# The Dice Game

1. Feb 18, 2012

### kirman

Hello everyone,

Kind regards,
Kirman.

The aim of this task is to create a dice game in a casino and model this using
probability. It is important to examine how best to run the game from both the
perspective of a player and the casino. In doing so analyse the game to consider the
optimal payments by the player and payouts by the casino.
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1. Consider a game with two players, Ann and Bob. Ann has a red die and Bob a white die. They roll their dice and note the number on the upper face. Ann wins if her score is higher than Bob’s (note that Bob wins if the scores are the same). If both players roll their dice once each what is the probability that Ann will win the game?
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2. Now consider the same game where Ann can roll her die a second time and will note the higher score of the two rolls but Bob rolls only once. In this case what is the probability that Ann will win?
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3. Investigate the game when both players can roll their dice twice, and also when both players can roll their dice more than twice, but not necessarily the same number of times. Consider the game in a casino where the player has a red die and the bank has a white die. Find a model for a game so that the casino makes a reasonable profit in the case where the player rolls the red die once and the bank rolls the white die once. (When creating your model you will need to consider how much a player must pay to play a game and how much the bank will pay out if the player wins. Do this from the perspective of both the player and the casino and consider carefully the criteria for whether the game can be considered worthwhile for both the player and the casino.)

4. Now consider other models for the game including cases such as where the player or the bank rolls their dice multiple times, or where multiple players are involved in the game.

2. Feb 19, 2012

### HallsofIvy

Staff Emeritus
This looks like homwork and no effort is shown. I am moving this to the "precalculus homework" section. kirman, if you do not show some effort by tomorrow, I will delete it.

3. Feb 22, 2012

### kirman

Hi HallsofIvy,

The Dice Game
Table of rolling the two dice

A = event where Ann is winning
B = event where Bob is winning
L = event neither Ann and Bob is winning
P(A)=15/36, P(B)=6/36, P(L)=15/36

(point number one) Case where Ann is winning has probability of P(A)=15/36
(point number two) Case where Ann rolling for the second time is only happened when both Ann and Bob didn’t win the first roll. The probability is:
P(L)∙P(A)=15/36∙15/36
(point number three)
Case where both player can roll two times. There are two situations:
Ann is winning: P(L)∙P(A)=15/36∙15/36
Bob is winning: P(L)∙P(A)=15/36∙6/36
Case where they can roll to infinity
And Ann finally win the game
Winning the 1st roll = 15/36
Winning on the 2nd roll =15/36∙15/36
Winning on the 3rd roll =15/36∙15/36∙15/36

Sum of all (using geometric progression)
S=t_1/(1-r)
S=(15/36)/(1-15/36)
S=15/21
And Bob win the game
Winning the 1st roll = 6/36
Winning on the 2nd roll =15/36∙6/36
Winning on the 3rd roll =15/36∙15/36∙6/36

Sum of all (using geometric progression)
S=t_1/(1-r)
S=(6/36)/(1-15/36)
S=6/21
(point number four) In a casino, whenever a player wants to roll the die, he/she have to pay the bet. In this case, the situations where neither the player or bank are winning add favor to the bank’s side.
A = event where player is winning. P(A)=15/36
A’ = event where bank is winning. P(A')=21/36
P is pay out, the amount of the bank pays if the player win. B is betting value, the amount of money the player has to pay each time she/he wants to play.
Now, let’s calculate for a fair game (the expectancy is zero)
E=P(A)∙P-P(A')∙B
0= 15/36∙P-21/36∙B, thus we have the ratio
P=21/15 B
Case where they can roll to infinity
And the player finally win the game
Winning the 1st roll = 15/36
Winning on the 2nd roll =21/36∙15/36
Winning on the 3rd roll =21/36∙21/36∙15/36

Sum of all (using geometric progression)
S=t_1/(1-r)
S=(15/36)/(1-21/36)
S=15/15
This can't be right since the calculated probability is 1.
And Bank win the game
Winning the 1st roll = 21/36
Winning on the 2nd roll =15/36∙21/36
Winning on the 3rd roll =15/36∙15/36∙21/36

Sum of all (using geometric progression)
S=t_1/(1-r)
S=(21/36)/(1-15/36)
S=21/21
This can't be right since the calculated probability is 1.