# Baccarat - FAQ

^{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.

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.

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.

^{2}/combin(416,2))* (7

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

^{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%.

^{8}= 0.004353746 . The probability of the same thing on the player is 0.49317517

^{8}= 0.003499529.

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.

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.

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?

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%.

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.

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.

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%.

- 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.

- 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.

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.

Session I

Player wins: 282

Banker wins: 214

Session II

Player wins: 879

Banker wins: 831

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.

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.

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?

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

- 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

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

^{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.

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 |
---|---|---|---|

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 |

- 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.

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?

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.

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 |
---|---|---|---|

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 |

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.