Baccarat, Sort Of

I recently received a letter from one of the readers of The New Chance and Circumstance about a variation of Baccarat; I have since learned that the game is called Bahamian Baccarat. I am responding to his letter on this web page because I think the analysis illustrates some issues that we have looked at before in this column and it is an interesting game to analyze. Here is the letter:

Professor Don:

Last summer while in Las Vegas, I saw a Baccarat variation while at Bally's which I would like to have your expert analysis of. I figured it would cost the player about two units per shoe, compared to the regular game. As I recall, it works like this:

1. No commission charged on any winning bet.
2. Any winning hand--either on Bank or Player--that has a total of 2 or 3 is a push.
3. The stand/draw rules are exactly the same for both Player and Banker:
• No draw if either side has a natural 8 or 9
• If there is no natural, both sides stand with a 6 or 7.
• Both sides take a third card with a two-card total of 0 - 5.
4. As the dealer removes the cards from the shoe they are passed under some kind of reader which inputs the card values to a computer which tells the dealer whether to draw a third card; as a result there are no mistakes in either draws or commissions.

Since both sides have identical drawing rules, would Player and Banker have the same number of wins in the long (statistically relevant) run? How much do you calculate the player is giving up to play this game compared to the traditional one?

Thanks,
Darrell
Eureka, California

Well, the answer to the first question is clearly 'yes'; the Player could be called Mutt and the Banker Jeff, their situation is symmetrical. The answer to the second one takes some calculating. Let's see what we can do.

Suppose that we play this game by putting a one unit bet on both the Player and the Banker each time. From the above rules it is clear that from the casino's perspective every situation is a push unless one side beats the other with a 2 or 3 value hand. In other words, the house's expectation per game is

Since the amount at risk is 2 units, the number calculated in (1) must be divided by 2 to get the house edge (profit per unit risked). Clearly, both of the probabilities in (1) are the same, so the house advantage in this game is simply the probability that the Player beats the Banker with a 2 or 3 (or the other way around).

It is clear from the rules that for the situation wherein the Player beats the Banker with a 2 or 3 hand, both Player and Banker must start with two cards in the 0 to 5 range. Our job is to calculate the probability that starting with such hands the Player draws into a 3 and the Banker draws into a 0, 1, or 2 or the Player draws into a 2 and the Banker draws into a 0 or 1.

To calculate the aforementioned probabilities taking account of the effects of card removal from the dealing shoe is an arduous task. For example one would have to look at all of the combinations of hands making up the total of 5, there are 10 of them, and for each of them calculate the probability of such a hand, then calculate the probability of drawing an 8 for each of them. Adding all of these numbers would be the probability that the Player ends up with a total of 3 starting with 5. This same calculation would have to be carried out for 0, 1, 2, 3, and 4 as well and the results added. The same type of calculations for the Banker would have to be done too. Each one of them is simple but there are a lot of them. If I were to carry out this task, I would write a computer program to do it for me.

There is a simple approximation that one can make and it works quite well when multiple decks are used. We just assume that the deck is infinite. This means that the probabilities of drawing each card remains unchanged as cards are removed from the deck. In other words the probability of drawing a 10 is 4/13 at all times. It isn't true, but for multiple decks it is approximately true. (See Epstein's Theory of Gambling and Statistical Logic, Academic Press, 1977, p 193)

Under the infinite deck assumption, the events of drawing a particular card on the first draw is causally independent from the draw of a particular card on the second draw. Recalling my November 1999 article Independence Day, this means that we can calculate the probability of a particular card on the first draw and a particular card on the second draw by multiplying their respective probabilities. For example, the probability of a 3 on the first draw and a 0 on the second is just 1/13 x 4/13 or 4/169. We can make a simple table showing the probabilities for each of the first and second card combinations:

From the probabilities in Figure 1, it is easy to calculate the probability of obtaining any particular Baccarat total on the first two cards. You simply add the numbers in the appropriate diagonals from lower left to upper right. For example, for a total of 4 there are 5 numbers in the diagonal running from 4 on the left to 4 on top and 5 numbers in the diagonal running from (9,5) to (5,9) in the lower right side of the figure; adding these ten numbers produces 16/169. In fact, every total except 0 has a 16/169 probability of occurring; 0 occurs with probability 25/169.

Very well, let us calculate the probability that the Player draws into a 3 (get ready for a long sentence). This is just the probability that he has a 0 times the probability that he draws a 3 (25/169 x 1/13) plus the probability that he has a 1 times the probability that he draws a 2 (16/169 x 1/13) plus the probability that he has a 2 times the probability that he draws a 1 (16/169 x 1/13) plus the probability that he has a 3 times the probability that he draws a 0 (16/169 x 4/13) plus the probability that he has a 4 times the probability that he draws a 9 (16/169 x 1/13) plus the probability that he has a 5 times the probability that he draws an 8 (16/169 x 1/13). Whew! If you calculate this number it is 153/2197. It turns out that either Player or Banker drawing 1 or 2 is this same number, the probability of drawing a zero is somewhat higher at 180/2197. I'll leave it to you to check these out.

Using the above probabilities, the probability that the Banker draws into 0, 1, or 2 is just 180/2197 + 153/2197 + 153/2197 = 486/2197. The probability that the Player has a 3 is 153/2197 so the probability that the Player has a 3 and the Banker has 0, 1, or 2 is just 153/2197 x 486/2197 or 125,307/4,826,809. A similar calculation for the probability that the Player has a 2 and the Banker has a 0 or 1 produces the probability 50,949/4,826,809. Adding these we obtain 125,307/4,826,809, which is approximately 0.02596. Thus the approximate house advantage in this game is 2.596%. Again, this is using infinite deck approximation; the actual answer for six or eight decks will be different from but quite close to this number.

One final comment. If I were going to write a program to exactly calculate the house edge or if I were going to write a simulation to estimate the house advantage by exhaustive trials, I would first do the analysis above so that I could check the accuracy of my programs. That is to say, if my programs produced a number much different from 2.6%, I would suspect that I had bugs in my programs. This actually happened to me this fall. At the recent WGBE meeting in Las Vegas, Derek Webb (inventor of Three Card Poker^TM) was showing four new games (I'll be discussing some of these in a future article). I was hired to do the analysis for each. The distinguished gaming analyst Stanley Ko was also hired by Derek to do an analysis independent from mine; this gives you an indication as to why Derek is such a successful developer--he does his homework. Anyway, in one of the games, I had first done an infinite deck analysis and then wrote a simulator for finite decks. The simulator was way off from my infinite deck calculation. Sure enough, after a week of searching (more like tearing my hair out!) I found the bug.

I hope this reply suits you, Darrell; thanks for your letter. See you all next month.

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