# Langoor Burja Analysis

* Assuming fair (unbiased) dices

# Rules

* total places to bet = 6

* total no. of dces = 6

* player wins = n * bet

n = no. of dices rolled to bet position

*Rules source: https://www.techlekh.com/langur-burja-android-app-3d/#:~:text=In%20this%20game%2C%20symbols%20named,host%20will%20roll%20the%20dices.

# calculation:

* probability of loosing = (5/6)^6 = 0.335

* because each die has a 5/6 chance of not showing the chosen symbol, and we want all six dice to not show the chosen symbol.

* probability of draw (n=1):

6 * (1/6) * (5/6)^5 = 0.402

. Here, we multiply by 6 because there are six possible dice that could show the chosen symbol.

The term (1/6) represents the probability of a single die showing the chosen symbol, and (5/6)^5 represents the probability of the remaining five dice not showing the chosen symbol.

* probability of getting twice the bet(n=2):

C(6, 2) * (1/6)^2 * (5/6)^4

C(6, 2): represents the combination formula, specifically the number of ways to choose 2 items from a set of 6. In this case, we have 6 dice, and we want to calculate the number of ways to select 2 dice to show the chosen symbol. The formula C(6, 2) calculates this value, which is 15. So, there are 15 different combinations of 2 dice out of 6.

(1/6)^2: represents the probability of a single die showing the chosen symbol. Since we have assumed fair and unbiased dice, each die has a 1/6 chance of showing the chosen symbol. Multiplying this probability by itself (squared) accounts for the fact that we want two dice to show the chosen symbol simultaneously.

(5/6)^4: represents the probability of the remaining 4 dice not showing the chosen symbol. Since we assume fair dice, each of the other 4 dice has a 5/6 chance of not showing the chosen symbol. Multiplying this probability by itself (four times) accounts for the fact that we want all four remaining dice to not show the chosen symbol.

* probability of getting thrice the bet (n=3):

C(6, 3) * (1/6)^3 * (5/6)^3

* probability of getting four times the bet(n=4):

C(6, 4) * (1/6)^4 * (5/6)^2

* probability of getting 5 times the bet(n=5):

C(6, 5) * (1/6)^5 * (5/6)^1

* probability of getting 6 times the bet(n=6):

C(6, 6) * (1/6)^6 * (5/6)^0

* probability of winning:

sum(n=2,3,4,5,6) = 0.263

Alternatively: 1 - p_draw - p_loosing = 1 - .402 - .335 = .263

* win amount:

= [x * p(n=1)] + [2x * p(n=2)] + [3x * p(n=3)] + [4x * p(n=4)] + [5x * p(n=5)] + [6x * p(n=6)] - p(loosing) * x

= x * 6 * (1/6) * (5/6)^5 + 2x * C(6, 2) * (1/6)^2 * (5/6)^4 + 3x * C(6, 3) * (1/6)^3 * (5/6)^3 + 4x * C(6, 4) * (1/6)^4 * (5/6)^2 + 5x * C(6, 5) * (1/6)^5 * (5/6)^1 + 6x * C(6, 6) * (1/6)^6 * (5/6)^0 - (5/6)^6

= 0.6651 x