**A Calculation of Odds in the Dice Game ****Langur Burja**

Gale Richard Walker

September 2017

Langur Burja, a dice game popular in Nepal and in the British navy (where it is known as Crown and Anchor), uses six identical dice, each with six different symbols on it. After bets are placed, the banker tosses all six dice at once. The betting rules and payouts to the players may be found on the Internet. Variations of the game are played in France, Vietnam, China, Belgium, and Bermuda. They are not described here.

Overall, the Langur Burja banker (commonly know as “The House”) who tosses the dice has a big edge — eight percent — over the players, but that is not the focus here. The purpose here is to show how to calculate the probability of each possible outcome of one toss of six dice. Any six symbols, pictures, colors, or numbers will do. In England, for example, the six symbols are a heart, a spade, a diamond, a club, a crown, and an anchor. Herein they are simply called colors, as any six colors on each die will do.

When all six dice are tossed the order of appearance of the six colors is not important. Thus, one need only calculate *combinations* — C(n:r) — and not permutations. Of course, the banker and players are interested in the odds of getting no matching colors, one pair of colors, two pair, three pair, three of a kind, etc., on a toss of the six dice. The equation for calculating combinations is:

**C(n:r) = n!/[(n-r)! x (r!)]**

where factorial symbol, !, in 4!, means = 4x3x2x1 = 24. Also note, by definition: 0! = 1.

To illustrate the interpretation of the equation shown above: How many ways can five people be randomly selected from a group of eight? Answer: C(8:5) = 8!/[(8 – 5)! x 5!] = (8x7x6x5x4x3x2x1)/[(3x2x1) x (5x4x3x2x1)] = 40,320/720 = 56 ways. Note that the order in which they are selected does not matter. Similarly, when tossed, the order in which colors appear on the six dice does not matter. Since Langur Burga uses six dice, the number of possible combinations is small (compared, say, to the combinations of five cards dealt from a 52-card deck). For those who want to do the calculations using the equation above, Table 1 presents frequently occurring combinations:

## Table 1

C(6:0) = 1 | |||||

C(6:1) = 6 | C(5:0) = 1 | ||||

C(6:2) = 15 | C(5:1) = 5 | C(4:0) = 1 | |||

C(6:3) = 20 | C(5:2) = 10 | C(4:1) = 4 | C(3:0) = 1 | ||

C(6:4) = 15 | C(5:3) = 10 | C(4:2) = 6 | C(3:1) = 3 | C(2:0) = 1 | |

C(6:5) = 6 | C(5:4) = 5 | C(4:3) = 4 | C(3:2) = 3 | C(2:1) = 2 | C(1:0) =1 |

C(6:6) = 1 | C(5:5) = 1 | C(4:4) = 1 | C(3:3) = 1 | C(2:2) = 1 | C(1:1) =1 |

Three examples illustrate the calculation of combinations shown above:

C(6:0) = 6!/[(6-0)!][0!] = 6!/[6!][1] = 6!/6! = 1

C(6:3) = 6!/[(6-3)!][3!] = 6!/[3!][3!] = (6x5x4x3x2x1)/[(3x2x1)(3x2x1)] = 5 x 4 = 20

C(6:5) = 6!/[(6-5)!][5!] = 6!/[1!][5!] = (6x5x4x3x2x1)/[1!][5x4x3x2x1] = 6

**ODDS**

The total possible outcomes (sampling space) is (6^{6}) = 46,656. Of those, the total number of outcomes without a pair or better (i.e., when all six colors occur) is 6! = (6x5x4x3x2x1) = 720. (See “No pair” below.) That — (720/46,656) — is only 1.5 percent of the tosses. Thus the odds of getting at least one pair of colors or better are 98.5 percent. The calculation of odds of all possible outcomes, from most to least probable, is shown below, then summarized and compared with actual trial results in Table 2 at the end.

odds = (combinations)/(46,656)

**TWO PAIR (Odds: 34.7%)**

[C(6:1)] [C(6:2) x C(6:2)] [C(4:2) x (2!)] =

6 x 15 x 15 x 6 x 2 = **16,200**

The first bracket: How many different colors are available on each of the six dice? Six. The second bracket represents the first two pairs. First pair: How many ways can any one of those six colors appear on two of six dice? Fifteen. Second pair: Likewise, fifteen. Third bracket: Since the four dice (two pair) utilize two colors, how many colors are left for the last two, nonmatching dice? Four. How many ways can the four remaining colors appear on the last two dice? Six. How many combinations of nonmatching colors are the last two dice free to display? 2.

An alternate logic and calculation are given in this footnote.[1]

**ONE PAIR (Odds: 23.2%)**

[C(6:1) x C(6:2)] [C(5:4)](4!)] =

6 x 15 x 5 x 24 = **10,800**

How many different colors are available on each of the six dice? Six. How many ways can one color appear on two of six dice? Fifteen. How many colors are left for the last four dice? Five. How many ways can five different colors appear on the last four dice? Five. How many nonmatching combinations of colors are the last four dice free to form? 24.[2]

**THREE OF A KIND (Odds: 15.3%)**

[C(6:1) x C(6:3)] [C(5:3)] [(3!)] =

6 x 20 x 10 x 6 = **7,200**

How many different colors are available on each of the six dice? Six. How many ways can one color appear on three of six dice? Twenty. How many colors are left for the last three dice? Five. How many ways can five colors appear on the last three dice? Ten. How many nonmatching combinations can three colors take? Six.

**THREE OF A KIND and TWO OF A KIND** (“full house”) **(Odds: 15.3%)**

[C(6:1) x C(6:3)] [C(5:1 x *C*(3:2)] [C(4:1) x 1!] =

6 x 20 x 5 x 3 x 4 x 1 = **7,200**

How many different colors are available on each of the six dice? Six. How many ways can one color appear on three of six dice? Twenty. How many colors are left for the remaining dice? Five. How many ways can one color appear on two (i.e., the pair) of the three remaining dice? Three. How many colors are left for the last die? Four.

As a check, reversing the order — TWO OF A KIND and THREE OF A KIND — gives the same result:

[C(6:1) x C(6:2)] [C(5:1 x C(4:3)] [C(4:1) x 1!] = 6 x 15 x 5 x 4 x 4 x 1 = 7,200

How many different colors are available on each dice? Six. One pair can be chosen from six dice 15 ways. That leaves five colors and four dice. Three dice can be chosen from four dice 4 ways. That leaves four colors for the last die.

**FOUR OF A KIND (Odds: 3.9%)**

[C(6:1) x C(6:4)] [C(5:2) x (2!)] =

6 x 15 x 10 x 2 = **1,800**

How many different colors are available on each of the six dice? Six. How many ways can one color appear on four of six dice? Fifteen. How many colors are left for the last two dice? Five. How many ways can five colors appear on the last two dice? Ten. How many nonmatching combinations can the last two dice make? Two.

**THREE PAIRS (Odds: 3.9%)**

[C(6:3) x C(6:2) x C(6:1)] =

20 x 15 x 6 =** 1,800**

Think in terms of pairs of colors (because there are no singletons). How many different ways can three pairs of colors be selected from six colors? Twenty. That leaves two pairs of colors yet to be selected. How many ways can two pairs of colors be selected from six colors? Fifteen. That leaves one pair of colors to be selected. How many ways can one pair be selected from six colors? Six.[3]

**NO PAIR** (“nothing hand”) **(Odds: 1.5%)**

[C(0:0) x C(6:0)] x (6!) =

1 x 1 x 720 = **720**

Six colors are available and all six colors appear; i.e., no matches were cast. The first die could be any one of six colors, the second die could be any of the five remaining colors, the third could be any of the four remaining colors, etc.

**FOUR OF A KIND and a PAIR (Odds: 1.0%)**

[C(6:1) x C(6:4)] [C(5:1) x C(2:2)] =

6 x 15 x 5 1 = **450**

How many different colors are available on each of the six dice? Six. How many ways can one color appear on four of six dice? Fifteen. How many colors are left for the last two dice? Five. How many ways can one of five colors appear on the last two dice? Five. How many ways can two of two dice appear? One.

**THREE OF A KIND and THREE OF A KIND (Odds: 0.6%)**

[C(6:3) x C(6:2)] =

20 x 15 = **300**

Think in terms of triplets (because there are no pairs and no singletons). First, how many different colors are available? Six. How many ways can any triplet appear on six dice? Twenty. Second, how many ways can two triplets appear on six dice? Fifteen.

**FIVE OF A KIND (Odds: 0.4%)**

[C(6:1) x C(6:5)] [C(5:1) x (1!)] =

6 x 6 x 5 x 1 = **180**

How many different colors are available on each of the six dice? Six. How many ways can one color appear on five of six dice? Six. How many colors can appear on the last die? Five. How many dice are left? One.

**SIX OF A KIND** **(Odds: 0.0%)**

[C(6:1) x C(6:6)] =

6 x 1 = **6**

How many different colors are available on each of the six dice? Six. How many ways can one color appear on six of six dice? One.

Predicted Results: Pairs | Combinations | % | Experimental Results: n = 500 | + 100 | 1500 | % |

2 - 2 | 16,200 | 34.72 | 165 | 360 | 525 | 35.0 |

one pair | 10,800 | 23.15 | 133 | 218 | 351 | 23.4 |

3 - kind | 7,200 | 15.43 | 79 | 162 | 241 | 16.1 |

3 -2 | 7,200 | 15.43 | 73 | 142 | 215 | 14.3 |

4 - kind | 1,800 | 3.86 | 21 | 34 | 55 | 3.7 |

2 - 2 - 2 | 1,800 | 3.86 | 14 | 14 | 43 | 3.8 |

No pair | 720 | 1.54 | 8 | 8 | 21 | 1.9 |

4 - 2 | 450 | 0.96 | 10 | 10 | 12 | 0.8 |

3 - 3 | 300 | 0.64 | 4 | 4 | 5 | 0.3 |

5 - kind | 180 | 0.39 | 6 | 6 | 10 | 0.7 |

6 - kind | 6 | 0.01 | 0 | 0 | 0 | 0.0 |

11 | 46,656 | 100 | 500 | 1000 | 1500 | 100 |

Expected total = (6^{6}) = 6 x 6 x 6 x 6 x 6 x 6 = 46,656 = calculated combinations

[1] An alternate logic yields the same result:

[C(6:2)] [C(6:1) x C(6:1)] [(6:4) x (2!)] =

** ** 15 x 6 x 6 x 15 x 2 = **16,200 **

In this not-as-easy-to-explain case, the first bracket refers to the two pair of dice. Thereby this first step, in calculating possible pairs, automatically eliminates redundancy of colors later. How many ways can any two pairs of colors appear on six dice? Fifteen. How many different colors are available on each die? Six. How many colors are available for the first pair? Six. The second pair? Six. Said again, the first two brackets first establish the number of possible pairs without risk of reusing the same colors. So far, two colors and four dice have been utilized. How many different colors are left for the last two dice?[1] Four of six. And how many nonmatching combinations of colors are those last four colors free to form? 2.

[2] An alternate way: [C(6:2) x C(5:2)] [C(4:2) x C(4:2)] [(2!)] =

15 x 10 x 6 x 6 x 2 = 10,800

[3] An alternate way: [C(6:3)] [C(6:2) x C(4:2) x C(2:2)] = 20 x 15 x 6 x 1 = 1,800.

First bracket: Three pairs can be selected from six colors 20 ways. Second bracket: The first pair can be any of six colors, the second pair any of the four remaining colors, and the third pair can be any of the two remaining colors.