Baccarat - FAQ
Jon from Danville, New Hampshire, USA
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.
Gavin P. from Bury, St Edmunds, England
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.
Michael from Sand Springs, Oklahoma
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.
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.
Steven from Miri, Malaysia
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%.
Chris from Gaithersburg, Maryland
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.
Herman from Manila, Philippines
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.
Gil from Fullteron, California
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.
Brantley from Waycross, Georgia
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.
Michael from Fort Worth, Texas
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.
Mandy from Gold Coast, Australia
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.
Marcio from Sau Paulo, Brazil
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.
Richard from Glendora, USA
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|
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.
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%.
Ted from Las Vegas, USA
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.
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.
Dave from Port St. Lucie
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.
Emi from Manila, Philippines
(82/combin(416,2))* (72/combin(414,2)) = 0.00000043, or 1 in 2308093
Emi from Manila, Philippines
There are lots of them, and they are all worthless.
Andrew from Maitland, Canada
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.
Charline from Las Vegas
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.
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) =
So the answer is 92.49%.
Clint from Singapore
The house edge on the banker bet is 4.07%.
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.
Thefamousv from Manila
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%.
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 0.506824838 = 0.004353746 . The probability of the same thing on the player is 0.493175178 = 0.003499529.
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.
My webmaster Michael Bluejay is a loyal Mac user and has a helpful page about Macintosh casino games.
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%.
Player 44 (64.7%)
banker 19 (27.9%)
Tie 5 (7.4%)
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.
So the variance on a single wager is .4596*(.95)2 + .4468*(-1)2 +.0936*02 - (-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 58.579883611/2 = 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.
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.
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%.
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.
The house edge is 0.93%. More details can be found in my baccarat appendix 6.
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.
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.
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.
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.
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
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%.
Bob from Largo
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!).
Phil from Yonkers
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.
Gary from Albuquerque
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%.
Davis Q from San Diego
I hope you’re happy, I just made an entire section to answer this question on flashing dealers in baccarat.
Bill from Las Vegas
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
|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.
Darryl from Longueuil, QC
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.
Bryan from Mill Valley
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%.
Ray from Egg Harbor Township
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.
William R. from Las Vegas
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.
Frank from Copenhagen
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.
Al from Mississauga, Ontario Canada
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.
Player wins: 282
Banker wins: 214
Player wins: 879
Banker wins: 831
Arthur from Wayne, New Jersey
From my baccarat page, we see the probabilities in the usual 8-deck game are:
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.
Jim from Las Vegas
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!).
Jerry T. from Hertford
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:
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.
Mike S. from Michigan City
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.
Matt from Fort Myers, FL
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.
Brian from Las Vegas
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.
Raul from Manila, Philippines
J.J. from Oceanside, CA
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.
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?
I find this hard to believe. What are your thoughts?
Andrew from Fort Wayne, IN
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.
Assuming eight decks, the house edge is 4.5%. For more information, visit my page on baccarat side bets.
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.
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.
Tom from Hong Kong
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.
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|
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.
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?
Mdawg from California
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.
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|
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|