Candidate number:004437-028
Candidate name:Marat Omargaliyev
In 2013 in the most flashing Megapolis of the World-Las Vegas will happen an opening of one of the largest and competitive casinos in its market, “The Gloruius”. Me a person holding Phd in Statistics had been contacted by the shareholders and the employees of the project “The Glorius” and I’ve been requested to design carefully the criteria for the dice game to be worthwhile for both player and the casino and then give the suggestion on whether the dice game will be worth having it in the casino. I’ve been given unlimited amount of freedom on the different aspects of the rules for a player as well as for the percentage profit margin that will be sensible for “The Glorius”. The best approach for this task in my opinion was using probability on which the management has given me a green light. In order to make understandable for my supervisor and for the board of directors I’ve started with the lowest number of throws allowed for both player and the dealer where only one and important condition was kept at all times- If a dealer gets the same number on the face of the dice as the player, dealer wins.
As a starting point the probability of Ann winning if both her and Bob throw the die at a given condition only once will be calculated. Given condition: Ann wins if her score is higher than Bob’s and Bob wins if his score is higher or the same as Ann’s. Therefore the probability is calculated with the help of the table where in each cell of the table the first number is the number on the upper face of the die that Ann gets and the second number is the number that Bob gets.
6,1 |
6,2 |
6,3 |
6,4 |
6,5 |
6,6 |
5,1 |
5,2 |
5,3 |
5,4 |
5,5 |
5,6 |
4,1 |
4,2 |
4,3 |
4,4 |
4,5 |
4,6 |
3,1 |
3,2 |
3,3 |
3,4 |
3,5 |
3,6 |
2,1 |
2,2 |
2,3 |
2,4 |
2,5 |
2,6 |
1,1 |
1,2 |
1,3 |
1,4 |
1,5 |
1,6 |
Table 1 Combinations highlighted in green are the winning combinations for Ann and combination highlighted in red are the winning combination for Bob.
Therefore it can be observed from the table that there are 15 combinations out of 36 possible when Ann wins.
Hence, the probability that Ann will win the game if both players roll their die once each at a condition mentioned above is represented by substituting the values into the equation.
Where
is the number of occurrence of the event
and
is the total number of possible outcomes.
Hence the probability of Ann winning is:
Meaning that the
probability of Bob winning is:
Now the case is that Ann throws her die twice and then chooses the best out of two numbers against the number that Bob has after throwing his die once. The same condition used in the first case applies to this case as well. However now I will investigate the probability of Bob winning or in other words all the possible combinations that Ann gets in two of her throws that will lead to her loss. I think approaching the issue from Bob’s perspective will be easier to come up with the pattern. Therefore the table below was created.
Number Bob gets on the face of his die |
Possible combination of Ann losing |
Number of the combinations(c) |
1 |
(1,1) |
1 |
2 |
(1,1), (1,2),(2.1),(2,2) |
4 |
3 |
(1,1),(1,2),(2.1),(2,2),(3,1),(3,2),(3,3),(1,3),(2,3) |
9 |
4 |
(1,1),(1,2),(2.1),(2,2),(3,1),(3,2),(3,3),(1,3),(2,3),(1,4),(4,1),(2,4),(4,2),(3,4),(4,3),(4,4) |
16 |
5 |
(1,1),(1,2),(2.1),(2,2),(3,1),(3,2),(3,3),(1,3),(2,3),(1,4),(4,1),(2,4),(4,2),(3,4),(4,3),(4,4),(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4),(5,5) |
25 |
6 |
(1,1),(1,2),(2.1),(2,2),(3,1),(3,2),(3,3),(1,3),(2,3),(1,4),(4,1),(2,4),(4,2),(3,4),(4,3),(4,4),(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4),(5,5),(1,6),(6,1),(2,6),(6,2),(3,6),(6,3),(4,6),(6,4),(5,6),(6,5),(6,6) |
36 |
Table 2, which represents the values to be used to calculate probability of Bob winning when Ann throws twice and Bob once
Since Ann and Bob
have their own private die and they throw it one after another then
these can be considered as Independent events. Since they are
Independent events the formula for calculating the probability of Bob
winning for each face of his die will be:
.
Where
is the probability of Ann getting a losing number combinations which
will however vary for each number Bob gets on the face of his die and
then each will be multiplied by
which is constant
as Bob only rolls once, and finally they will be added up to find
probability that Bob wins.
Now these values are added up:
Is the probability that Bob wins if Ann throws her die twice and then chooses the best out of two numbers against the number that Bob has after throwing his die once.
Probability
of Ann winning if she throws her die twice and then chooses the best
out of two numbers against the number that Bob has after throwing his
die once is:
Since I’ve decided to shift onto the finding of the probability of Bob winning I will now do the same table but for Ann and Bob throwing their die once only.
Number Bob gets on the face of his die |
Possible combin6ation of Ann losing |
Number of the combinations(c) |
1 |
(1) |
1 |
2 |
(1),(2) |
2 |
3 |
(1),(2),(3) |
3 |
4 |
(1),(2),(3),(4) |
4 |
5 |
(1),(2),(3),(4),(5) |
5 |
6 |
(1),(2),(3),(4),(5),(6) |
6 |