1 + 0.095156*0 = -0.010579
- Player: 0.458597*-1 + 0.446274*1 + 0.095156*0 = -0.012351
- Tie: 0.458597*-1 + 0.446274*-1 + 0.095156*8 = -0.143596

So I get a house edge of 1.24% on the player, 1.06% on the banker, and 14.36% on the tie.

Other gambling writers prefer to think of ties as a non-event, in other words leaving the bet up until it is resolved. The probability of a banker or player win is 45.8597% + 44.6274% = 90.4844%. The probability the next bet resolved will be a player win is 44.6274%/90.4844% = 49.3175%. The probability the next bet resolved will be a banker win is 45.8597%/90.4844% = 50.6825%.

The way the other camp would calculate the expected return on the player bet is 49.3175%*1 + 50.6825%*-1 = -1.3650%. The expected return on the banker bet, ignoring ties, is 49.3175%*-1 + 50.6825%*0.95 = -1.1692%. Thus the house edge ignoring ties is 1.36% on the player and 1.17% on the banker.

One reason I think counting ties is appropriate is that it gives the player an accurate measure of expected losses over time. For example if a player bet $100 a hand on the banker in baccarat for 4 hours, and the casino’s average rate of play was 80 hands per hour, then the expected player loss is $100*4*80*0.0106=$339.20. No need to worry about the probability of a tie in the calculation. If a casino used the 1.17% house edge for the banker it would be overestimating expected loss, and perhaps over-comp the player as a result.

Another reason I count ties is all the major blackjack and video poker experts count ties in the analysis of those games. For example if you ignored ties in 9/6 Jacks or better, when getting a pair of jacks to aces, then the return would be 99.4193%. Never once have I seen such a figure quoted for 9/6 jacks; it is firmly held that it is 99.5439% with optimal strategy.

Finally, here is a table of some gambling books and the figures used for baccarat.

### House Edge in Baccarat

Book | Author | Copyright | Player | Banker |

Casino Operations Management | Jim Kilby & Jim Fox | 1998 | 1.24% | 1.06% |

The Casino Gambler’s Guide | Allan N. Wilson | 1965, 1970 | 1.23% | 1.06% |

Smart Casino Gambling | Olaf Vancura, Ph.D. | 1996 | 1.24% | 1.06% |

The American Mensa Guide to Casino Gambling | Andrew Brisman | 1999 | 1.24% | 1.06% |

Casino Gambling for Dummies | Kevin Blackwood | 2006 | 1.24% | 1.06% |

Scarne’s New Complete Guide to Gambling | John Scarne | 1961, 1974 | 1.34% | 1.19% |

The New American Guide to Gambling and Games | Edwin Silberstang | 1972, 1979, 1987 | 1.36% | 1.17% |

Casino Gambling: Play Like a Pro in 10 Minutes or Less | Frank Scoblete | 2003 | 1.36% | 1.17% |

Beating the Casinos at Their Own Game | Peter Svorboda | 2001 | 1.36% | 1.17% |

The Complete Idiot’s Guide to Gambling Like a Pro | Stanford Wong & Susan Spector | 1996 | 1.36% | 1.17% |

Casino Math by Robert C. Hannum and Anthony N. Cabot lists the house edge both ways.

I have just returned from a gambling trip to Niagara Falls, Ontario. Interestingly, Casino Niagara has a Mini-Baccarat table (a 9-seater) where the bank wager is rounded down to the nearest $20 for the purpose of calculating the commission. Thus, a $35 winning bet is only charged a $1 commission. This places the commission percentage on that wager at 2.86%! If I am not mistaken, this means that there is no house edge on bank wagers, but actually a player edge! Do you agree?

As I quoted in an earlier question the probabilities in 8-deck baccarat are:

- Banker wins: 45.8597%
- Player wins: 44.6274%
- Tie wins: 9.5156%

So the expected value on the banker bet is 45.8597%*(1-(1/35)) + 44.6274%*-1 = -0.00075. So the house still has an edge of 0.075%. The breakeven commission on the banker bet is 2.693%. If you could bet $37.14 the odds would swing to your favor.

Some of the online casinos such as Bodog pay 9 to 1 for the tie bet in baccarat. What is the house edge for the tie bet with the 9 to 1 payout?

Yes, Bodog does indeed pay 9 to 1 on the tie. Assuming eight decks, that lowers the house edge from 14.360% to 4.844%.

First of all, thanks for the wonderful site. I recently saw a set of bonus bets in Baccarat called 4-5-6, on the total number of cards between the player and banker hands. The odds they offer at the Atlantic City Hilton are 3 to 2 for 4 cards, 2 to 1 for 5 cards, and I believe 3 to 1 for 6 cards if I remember correctly. That means we should see more hands that end with 4 cards. What are the odds on all three bets?

You’re welcome, thanks for the compliment. Without knowing anything about the probabilities, if those were the payoffs, then there would exist a player advantage on at least one bet. The way you can tell is to take the sum of 1/(1+x), where x is what the bet pays on a "to one" basis, over all the bets. If this sum is less than 1, then at least one bet has a player edge. In this case, according to your odds, this sum would be 1/2.5 + 1/3 + 1/4 = 0.9833. This trick may come in handy, for example, if you see an amateur putting up sports betting futures.

What is probably the case here is that six cards pays 2 to 1. Based on that assumption, and six decks, the house edge is 5.27% on four cards, 8.94% on five cards, and 4.74% on six cards. For more information see my baccarat appendix 5.

For baccarat, if you found a casino that allowed a player to bet both player and banker at the same time, is there any advantage to do that? How about if they rated you for the total of both bets? (I.E. - bet $25 on Banker, $25 on Player and be rated for $50)

I asked Barney Vinson this question, author of Ask Barney: An Insider’s Guide to Las Vegas. He said the casino would likely only rate one of the bets, in your case $25. An advantage to doing that is that it certainly lowers risk. This might be a good play if you needed to put in a lot of action, for example to qualify for an event you were invited to, and didn’t have much money to lose. However I think that if large bets were involved ($100 or over) it would set off a red flag, and you probably wouldn’t be invited to the next event.

Great site! My question is related to the game of baccarat. In baccarat the casino typically offers somewhere in the region of 20 to 40 times between the minimum and the maximum bet. If a casino, for promotion purposes, decides to offer a bigger spread between min/max and for example offer 100 times between the min/max, what advantage does the player get from such promotion and what mathematical disadvantage or risk, if any, is the Casino taken from such promotion?

I have answered this about roulette, and my answer is the same in baccarat. The house edge is exactly the same regardless of the bet spread allowed. I have asked casino executives a few times why they keep the spread as small as they do. For example, if the casino will take a $150,000 baccarat bet in the high-limit room, why a $5,000 maximum in the main casino? The consensus answer is that casinos like to corral their big players into the high limit areas. The reason I get for this is the service and game security is better in those areas. The reason is definitely not to foil system players.

All Books I have seen regarding Baccarat say the banker wins more often than the player. None of them explain why this is. Common sense would make it seem that each side has an equal chance over the long run. Your explanation will be appreciated.

Briefly, it is because the banker gets to act last. If the player got a third card that is likely to help, the banker will hit. If the player’s third card is likely to make the player hand worse, the banker will stand.

On two recent visits to the baccarat tables the results were definitely player biased. Please tell me if these results would be considered within two standard deviations of the expected results for bank and player. I have eliminated tie hands.

Session I

Player wins: 282

Banker wins: 214

Session II

Player wins: 879

Banker wins: 831

From my baccarat page, we see the probabilities in the usual 8-deck game are:

Banker: 45.86%

Player: 44.62%

Tie: 9.52%

Skipping the ties, the probabilities for the banker and player are:

Banker: 45.68%/(45.68%+44.62%) = 50.68%.

Player: 44.62%/(45.68%+44.62%) = 49.32%.

The total number of hands in session I was 282+214 = 496. In session I the expected number of player wins is 49.32% × 496 = 244.62. The actual total of 282 exceeds expectations by 282-244.62 = 37.38.

The variance for a series of win/lose events is n × p × q, where n is the number is the sample size, p is the probability of winning, and q is the probability of losing. In this case, the variance is 496 × 0.5068 × 0.4932 = 123.98. The standard deviation is the square root of that, which is 11.13. So, the total player wins exceeded expectations by 37.38/11.13 = 3.36 standard deviations. The probability of results that skewed, or more, is 0.000393, or 1 in 2,544.

Using the math method for sample II, the probability is 0.042234. If you combine the two samples into one, the probability is 0.000932. About 0.1% is not enough to be "definitely player biased." If you still think the game isn’t fair, I would collect more data, for a larger sample size.

I was playing a baccarat game in Asia, where the house paid 150 to 1 for bets on a 1 to 1 tie. What are the odds on that bet? Your site rocks, thanks for all the great work.

Thank you for the kind words. My baccarat appendix shows the probability of a 1-1 tie is 0.004101. That is the most infrequent outcome in baccarat, I might add. Fair odds would be (1/p)-1 to 1, where p is the probability of winning, which comes to 242.84 to one. An easy formula for the house edge is (t-a)/(t+1), where t is the true odds, and a is the actual odds. At an actual win of only 150 to 1, the house edge is (242.84-150)/(242.84+1) = 38.1% (ouch!).

Hi, Wizard. Let’s say I have $300 to gamble with, and can accept a 25% risk of ruin. What should I do to maximize my upside? Thanks!

I would make the banker bet in baccarat. My betting advice would be to do what is known as a two-step progression. First, bet 1/3 of your bankroll. If that wins, walk away. If that loses, then bet the other 2/3. Again, if you win, walk. With any tie, just bet again until the bet is resolved. Here are the probabilities in baccarat:

Banker: 45.86%

Player: 44.62%

Tie: 9.52%

The probability of a banker win, given that the bet is resolved is 45.86%/(45.86%+44.62%) = 50.68%. The probability of losing both steps of the progression is (1-0.5068)^{2} = 24.32%. The banker bet pays 19 to 20, so you will have a 75.68% chance of winning $95 or $90 (depending on whether you win on the first or second bet), and a 24.32% chance of losing $300.

In baccarat, the cut card is placed in front of the last 13 cards in the shoe, and one hand is dealt following the hand the cut card came out on. If the cut card came out after the first player card was dealt, and both the player and banker draw a card, only 8 cards will remain in the shoe for the last hand. If you are tracking the cards, and know the last 8 cards are all 0-value cards, a table max bet on tie would net you a huge profit. My question is, what are the odds that the last 8, 9, or 10 cards in an 8 deck shoe are all 10 value? Also if you knew exactly what the last 8 cards were, could you use a formula or program to figure the odds that the next hand will be banker, player, or tie?

To answer your first question, the probability that the last 8 cards in an 8-deck shoe are all 0-valued cards is combin(128,8)/combin(416,8) = 0.0000687746. So, it isn't something to wait around for. I know of no easy formula for what to bet in other situations. If you could find a casino that would allow you to use a computer, the advantages would sometimes be huge towards the end of the shoe, especially on the tie.

Why do casinos burn cards in blackjack and baccarat?

A minor reason is to foil card counters. However, instead of burning x cards, the dealer could move the cut card x cards forward, and achieve the same purpose. The major reason is game protection. For one, the player might catch a glimpse of the top card, and alter his bet and strategy, based on this information. Such a tactic would not be cheating, I might add. The top card is also vulnerable to lots of cheating schemes. It could be marked, the dealer could peek at it, or force a desired card to the top. If for any reason the dealer knew what the top card was, he could signal that information to a confederate player, giving him a huge advantage.

I know the commandment to not make side bets. However, I have seen a side bet in blackjack that pays 11 to 1, if the player has a pair in his first two cards. Would it be possible using a count system to gain an advantage?

It sounds like you are talking about Lucky Pairs, a side bet that wins if the player’s first two cards are a pair. Many baccarat tables also offer this bet. As I show in my baccarat page, the house edge is 10.36%, assuming eight decks. In either game, you would pretty much need to eliminate all cards of at least one rank to have an advantage. To know that, you would need to keep 13 different counts. In baccarat, this could be done, since you are allowed to take notes while you play. However, based on some very extensive analysis, profitable opportunities don’t happen often enough for this to be a practical use of time.

There is a no-commission baccarat game here that pays 1 to 2 on every banker win of seven, except if the player also has four points, it is paid 2 to 1. Do I get better odds if I play this game, or the no-commission baccarat that pays 1 to 2 on banker win on six?

The house edge under the first set of rules is 1.23%. The house edge of the second set of rules is 1.46%. So the first version is the better bet.

I was wondering if there is a way to calculate on average how many trials it would take to lose ten units from any given point, betting on Player in baccarat.

The house edge on the Player bet is 1.2351%, assuming eight decks. The expected number of hands it takes to have a loss of ten units is 10/0.012351 = 809.66.

A 60 Minutes interview of Steve Wynn featured the following exchange:

Charlie Rose: You have never known, in your entire life, a gambler who comes here and wins big and walks away?

Steve Wynn: Never.

CR: You know nobody, hardly, who over the stretch of time, is ahead?

SW: Nope.

I find this hard to believe. What are your thoughts?

I personally know lots of professional gamblers up on the Wynn. However, I’m sure that none of them have met Steve Wynn personally. I would imagine that only the super whales are granted an audience with him, and such whales are usually superstitious (i.e. losing) baccarat players. Most heavy recreational gamblers do lose over the long run. However, if Mr. Wynn believes that nobody is up on him, I would invite him to repeat the triple-points promotion he ran Labor Day weekend 2007. Even if the promotion loses money, surely the foolish players will give it back eventually.

In London, there is a royal match side bet in baccarat. It pays if the Banker or Player get a king and queen in the first two cards. Do you have any odds on it?

Assuming eight decks, the house edge is 4.5%. For more information, visit my page on baccarat side bets.

How high can a person off the street bet in the big high roller lounges?

Las Vegas casinos are surprisingly risk averse; they don’t like taking big bets. For customers off the street, the biggest bet a nice casino will take is usually $150,000 in baccarat, on player or banker. In other traditional table games, the limit is usually $10,000. Limits can be raised upon request by known customers.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

I was recently playing in a casino in London where they offer the "Egalite" total side bets in baccarat. These are bets that the banker and player tie at a specific total. Here are the payoffs:

- 0=150-1
- 1=215-1
- 2=225-1
- 3=200-1
- 4=120-1
- 5=110-1
- 6=45-1
- 7=45-1
- 8=80-1
- 9=80-1

The house edge ranges from a low of 6.39% on a 7 tie, to a high of 12.45% for a 0 tie. I show the probability of winning and expected return for all ten bets on my baccarat page.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

There is a well-known story about a freeze-out competition between a Japanese high roller, Kashiwagi, and Donald Trump, that took place 20 years ago. Kashiwagi was not allowed to play more than $200K per hand at baccarat. The game would be over when either the casino or the player was ahead by $12 millon. Assume that Kashiwagi always bet the maximum on Banker. What is the probability that Kashiwagi will win?

The math works out more easily if he bet on the Player. I work out a similar problem in roulette at my mathproblems.info site, problem number 116. For even money bets, the general formula is ((q/p)^{b}-1)/((q/p)^{g}-1), where:

b = starting bankroll in units.

g = bankroll goal in units.

p = probability of winning any given bet, not counting ties.

q = probability of losing any given bet, not counting ties.

Here the player starts with $12 million, or 60 units of $200,000, and will play until reaches 120 units or goes bust. So in the case of the Player bet the equation values are:

b = 60

g = 120

p = 0.493175

q = 0.506825

So the answer is ((0.506825/0.493175)^{60}-1)/(( 0.506825/0.493175)^{120}-1) = 16.27%.

It is much more complicated on the Banker bet, because of the 5% commission. That would result in the distinct possibility of the player overshooting his goal. If we add a rule that if a winning bet would cause the player to achieve his goal, he could bet only what was needed to get to $12 million exactly, then I estimate his probability of success at 21.66%.

A simpler formula for the probability of doubling a bankroll is 1/[1+(q/p)^{b]. }

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

I heard that blackjack pioneer Ed Thorp also had a card counting strategy for beating baccarat. What do you know about it?

I found two sources online that address your question. The first is a quote from an article I found:

But Edward Thorp and his computer are not done with Nevada yet. The classiest gambling game of all — just ask James Bond — is that enticing thing called baccarat, or chemin de fer. Its rules prevent a fast shuffle, and there is very little opportunity for hanky-panky. Thorp has now come up with a system to beat it, and the system seems to work. He has a baccarat team, and it is over $5,000 ahead. It has also been spotted and barred from play in two casinos. Could it be bye-bye to baccarat, too? —Sports Illustrated, January 13, 1964 issue

Thorp also addresses the vulnerability of baccarat to card counters in his book The Mathematics of Gambling. The link goes to a free online copy. Thorp concludes by saying:

Practical card counting strategies are at best marginal, and at best precarious, for they are easily eliminated by shuffling the deck with 26 cards remaining.

Interestingly, Thorp also says the tie bet pays 9 to 1. Perhaps that rule was more common in 1985, when the book was published. If memory serves me correctly, Binion’s paid 9 to 1 until the late 90’s.

My own analysis points to the same conclusion, although I studied the tie bet with an 8 to 1 win. I find the pair bets that some casinos now offer have the greatest vulnerability, but are still not a practical advantage play.

I asked Don Schlesinger about the apparent contradiction and Thorp’s baccarat team. Don said that he believed that Thorp did indeed have a team trying to exploit the tie bet. Either Thorp’s team found games with a cut deeper than 26 cards, or he had a change of opinion about it sometime between 1964, the date of the SI article, and 1985, when The __Mathematics of Gambling__ was published.

I follow Macau casino stocks and they often quote the casino's theoretical win percentage at VIP Baccarat to be 2.85% of dollars wagered. This number is used by pretty much everyone to forecast earnings for the companies. I was wondering how they calculate this number, and whether this is accurate in your opinion.

That is likely a weighted average of all four types of bets on the table. Most of the money is bet on the Player and Banker, with a house edge of 1.24% and 1.06% respectively. However, the Tie and Pair bets carry much higher house edges of 14.36% and 10.36% respectively. Players apparently are betting a little on this to increase the overall win percentage to 2.85%.

The table below shows a hypothetical mix of bets that arrive at the overall Macau Win Percentage, ignoring the issue of Dead Chips.

### Macau Baccarat — Weighted House Edge

Bet | House Edge | Ratio of Bets | Expected House Edge |
---|---|---|---|

Player | 0.012351 | 43.25% | 0.005342 |

Banker | 0.010579 | 43.25% | 0.004575 |

Tie | 0.143596 | 11.50% | 0.016514 |

Pairs | 0.103614 | 2.00% | 0.002072 |

Total | 100.00% | 0.028503 |

Do you have any tips to remember the drawing rules for the Banker in baccarat?

Yes. Remember these numbers: 8, 27, 47, 67. Here is what they mean.

- If the Banker's total is 3, and the Player draws anything except an 8, then Banker draws.
- If Banker's total is 4, then the Banker draws against a Player third card of 2 to 7.
- If Banker's total is 5, then the Banker draws against a Player third card of 4 to 7.
- If Banker's total is 6, then the Banker draws against a Player third card of 6 to 7.

I think I have a winning betting system. However, I need more than the 3,000 baccarat shoes that you have on your baccarat page to test it. Can you make more?

I hope you're happy; I just created a quarter million of them. Better to find out your system will fail eventually, which they all do, for free than with real money in the casino.

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

According to your review of Gamesys N.V. software, the Player bet in baccarat pays 1.0282 to 1. You note the player advantage is 0.02%. If we ignore the 24-hour time limit rule, is there a way to have an advantage on this bet, even after the 10% commission on net gambling session wins?

Yes! Keep playing, betting the same amount every time, until you are up any amount of money. Then quit, wait 24 hours, and repeat.

To be specific, the advantage per bet is 0.0233341%. The overall advantage following this strategy is 90% of that, or 0.0210007%.

There may be other equally good strategies but if anyone has a superior strategy, I'm all ears.

I once saw 49 consecutive baccarat hands with 48 Player wins, not counting ties. What is the probability of that per shoe?

The average shoe has 80.884 total hands. The probability of a Tie is 0.095156, so if we take those out we can expect 73.18740 hands per shoe, not counting ties. /p>

The probability of any 49 consecutive hands, not counting ties, having 48 Player wins is 1 in 21,922,409,835,345. However, there are 25.1874 possible starting points for these 49 hands, to make estimate. Thus, the probability of seeing the aforementioned event in a shoe is 1 in 870,371,922,467. This is not a hard and fast answer, but what I feel is a very good estimate.

If a casino increased the win on the Tie bet to 9 to 1, above the usual 8 to 1, how much additional wagering would it need on the Tie to have the same expected win?

The probability of a tie in baccarat is 0.095155968.

At the usual win of 8 to 1, the expected return to the player is 0.095156 × (8+1) - 1 = -0.143596.

At the a win of 9 to 1, the expected return to the player is 0.095156 × (9+1) - 1 = --0.048440.

The expected player loss is 0.143596/0.048440 = 2.9643960 times higher at a win of 8 to 1. Thus, the casino would need 2.9643960 times as much action on the Tie if they increased the win to 9 to 1 for the expected casino win to be the same.

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

What is the probability of a four-, five-, and six-of a kind in baccarat, combining all the cards that were dealt between the Player and Banker hand, for all 13 ranks?

The following table shows the number of permutations for all four-, five-, and six-of a kinds in baccarat by rank out of a possible 4,998,398,275,503,360 permutations.

### Keno 4-6 of a Kind Permutations in Baccarat

Rank | 4 of a Kind | 5 of a Kind | 6 of a Kind |
---|---|---|---|

Ace | 1,174,231,511,040 | 40,210,759,680 | 652,458,240 |

2 | 1,130,651,443,200 | 36,344,340,480 | 652,458,240 |

3 | 840,162,535,680 | - | - |

4 | 431,482,026,240 | - | - |

5 | 1,201,241,210,880 | 43,303,895,040 | 652,458,240 |

6 | 1,079,228,067,840 | 40,210,759,680 | 652,458,240 |

7 | 986,765,414,400 | 30,158,069,760 | 652,458,240 |

8 | 502,955,546,880 | - | - |

9 | 230,538,696,960 | - | - |

10 | 1,174,176,276,480 | 40,210,759,680 | 652,458,240 |

Jack | 1,174,176,276,480 | 40,210,759,680 | 652,458,240 |

Queen | 1,174,176,276,480 | 40,210,759,680 | 652,458,240 |

King | 1,174,176,276,480 | 40,210,759,680 | 652,458,240 |

Total | 12,273,961,559,040 | 351,070,863,360 | 5,872,124,160 |

The following table shows the probability for all four-, five-, and six-of a kinds in baccarat by rank.

### Keno 4-6 of a Kind Probabilities in Baccarat

Rank | 4 of a Kind | 5 of a Kind | 6 of a Kind |
---|---|---|---|

Ace | 0.0002349216 | 0.0000080447 | 0.0000001305 |

2 | 0.0002262028 | 0.0000072712 | 0.0000001305 |

3 | 0.0001680864 | 0.0000000000 | 0.0000000000 |

4 | 0.0000863241 | 0.0000000000 | 0.0000000000 |

5 | 0.0002403252 | 0.0000086636 | 0.0000001305 |

6 | 0.0002159148 | 0.0000080447 | 0.0000001305 |

7 | 0.0001974163 | 0.0000060335 | 0.0000001305 |

8 | 0.0001006233 | 0.0000000000 | 0.0000000000 |

9 | 0.0000461225 | 0.0000000000 | 0.0000000000 |

10 | 0.0002349105 | 0.0000080447 | 0.0000001305 |

Jack | 0.0002349105 | 0.0000080447 | 0.0000001305 |

Queen | 0.0002349105 | 0.0000080447 | 0.0000001305 |

King | 0.0002349105 | 0.0000080447 | 0.0000001305 |

Total | 0.0024555789 | 0.0000702367 | 0.0000011748 |

What is the probability of seeing eight 30-1 wins on the Dragon Bonus in baccarat in a single shoe?

For the benefit of those not familiar with the Dragon Bonus, it is a side bet that pays based on the margin of victory in baccarat. It pays 30 to 1 for a margin of victory of 9 and the nine-point hand was not a natural.

The average number of hands in a baccarat shoe is 80.884, but let's just say 81 to make the math easier.

Using the binomial formula, the following table shows the probability of 0 to 10 30-1 wins in 81 hands.

### Dragon Bonus 30-1 Wins per Shoe

Wins | Probability |
---|---|

10 | 0.0000000002 |

9 | 0.0000000047 |

8 | 0.0000000857 |

7 | 0.0000013607 |

6 | 0.0000186536 |

5 | 0.0002163066 |

4 | 0.0020630955 |

3 | 0.0155401441 |

2 | 0.0866801257 |

1 | 0.3182950376 |

0 | 0.5771851856 |

Total | 1.0000000000 |

The table shows the probability of exactly eight 30-1 wins is 0.0000000857, or 1 in 11,670,083.

The probability of eight or more wins is 0.0000000907, or 1 in 11,029,777.

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

I have seen rolling-chip commissions as high as 2.4% in Asian casinos. What is the house edge at that commission?

As a reminder, the way these programs work is the player buys non-negotiable chips with cash. These are the use-until-you-lose type of chips, which are usually called "dead chips" in Macau. Wins are paid in cashable chips. After all the non-negotiable chips have been played through, the player will be given a commission, based on the original buy in. I assume the commission itself is paid in non-negotiable chips as well. The commission can also be paid up-front, for which the math is still the same.

To answer your question, let's review baccarat probabilities. Here is the probability of each outcome.

- Banker wins = 0.458597423
- Player wins = 0.446246609
- Tie wins = 0.095155968

Let's look at the Banker bet. The number of times the player must wager on the Banker bet, on average, before a loss is 1/0.446246609 = 2.240913385 bets.

The expected win the Banker bet is 0.95*0.458597423 - 0.446246609 = -0.01057900.

The expected cost to play off a non-negotiable chip is 0.01057900 × 2.240913385 = 0.02370675 units.

Assuming the player gets an extra 2.4%, the value of the chip is (1+0.024) × (1-0.02370675) = 0.02343104.

Overall, the house edge to the player is the cost to play less the expected value of the promotion. This is 0.02370675 - 0.02343104 = 0.00012304.

So, the house edge is 0.01%.

Using the same logic, the house edge on the Player bet with a 2.4% commission is 0.164089%.

The break-even commission on the Banker bet is 2.4282409%.

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

I hear some casinos in Singapore are using ten decks, as opposed to eight, in Singapore. How does that affect the odds?

Going from eight to ten decks is very marginally good for the Player bet, decreasing the house edge by 0.0014%, from 1.2351% to 1.2337%. It is slightly bad for the Banker bet, increasing the house edge by 0.0012%, from 1.0579% to 1.0591%.

It has a bigger benefit on the Tie bet, decreasing the house edge by 0.0477%, from 14.3596% to 14.3119%.

However, the biggest benefit is on the Player and Banker Pair bets, decreasing the house edge by 0.5349%, from 10.3614% to 9.8266%.

For more information, please see my page on ten-deck baccarat.