Sic Bo Appendix

Last update: Feb 5, 2007

Image take from the Claridge Hotel/Casino rule book. Click on the image to see the rules on the next page.

The purpose of this appendix is to derivate the player's edge for the various betting options in sic-bo. To make things simple I use the player's edge, as opposed to the house edge, because it is easier to think of things from the player's perspective. The player's edge will always be negative, to get the house edge just multiply by -1. The general formula for the player's edge is the dot product of all possible returns and their respective probability. Note that there are 6³=216 possible combinations of the throw of three dice.

Low:
This bet would have no house edge if it were not for the triple exception. The probability of a triple 1, 2, or 3 is 3/216. The probability of any total between 3 and 10 is 1/2, or 108/216. So the probability of a winning triple is 108/216 - 3/216 = 105/216. The player's edge is thus (105/216)*(+1) + (111/216)*(-1) = -6/216 =~ -2.78%.
High: See low.
Specific Number:
The probability of rolling zero of a specific number is (5/6)³ = 125/216.
The probability of rolling one of a specific number is 3*(1/6)¹*(5/6)² = 75/216.
The probability of rolling two of a specific number is 3*(1/6)²*(5/6)¹ = 15/216.
The probability of rolling three of a specific number is (1/6)³ = 1/216.
The player's edge is thus (125/216)*(-1) + (75/216)*(+1) + (15/216)*(+2) + (1/216)*(+3) = -17/216 =~ -7.780%.
Total of 4:
There are 3 ways to roll a 4: (1+1+2, 1+2+1, 2+1+1). The player's edge is thus 3/216*(+60) + (213/216)*(-1) = -33/216 = -15.278%.
Total of 5:
There are 6 ways to roll a 5: (1+1+3, 1+3+1, 3+1+1, 1+2+2, 2+1+2, 2+2+1). The player's edge is thus 6/216*(+30) + (210/216)*(-1) = -30/216 = -13.889%.
Total of 6:
There are 10 ways to roll a 6: (1+1+4, 1+4+1, 4+1+1, 1+2+3, 1+3+2, 2+1+3, 2+3+1, 3+1+2, 3+2+1, 2+2+2). The player's edge is thus 10/216*(+17) + (206/216)*(-1) = -36/216 = -16.667%.
Total of 7:
There are 15 ways to roll a 7: (1+1+5, 1+5+1, 5+1+1, 1+2+4, 1+4+2, 2+1+4, 2+4+1, 4+1+2, 4+2+1, 1+3+3, 3+1+3, 3+3+1, 2+2+3, 2+3+2, 3+2+2). The player's edge is thus 15/216*(+12) + (201/216)*(-1) = -21/216 = -9.722%.
Total of 8:
There are 21 ways to roll a 8: (1-6-6 * 3 ways, 1-2-5 * 6 ways, 1-3-4 * 6 ways, 2-2-4 * 3 ways, 2-3-3 * 3 ways). The player's edge is thus 21/216*(+8) + (195/216)*(-1) = -27/216 = -12.500%.
Total of 9:
There are 25 ways to roll a 9: (1-2-6 * 6 ways, 1-3-5 * 6 ways, 1-4-4 * 3 ways, 2-2-5 * 3 ways, 2-3-4 * 6 ways, 3-3-3 * 1 way). The player's edge is thus 25/216*(+6) + (191/216)*(-1) = -41/216 = -18.982%.
Total of 10:
There are 27 ways to roll a 10: (1-3-6 * 6 ways, 1-4-5 * 6 ways, 2-2-6 * 3 ways, 2-3-5 * 6 ways, 2-4-4 * 3 ways, 3-3-4 * 3 ways). The player's edge is thus 27/216*(+6) + (189/216)*(-1) = -27/216 = -12.500%.
Total of 11: See total of 10
Total of 12: See total of 9
Total of 13: See total of 8
Total of 14: See total of 7
Total of 15: See total of 6
Total of 16: See total of 5
Total of 17: See total of 4
Two numbers:
Lets suppose the two numbers chosen are 1 and 2. There are 30 combinations featuring a 1 and a 2: 1-2-1 * 3 ways, 1-2-2 * 3 ways, 1-2-3 * 6 ways, 1-2-4 * 6 ways, 1-2-5 * 6 ways, 1-2-6 * 6 ways. The player's edge is thus 30/216*(+5) + (186/216)*(-1) = -36/216 = -16.667%.
Specific triplet:
There is only 1 way to roll a specific triplet. The player's edge is thus 1/216*(+180) + (215/216)*(-1) = -35/216 = -16.20%.
Any triplet:
There are 6 ways to roll a triplet. The player's edge is thus 6/216*(+30) + (210/216)*(-1) = -30/216 = -13.889%.
Specific pair:
Lets assume the pair chosen is ones. There are 16 ways two or three of that number can be rolled: 1+1+1, 1+1+2 * 3 ways, 1+1+3 * 3 ways, 1+1+4 * 3 ways, 1+1+5 * 3 ways, 1+1+6 * 3 ways. The player's edge is thus 16/216*(+10) + (200/216)*(-1) = -72/216 = -18.52%.

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.

For example, let's look at 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) ] =
1/216 * [1*1*combin(10,2) + -1*3*combin(4,2)] =
1/216 * [1*1*45 + -1*3*6] =
1/216 * [45-18] = 27/216 = 12.5%

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