It is not surprising fact that most of the casino games like Blackjack, Roulette, lottery games, and Sic Bo are considered to be a game of probability. There are players who use their mind scientifically and take a logic strategy to play many of the casino games. Ordinary players like most of us would not bother to go much deeper into the Genetic Algorithms application to a specific data set of such strategies while playing Sic Bo or any other game. But the professional and mind blowing players detect the game and evaluate them in different measures and conditions before they start any casino game online or at land casinos.

Let us understand here how a mathematical application is applied to Sic Bo to calculate its bets and odds logically and applied to win the game.

## Firstly understand what is Sic Bo

Sic Bo as you know is Chinese traditional game to play with a dice pair. The game uses three dice and a table with a variety of betting options on the roll of those dice. The odds and table layout may also vary from place to place.

## Stick to only Low and High Bets as the perfect strategy to play Sic Bo

Each online casino has its own odds and it may vary because casinos are a commercial business they have to look to their revenue and profits while offering responsible gaming to the players.

Here is an example quoted by Michael Shackleford in his Sic Bo mathematical calculations for high and low bets. He has taken the list of the bets and their payoffs for Atlantic City and the Mirage in Las Vegas as following:

Small: Wins on total of 4-10, except for a three of a kind. Pays 1 to 1.

Big: Wins on total of 11-17, except for a three of a kind. Pays 1 to 1.

4: Wins on total of 4. Pays 60 to 1.

5: Wins on total of 5. Pays 30 to 1.

6: Wins on total of 6. Pays 17 to 1.

7: Wins on total of 7. Pays 12 to 1.

8: Wins on total of 8. Pays 8 to 1.

9: Wins on total of 9. Pays 6 to 1.

10: Wins on total of 10. Pays 6 to 1.

11: Wins on total of 11. Pays 6 to 1.

12: Wins on total of 12. Pays 6 to 1.

13: Wins on total of 13. Pays 8 to 1.

14: Wins on total of 14. Pays 12 to 1.

15: Wins on total of 15. Pays 17 to 1.

16: Wins on total of 16. Pays 30 to 1.

17: Wins on total of 17. Pays 60 to 1.

- Two of a kind: Player may bet on any of the 15 possible two dice combinations (for example a 1 and 2). Bet wins if both numbers appear. Probability of winning is 13.89%. Pays 5 to 1.
- Double: Player may bet on any specific number (for example a 1). Player wins if at least 2 of the 3 dice land on that number. Probability of winning is 7.41%. Pays 10 to 1.
- Triple: Player may bet on any specific number (for example a 1). Player wins if all 3 dice land on that number. Probability of winning is 0.46%. Pays 180 to 1.
- Any Triple: Wins on any three of a kind. Pays 30 to 1.
- Individual Number: Player may bet on any specific number from 1 to 6. If chosen number appears 1 time bet pays 1 to 1, if it appears 2 times bet pays 2 to 1, and if it appears 3 times it pays 3 to 1.

## Formula for “s” spots over “n” dice

The critical step in calculating the odds in Sic Bo is to find the probability of any given total in the throw of three dice. Following is a formula for s spots over n dice, taken from The Theory of Gambling and Statistical Logic by Richard A. Epstein, formula 5-14.

An Example to show the number of ways to get 11 spots over 3 dice.

int[(s-n)/6] = int[(11-3)/6] = int[1.33] = 1

The total would be 6-3 * [-10*combin(3,0)*combin(11-6*0-1,3-1) + -11*combin(3,1)*combin(11-6*1-1,3-1) ] =

1/218 * [1*1*combin(10,2) + -1*3*combin(4,2)] =

1/218 * [1*1*45 + -1*3*6] =

1/218 * [45-18] = 27/216 = 12.50%

## The formula in a computer program with C++

You can program a computer that would probably be the fastest way to get the results.

Here is a simple function in C++.

void SicBo(void)

{

int i,d1,d2,d3,tot,tot_array[19];

for (i=0; i<=18; i++)

tot_array[i]=0;

for (d1=1; d1<=6; d1++)

{

for (d2=1; d2<=6; d2++)

{

for (d3=1; d3<=6; d3++)

{

tot=d1+d2+d3;

tot_array[tot]++;

}

}

}

cout << “Total\tCombinations\tProbability\n”;

for (i=3; i<=18; i++)

cout << i << “\t” << tot_array[i] << “\t” << (float)tot_array[i]/216 << “\n”;

}

### The output of the function

Total | Permutations | Probability |

3 | 1 | 0.00463 |

4 | 3 | 0.013889 |

5 | 6 | 0.027778 |

6 | 10 | 0.046296 |

7 | 15 | 0.069444 |

8 | 21 | 0.097222 |

9 | 25 | 0.115741 |

10 | 27 | 0.125 |

11 | 27 | 0.125 |

12 | 25 | 0.115741 |

13 | 21 | 0.097222 |

14 | 15 | 0.069444 |

15 | 10 | 0.046296 |

16 | 6 | 0.027778 |

17 | 3 | 0.013889 |

18 | 1 | 0.00463 |

If you don’t know how to program you’re going to have to do this the hard way.

## List every combination of 3 dice

To avoid the list being 63=216 items long do not repeat the same combinations in different orders. In the interests of not listing the same number twice always order each combination from lowest to highest, not forgetting combinations with a pair or three of a kind.

Start with 1,1,1.

Next would be 1,1,2.

Then 1,1,3; 1,1,4; 1,1,5; and 1,1,6.

Obviously you can’t roll a 7 with one dice so next we increment the second die.

1,2,?

The third die must be greater or equal to the second die so the next combination in full would be 1,2,2.

Next comes 1,2,3; 1,2,4; 1,2,5; and 1,2,6.

Then comes 1,3,3.

The whole list would look like the following pattern

Low die | Medium Die | High Die |

1 | 1 | 1 |

1 | 1 | 2 |

1 | 1 | 3 |

1 | 1 | 4 |

1 | 1 | 5 |

1 | 1 | 6 |

1 | 2 | 2 |

1 | 2 | 3 |

1 | 2 | 4 |

1 | 2 | 5 |

1 | 2 | 6 |

1 | 3 | 3 |

1 | 3 | 4 |

1 | 3 | 5 |

1 | 3 | 6 |

1 | 4 | 4 |

1 | 4 | 5 |

1 | 4 | 6 |

1 | 5 | 5 |

1 | 5 | 6 |

1 | 6 | 6 |

2 | 2 | 2 |

2 | 2 | 3 |

2 | 2 | 4 |

2 | 2 | 5 |

2 | 2 | 6 |

2 | 3 | 3 |

2 | 3 | 4 |

2 | 3 | 5 |

2 | 3 | 6 |

2 | 4 | 4 |

2 | 4 | 5 |

2 | 4 | 6 |

2 | 5 | 5 |

2 | 5 | 6 |

2 | 6 | 6 |

3 | 3 | 3 |

3 | 3 | 4 |

3 | 3 | 5 |

3 | 3 | 6 |

3 | 4 | 4 |

3 | 4 | 5 |

3 | 4 | 6 |

3 | 5 | 5 |

3 | 5 | 6 |

3 | 6 | 6 |

4 | 4 | 4 |

4 | 4 | 5 |

4 | 4 | 6 |

4 | 5 | 5 |

4 | 5 | 6 |

4 | 6 | 6 |

5 | 5 | 5 |

5 | 5 | 6 |

5 | 6 | 6 |

6 | 6 | 6 |

## How to determine the number of permutations for each combination

A combination is a set without regard to order and a permutation is a set with regard to order.With a three of a kind there is only one way permutation. For example if the three dice are 1,1,1 there is only one way to roll that a 1 each time.

If the combination is 1,1,2 there are three ways to roll that: 1,1,2; 1,2,1; and 2,1,1.

If all three dice are 1,2,3 there are six possible permutations: 1,2,3; 1,3,2; 2,1,3; 2,3,1; 3,1,2; 3,2,1

The general formula is that if you have a total of d dice and the totals of each number are x1, x2, x3…xn then the number of permutations are d!/(x1!*x2!*x3*…*xn). So the number of ways to get a three of a kind would be 3!/3! = 6/6 = 1. The number of ways to get a pair would be 3!/(2!*1!) = 6/(2*1) = 3. The number of ways to get three different numbers would be 3!/(1!*1!*1!) = 6/(1*1*1) = 6.

Low die | Medium die | High die | Total | Permutations |

1 | 1 | 1 | 3 | 1 |

1 | 1 | 2 | 4 | 3 |

1 | 1 | 3 | 5 | 3 |

1 | 1 | 4 | 6 | 3 |

1 | 1 | 5 | 7 | 3 |

1 | 1 | 6 | 8 | 3 |

1 | 2 | 2 | 5 | 3 |

1 | 2 | 3 | 6 | 6 |

1 | 2 | 4 | 7 | 6 |

1 | 2 | 5 | 8 | 6 |

1 | 2 | 6 | 9 | 6 |

1 | 3 | 3 | 7 | 3 |

1 | 3 | 4 | 8 | 6 |

1 | 3 | 5 | 9 | 6 |

1 | 3 | 6 | 10 | 6 |

1 | 4 | 4 | 9 | 3 |

1 | 4 | 5 | 10 | 6 |

1 | 4 | 6 | 11 | 6 |

1 | 5 | 5 | 11 | 3 |

1 | 5 | 6 | 12 | 6 |

1 | 6 | 6 | 13 | 3 |

2 | 2 | 2 | 6 | 1 |

2 | 2 | 3 | 7 | 3 |

2 | 2 | 4 | 8 | 3 |

2 | 2 | 5 | 9 | 3 |

2 | 2 | 6 | 10 | 3 |

2 | 3 | 3 | 8 | 3 |

2 | 3 | 4 | 9 | 6 |

2 | 3 | 5 | 10 | 6 |

2 | 3 | 6 | 11 | 6 |

2 | 4 | 4 | 10 | 3 |

2 | 4 | 5 | 11 | 6 |

2 | 4 | 6 | 12 | 6 |

2 | 5 | 5 | 12 | 3 |

2 | 5 | 6 | 13 | 6 |

2 | 6 | 6 | 14 | 3 |

3 | 3 | 3 | 9 | 1 |

3 | 3 | 4 | 10 | 3 |

3 | 3 | 5 | 11 | 3 |

3 | 3 | 6 | 12 | 3 |

3 | 4 | 4 | 11 | 3 |

3 | 4 | 5 | 12 | 6 |

3 | 4 | 6 | 13 | 6 |

3 | 5 | 5 | 13 | 3 |

3 | 5 | 6 | 14 | 6 |

3 | 6 | 6 | 15 | 3 |

4 | 4 | 4 | 12 | 1 |

4 | 4 | 5 | 13 | 3 |

4 | 4 | 6 | 14 | 3 |

4 | 5 | 5 | 14 | 3 |

4 | 5 | 6 | 15 | 6 |

4 | 6 | 6 | 16 | 3 |

5 | 5 | 5 | 15 | 1 |

5 | 5 | 6 | 16 | 3 |

5 | 6 | 6 | 17 | 3 |

6 | 6 | 6 | 18 | 1 |

Total | 216 |

Next go through the tedious process of adding the number of permutations for each total. For example a total of 6 has the following combinations with the corresponding number of permutations.

Combinations | Number of Permutations |

1,1,4 | 3 |

1,2,3 | 6 |

2,2,2 | 1 |

Total | 10 |

The final table will look like this, not unlike the result of the computer function earlier.Now add a column to list for the number of combinations of each set. Let’s also add a total for the three dice.

Total | Permutations |

3 | 1 |

4 | 3 |

5 | 6 |

6 | 10 |

7 | 15 |

8 | 21 |

9 | 25 |

10 | 27 |

11 | 27 |

12 | 25 |

13 | 21 |

14 | 15 |

15 | 10 |

16 | 6 |

17 | 3 |

18 | 1 |

Total | 216 |

Now you can divide each total number permutations by the total number 3-dice permutations (216) to get the probability of each total.

Total | Permutations | Probability |

3 | 1 | 0.00463 |

4 | 3 | 0.013889 |

5 | 6 | 0.027778 |

6 | 10 | 0.046296 |

7 | 15 | 0.069444 |

8 | 21 | 0.097222 |

9 | 25 | 0.115741 |

10 | 27 | 0.125 |

11 | 27 | 0.125 |

12 | 25 | 0.115741 |

13 | 21 | 0.097222 |

14 | 15 | 0.069444 |

15 | 10 | 0.046296 |

16 | 6 | 0.027778 |

17 | 3 | 0.013889 |

18 | 1 | 0.00463 |

Total | 216 | 1 |

Finally, you are ready to evaluate the expected value of each bet. The expected value is the ratio of the amount the player can expect to win to the amount he bets on any given bet. So a fair bet would an expected value of zero. A positive expected value would mean the player has the advantage. A negative expected value would mean the dealer has the advantage.

Let’s start with the 4 bet. This wins with a total of 4 and pays 60 to 1. For those who don’t know, “60 to 1” means if the player wins he wins 60 times his bet and KEEPS his original wager. Had the odds paid “60 for 1” the player would not keep his original bet. Most table games pay on a “to 1” basis.

The probability of a total of 4 is 3/216 = 0.013889. Thus the probability of losing is 1-(3/216) = 1-0.013889 = 0.986111.

The expected value of any bet with only two possibilities, winning or losing, is:

(Probability of winning)*(Amount of win) + (Probability of losing)*(Amount of loss).

For the 4 bet the expected value is

0.013889 * 60 – 0.986111*-1 = -0.15278.

So, this tells us that for every dollar the player bets on a total of 4 he can expect to lose 15.278 cents on average. Or, the house edge is 15.278%.

## Table showing the expected value and how it was calculated for all bets of a total of 4 to 17

Total | Pays | Probability of Win | Probability of Losing | Formula of expected value | Expected Value |

4 | 60 | 0.013889 | 0.986111 | 0.0138889*60-0.986111*-1 | -0.15278 |

5 | 30 | 0.027778 | 0.972222 | 0.0277778*30-0.972222*-1 | -0.13889 |

6 | 17 | 0.046296 | 0.953704 | 0.0462963*17-0.953704*-1 | -0.16667 |

7 | 12 | 0.069444 | 0.930556 | 0.0694444*12-0.930556*-1 | -0.09722 |

8 | 8 | 0.097222 | 0.902778 | 0.0972222*8-0.902778*-1 | -0.125 |

9 | 6 | 0.115741 | 0.884259 | 0.115741*6-0.884259*-1 | -0.18981 |

10 | 6 | 0.125 | 0.875 | 0.125*6-0.875*-1 | -0.125 |

11 | 6 | 0.125 | 0.875 | 0.125*6-0.875*-1 | -0.125 |

12 | 6 | 0.115741 | 0.884259 | 0.115741*6-0.884259*-1 | -0.18981 |

13 | 8 | 0.097222 | 0.902778 | 0.0972222*8-0.902778*-1 | -0.125 |

14 | 12 | 0.069444 | 0.930556 | 0.0694444*12-0.930556*-1 | -0.09722 |

15 | 17 | 0.046296 | 0.953704 | 0.0462963*17-0.953704*-1 | -0.16667 |

16 | 30 | 0.027778 | 0.972222 | 0.0277778*30-0.972222*-1 | -0.13889 |

17 | 60 | 0.013889 | 0.986111 | 0.0138889*60-0.986111*-1 | -0.15278 |

### Two of a kind Bet

There are combinations(6,2)=6!/(4!*2!)=15 ways to choose two numbers out of six. Each of these combinations is listed on the table and the player bet on as many as he wishes. If both numbers appear on the roll of the three dice then the player wins and is paid 15 to 1.

Let’s assume the player picks a 1 and 2 as his two numbers. What is the probability that both a 1 and 2 occur in the roll of 3 dice? One way to do this would be to note all the possible winning permutations:

Dice | Number of Permutations |

1,2,3 | 6 |

1,2,4 | 6 |

1,2,5 | 6 |

1,2,6 | 6 |

1,1,2 | 3 |

1,2,2 | 3 |

Total | 30 |

Thus there are a total of 30 winning permutations. There are 63=216 total permutations, so the probability of winning is 30/216 = 1/36 = 0.1388889

The two of a kind bet pays 5 to 1. So the expected value is 0.1388889*5 + (1-0.1388889)*-1 = -0.16667. In other words the house edge is 16.67%.

### Double Bet

There are six double bets available, one for each number from 1 to 6. The player may be on any one or combination of bets. Any given bet wins if at least two of the three dice land on that number.

Let’s assume the player bets on the 1. One way to solve it would be to note all the winning permutations:

Dice | Number of Permutations |

1,1,2 | 3 |

1,1,3 | 3 |

1,1,4 | 3 |

1,1,5 | 3 |

1,1,6 | 3 |

1,1,1 | 1 |

Total | 16 |

Thus there are a total of 16 winning permutations. There are 63=216 total permutations, so the probability of winning is 16/216 = 0.0740741. The double bet pays 10 to 1. So the expected value is 0.0740741*10 + (1-0.0740741)*-1 = -0.18518. In other words the house edge is 18.52%.

### Triple Bet

Player may bet on any specific number (for example a 1). Player wins if all 3 dice land on that number. There is obviously only one way to win this bet, so the probability of winning is 1/216 = 0.0046296. The bet pays 180 to 1 so the expected value is 0.0046296*180 + (1-0.0046296)*-1 = -0.16204. So the house edge is 16.204%.

### Any Triple Bet

The Any Triple bet pays if any three of a kind is thrown. There are obviously six winning combinations (1,1,1; 2,2,2; 3,3,3; etc.). So the probability of winning is 6/216 = 0.027778. The bet pays 30 to 1 so the expected value is 0.027778*30 + (1-0.027778)*-1 = -0.13889. So the house edge is 13.89%.

### Low Bet

The low bet wins if the total of the three dice is 3 to 10, without being a three of a kind. The probability of any total 10 or less is exactly 50%. The average number on any one die is (1+2+3+4+5+6)/6 = 21/6 = 3.5. So the average of three dice is 3*3.5 = 10.5. It stands to reason that the probability of getting under or over 10.5 is 50%.

However the bet loses on a three of a kind. There are 3 three of a kinds that would turn a winner into a loser: 1,1,1; 2,2,2; and 3,3,3. So the probability of having a total of 10 or less as a three of a kind is 3/216 = 0.0188889. So the overall probability of winning is 0.5 – 0.188889 = 0.4861111. The bet pays 1 to 1 so the expected value is 0.4861111*1 + (1-0.4861111)*-1 = -0.02778. Thus the house edge is 2.78%

### High Bet

The high is just the opposite of the low bet, so it stands to reason the house edge would also be 2.78%.

### Individual Number Bet

Player may bet on any specific number from 1 to 6. If chosen number appears 1 time bet pays 1 to 1, if it appears 2 times bet pays 2 to 1, and if it appears 3 times it pays 3 to 1. Probability of 1 match is 34.72%, 2 matches is 6.94%, 3 matches is 0.46%.

Let’s assume the player picks the number one.

There is only one way to get three ones: 1,1,1. So the probability of three ones is 1/63 = 1/216.

Following are the ways to get two 1’s and the number of permutations of each.

Dice | Number of Permutations |

1,1,2 | 3 |

1,1,3 | 3 |

1,1,4 | 3 |

1,1,5 | 3 |

1,1,6 | 3 |

Total | 15 |

So the probability of two ones is 15/63 = 15/216.

Following are the ways to get one 1 and the number of permutations of each.

Dice | Number of Permutations |

1,2,2 | 3 |

1,2,3 | 6 |

1,2,4 | 6 |

1,2,5 | 6 |

1,2,6 | 6 |

1,3,3 | 3 |

1,3,4 | 6 |

1,3,5 | 6 |

1,3,6 | 6 |

1,4,4 | 3 |

1,4,5 | 6 |

1,4,6 | 6 |

1,5,5 | 3 |

1,5,6 | 6 |

1,6,6 | 3 |

Total | 75 |

So the probability of two ones is 75/63 = 75/216. Another way to arrive at the probability of one 1 would be find the probability that the first die is a one and the second and third are not:

Pr(one)*Pr(not one)*Pr(not one) = (1/6)*(5/6)*(5/6) = 25/216. However the one could appear in the first, second, or third position, so multiply by 3: 3*(25/216) = 75/216.

The probability of rolling zero ones is Pr(not one)*Pr(not one)*Pr(not one) = (5/6)*(5/6)*(5/6) = (5/6)3 = 125/216.

The following return table shows the possible outcomes, and the number of combinations, probability, and return of each. The return is the product of the probability and the win or loss to the player.

Event | Permutations | Probability | Pays | Return |

Player rolls 3 ones | 1 | 0.00463 | 3 | 0.013889 |

Player rolls 2 ones | 15 | 0.069444 | 2 | 0.138889 |

Player roll 1 one | 75 | 0.347222 | 1 | 0.347222 |

Player rolls 0 ones | 125 | 0.578704 | -1 | -0.5787 |

Total | 216 | 1 | -0.0787 |

So the total expected return is -0.0787, or the house edge is 7.87%. With all such solved calculations you can arrive at the house edge and the probability of your position in the game of Sic Bo. All the above examples and calculations have been taken from Michael Shackleford’s Sic Bo mathematical strategy for you to understand it in a logical and scientific way.

Online casinos like Royal Vegas Casino use online online gaming software to calculate a collection of probability applications that are used in game theory. To do so, this software uses different formulas, like the Fundamental Formula of Gambling. This formula calculates the number of things that must be done for an event of probability to appear with a degree of certainty. Now you can also try such formulas and calculations if you are good at numbers and calculus. Good Luck.