Baccarat Probability Question

Quote: gigolo2002ph

Hi Wizard,

I have a question with regards to the probability of winning for a Banker, Player and Tie.
Below are the combinations in each market. May I know how did you compute the no. of combinations in each event?
Is there any method or did you used a simulation software or a program to calculate all the possible combinations?
Thanks in advance and more power to your site. :)

Combinations Markets
2,292,252,566,437,888 - Banker
2,230,518,282,592,256 - Player
475,627,426,473,216 - Tie

Quote: teliot

...It's fairly trivial programming, less than 200 lines of code....takes about 3 milliseconds to run.

Quote: charliepatrick

Please forgive me for asking the obvious (also I haven't read the obvious book if there is one), but what programming language would you recommend (I'm guessing it's c or similar) and how does one run it on a PC. Personally I use scripting under HTML, so could write the code but haven't yet found the best engine to run such code quickly. Many thanks.

Quote: charliepatrick

Thanks

Quote: 98Clubs

About 40 is very high. The average number of total naturals dealt in a standard game of Baccarat is 30.7. Three of these will be a "double natural" that will either tie or win/lose. Considering 81 rounds per shoe average thats about a 38% occurance (call it 3/8).

Quote: charliepatrick

I worked out the values, based on 8 decks, as follows (for each deal):-
Both are naturals = 3.582 023%
One is a natural = 30.722 887%
Neither is a natural = 65.695 090%
So chances of any given hand being natural is 18.943% (2 * "Both"+1 * "One")/2


btw I wonder whether there's a bet paying (Evens for 1, 9/1 for two @ 97.27%) although it's fairly countable.

Quote: PhraterPhil

Hi Teliot

I am interested in mathematics generally, and lately have been looking at baccarat. I wanted to see if I could work out, on my own, the probabilities shown by Mike and confirmed by yourself. Rather than go all-in and work out everything at once, I decided to start small to see if my code was correct before expanding it further. What I have worked out thus far is the odds of each result given that both player and banker have a score of zero after four cards (of an eight-deck shoe) have been drawn.

I have a total of 169,332 possible combinations (412 x 411) for the fifth and sixth cards drawn. There are 416 cards in total - 4 have been drawn already.

Of these combinations, I have determined 72576 wins for the player, the same amount for the banker and 24180 ties.

This gives me percentages of 42.86018 for the player and banker and 14.27964 for ties. They are close to what Mike has but they do not match. I have thought about this for a while and tried a few things to find numbers to get the same percentages as Mike with no success. Since you have written code already I was hoping you would be able to suggest why my numbers might be different.

As I said, this is purely for my own mathematical curiosity.

And now I notice this thread is over four years old, but I think it is worth taking a shot at asking.

Kind regards
Phrater Phil

Quote: Tanko

Thank you for your response.

If the expected average of total naturals is 38%, then I can accept the 44% that I observed in my small twenty shoe sample.

I believed the average should be slightly less than 19%.

Any links to this information?

Quote: PhraterPhil

Hworked out for the player drawing 0,0 and the banker drawing 0,0 there are 256,032,000 ways this can happen with a starting shoe of eight decks and no burn cards.

Quote:

Both player and banker are forced to draw a third card. So all up there are 43,354,410,624,000 six-tuples.

  for (a = 1; a <= 10; a++) {
    for (b = 1; b <= 10; b++) { 
      for (c = 1; c <= 10; c++) { 
        for (d = 1; d <= 10; d++) { 
          for (e = 1; e <= 10; e++) { 
            for (f = 1; f <= 10; f++) {
              x = getPlayer(a,b,c,d,e,f);
              y = getBanker(a,b,c,d,e,f);
              gN = getNumber(a,b,c,d,e,f);

              totalHands += gN;

              if (x > y) 
                player += gN;
              else
              if (y > x) 
                banker += gN;
              else
              if (y == x) 
                tie += gN;
              else
                exit(1);

              if (x == 0 && y == 0) 
                total_0_0 += gN;
  
  } } } } } }

long long int getNumber(int a1, int b1, int c1, int d1, int e1, int f1) {

  long long int number = 1;
  int freq[11] = {0, 32, 32, 32, 32, 32, 32, 32, 32, 32, 128};   
  number *= freq[a1]--;          
  number *= freq[b1]--;  
  number *= freq[c1]--;  
  number *= freq[d1]--;  
  number *= freq[e1]--;                           
  number *= freq[f1];  
  return number;
}

Quote: charliepatrick

I assume you've worked out the combinations of cards and their perms by dragging and dropping the actual values from another master spreadsheet into this one

Quote: teliot

I just posted a spreadsheet that explains it all and allows you to do just about any other baccarat computation (that doesn't involve suits). No computers necessary. However, you will need to learn the Excel function "sumifs".

http://apheat.net/2014/05/16/baccarat-combinatorial-analysis-the-easy-way/