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BACCARAT CARD COUNTING
INTRODUCTION I. INTRODUCTION Much has been written on baccarat, but much of the published literature has been worthless or even harmful. "Money management" and progression betting systems do not work on baccarat any more than they work on craps or roulette; however, because baccarat is played with cards that are not shuffled between each trial, the trials are not truly independent, and so it may not be immediately apparent that such systems are doomed. This report debunks the systems by showing what happens with perfect card counting and by explaining the enormous variability in results, which sometimes confuses systems players into thinking they are playing with a winning system. II. THE BASICS Baccarat is a relatively simple game to describe how to play and to analyze the odds. i. The Rules [This subsection was lifted from a rec.gambling post, but I have lost the attribution. I did receive permission to use the text, however.] Your only choice in Baccarat is whether to bet on the bank, the player, the tie, or some combination, and how much to bet. There is a 5% commission on winning bank hands and ties pay 9 for 1, which is 8 to 1. The "banker" and the "player" are each dealt a hand of two cards; the value of each hand is determined by adding the two cards' value aces count for 1, tens and face cards for 0 -- and taking the last digit of the sum. A hand holding the 3 of clubs and 4 of diamonds has the value 7; a hand with the ace of hearts and jack of spades is worth 1. Now, if either or both side holds a score of 8 or 9 right off the top, it's called a natural, there is no draw, and the high value wins -- the hand is a tie if both have 8 or if both have 9. If there is no natural, play begins with the player hand. If the player hand holds a score of 0 through 5, the player draws a third card; if the player hand holds a score of 6 or 7, the player stands. Now turn to the bank hand. The bank hand always draws with a 0, 1, or 2. With a three, the bank draws unless the player drew and its third card was an 8. With a four, the bank draws unless the player drew and its third card was outside the range of 2 through 7. With a five, the bank draws unless the player drew and its third card was outside the range of 4 through 7. With a six, the bank stands unless the player drew and its third card was either a 6 or a 7. With a seven, the bank stands. After either, neither or both sides have drawn a third card according to the above rules, the final value of the hand is determined and the high hand wins -- or a tie is declared. A score of 0 is commonly called baccarat; the good scores, 8 and 9, are often called la petite and la grande, respectively. Baccarat is a negative expectation game which makes up a disporportionately high percentage of the casinos' intake, given the relatively low number of tables and players it attracts. The reason is that it is, as Silberstang says, the casino's only real "glamorous game," and the shills and tuxedoed dealers lend it an aura of elegance that attracts big, and sometimes huge, bettors. ii. The Odds I wrote a combinatorial analyzer that, not too surprisingly, output figures consistent with published figures: BACCARAT COMBINATORIAL
ANALYSIS As you can see, the bank wins more often than it loses (i.e., than the player wins.) To make up for this, a commission is charged on winning bank bets. "Expected value" doesn't mean we expect to lose 1.057906% of our bet on any particular bank bet, for example - that's impossible - we either win 0.95 units, lose 1.0 units, or push on any particular bank bet. Instead, expected value means we expect to lose that much on *average*. More on that in the next section. III. WHAT TO EXPECT Somebody played 2392 hands of real baccarat and recorded the following outcomes: 1103 bank hands How likely is this outcome? i. The Multinomial Distribution The multinomial distribution will tell us exactly (plus or minus a bit because of the very slight correlation between hands) how probable this particular outcome was: n! (r1)!(r2)!(r3)! Where r1, r2, r3 is the number of bank, player, and tie outcomes, p1, p2, p3 is the probability of bank, player, and tie and n = r1+r2+r3, the total number of hands Here r1=1103, r2=1075, r3=214, p1=.4596, p2=.4462, p3=.0952, and n=2392, so 2392! 1103!1075!214!
The point is simply that we can compute how likely a particular outcome, or set of outcomes, was. However, such computations are a bit awkward, so you can forget about that equation... I'm going to shift to a normal distribution approximation in a second here, discussing expected value and variance along the way. ii. Expected Value Expected value is the sum of the products of each possible outcome and its probability. For baccarat, the expected values are as follows for betting $20 a hand for 2392 hands: EV(banker) = 2392*$20*(.4596*.95 + .4462*(-1)) = -$458
^^^^^^^^^^^^^^^^^^^^^^^^ EV(player) = 2392*$20*(.4462*1.0 + .4596*(-1)) = -$641
^^^^^^^^^^^^^^^^^^^^^^^^ EV(tie) = 2392*$20*(.0952*8.0 + .9048*(-1)) = -$6851
^^^^^^^^^^^^^^^^^^^^^^^^ If you mix up your betting between banker and player, you will on average lose somewhere in between those amounts. If you raise and lower your bets, recompute your average bet in place of $20 and that's how much you will lose. For every $100 you bet on bank, $1.06 goes to the house on average, for every $100 bet on player, $1.24 goes to the house on average, and for every $100 bet on tie, $14.36 goes to the house on average. Period. Think of things that way, and you will soon be cured of any gambling problem you might have. iii. Variance Variance is the mean of the square of a random variable minus the square of the mean. It measures the variability of the outcome. The variances per hand for flat betting $20 a hand for 2392 hands are as follows: VAR(banker) = (2392)($20^2)(0.860969 - .0106^2) = $$822,848
^^^^^^^^^^^^^^^^^^^^ VAR(player) = (2392)($20^2)(0.905800 - .0124^2) = $$865,904
^^^^^^^^^^^^^^^^^^^^ VAR(tie) = (2392)($20^2)(6.9976 - .1436^2) = $$6,675,573 ^^^^^^^^^^^^^^^^^^^^ If you're ranging your bets with an average of $20, that will increase your variance relative to flat betting $20. The standard deviation is the square root of the variance. Here: STD(banker) = sqrt($$822,848) = $907.11 STD(player) = sqrt($$865,904) = $930.54 STD(tie) = sqrt($$6,675,573) = $2,583.71 iv. The Normal Distribution Because the sample size is more than 100 or so and the trials are very nearly independent, the Central Limit Theorem takes hold and the winnings will fall very nearly exactly into a normal distribution, that is a bell-shaped curve. Sometimes you do very well, sometimes very poorly, but most of the time you fall in the big bulge somewhere, and the center of this bulge is on the expected value. The normal distribution can be used to find out just how likely or unlikely it was for you to win at least as much as you did. In this case, the normal distribution is easier to use than the multinomial or binomial distributions, which are the exact models for coin-tossing type games. About 2/3's of the time, your normally-distributed result will be within +- one standard deviation of your expected value. So the $20/hand banker after 2392 hands will 2/3's of the time be within $907.11 (see the previous subsection) of his expected value of -$458 (see a couple subsections ago.) In general, don't get excited unless your results are outside 1.96 times the standard deviation relative to your expected value; values fall outside this range 5% of the time (loosely speaking). You can make a 95% confidence interval on your results by adding and subtracting 1.96 times the standard deviation of your result. So a 95% confidence interval on the $20/hand banker after 2392 hands is [-$458-$907.11, -$458+907.11] = [-$1365.11, $449.11]. So, loosely speaking, 95% of the time the results would be between losing $1365.11 and winning $449.11. The standardized normal variable z, which I need to consult the probability charts for a normal distribution, is: z = (x-EV)/sqrt(VAR) Here x is the point of interest. Suppose you won $401 during those 2392 hands, and you wanted to know how unusual that was. z(banker) = ($401-(-$468))/sqrt($$822,848) = 0.958 z(player) = ($401-(-$641))/sqrt($$865,904) = 1.120 z(tie) = ($401-(-$6851))/sqrt($$6,675,573) = 2.803 Those z's are the number of standard deviations from the norm. Consulting a normal distribution chart, I find that the probability of such a win or better when betting banker, player, or tie while never saying "presto" (or else all mathematical probability is null and void) is: Banker 17% So, this win of $401 would not be unusual, unless you were betting on the tie every time. Your results won't necessarily follow a normal distribution if you make your bets a function of your previous results. However, the expected value stuff still holds. You can use a system such as Martingale to give you a high chance of walking away a winner (up $1), but this is always exactly balanced out by some small chance of having a big loss - the average of all the possible outcomes produces the same house percentage as always. In a house game such as baccarat, you will lose a fixed percentage of the amount you put on the table, on average. For betting any one of bank, player, or tie, you can neither turn this house percentage into your favor, nor lessen it, nor worsen it. i. Theorems In _Theory of Gambling and Statistical Logic_, Epstein writes: | Theorem I: If a gambler risks
finite capital over a large number of plays | It follows that an unfavorable game
remains unfavorable regardless of the | The number of
"guaranteed" betting systems, the proliferation of myths | Theorem II: No advantage accrues to
the process of betting only on | Corollary: No advantage in terms of
mathematical expectation accrues | Theorem III: For n plays of the
general game, the mean or mathematical [Theorem IV omitted.] | Theorem V: A gambler with initial
fortune z, playing a game with the ii. Systems Given the theorems, for the baccarat player who does not count cards, the following are the *optimal* systems for the specified goals:
One might argue that one could do better by keeping track of the number of player, bank, and tie wins in a shoe. The casino gives you scorecards for this purpose. Hmmm, they wouldn't encourage you to do it if it actually were worthwhile, now would they? No, they would not. The number of player, bank, and tie wins is extremely weakly correlated to the cards that have been removed, and as you'll see in the next section, even perfect card counting reveals depressingly few favorable situations. iii. Ultimate (Non-Linear) Counting System In _Theory of Blackjack_ Peter Griffin presents the following table describing the "ultimate" point count values for baccarat: Denomination `Player Bet' `Bank Bet' `Tie Bet'
A
-1.86
1.82 5.37 These are the best linear estimates of the effects of removal from a full shoe. (Actually, Griffin made a small error: 2's should be weighted as 2.17, not 2.28.) Griffin gives an example of how to use this: Suppose the first hand out of the shoe uses a 3 and 4 for the Player and a 9 and Jack for the Bank. Our running count is 2.69+4.80-.96+.78
-1.05791 + 7.31/412 = -1.04016% so the shoe is not quite ready for us. I repeated the previous combinatorial analysis, but this time computed the opportunity and certainty equivalent for the above Ultimate Linear Count System for baccarat.
COMBINATORIAL ANALYSIS OF BACCARAT KELLY CARD COUNTER Note that the tie bet is often producing a negative certainty equivalent. This is because the linear count is either saying we have an advantage when we don't or vastly overestimating our advantage when we do have an advantage. To me, this says in black and white that the best linear estimate isn't the best estimate to use. First, one should use the linear estimate that maximizes one's profits, probably by erring on the side of conservatism; there is no excuse for having a *negative* certainty equivalent, as we could do better by not counting (and thus not betting.) Second, the tie bet is inherently nonlinear when you get down to small card subsets, so probably no linear count would do well on it. For the previous case of the Binion's Horseshoe 4% bank bet dealt down to less than 25 cards unseen, the ultimate linear baccarat count would produce a certainty equivalent of .00109% bankroll per 100 rounds, down considerably from the .00163% of the perfect counter case. A millionaire ultimate linear baccarat counter would earn just $10.90 per hour, roughly. Now Go play!!! remember the odds and enjoy your game. |