- •Table 1 Combinations highlighted in green are the winning combinations for Ann and combination highlighted in red are the winning combination for Bob.
- •Table 2, which represents the values to be used to calculate probability of Bob winning when Ann throws twice and Bob once
- •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.
- •Table 3, which represents the values to be used to calculate probability of Bob winning when each throws once only
- •In order to prove whether the formula works the values are going to be substituted into the formula. I am going to use the values for Ann losing when she throws her die twice and Bob once only:
- •Table 4 to show number of combinations for Bob winning and number of combinations for Ann losing
- •Instead.
- •It is evident that for just above 100,000,000 throws of a die the probability seems getting close to the theoretical value used throughout my investigation.
Table 3, which represents the values to be used to calculate probability of Bob winning when each throws once only
The same steps will be done to calculate the sum of 6 , which are different due to the Number Bob gets on the face of his die.
From both examples examined I can suggest that the number of the combinations that result in Ann losing is a number Bob gets on the face of his die with an exponent since for the case when Ann has two throws each value of (from the table 3) corresponding to the right number Bob gets on the face of his die (as in the table) 2. Therefore represents the number of throws that Ann has.
Since mentioned statement above is explained I’ve arrived onto the general formula:
Where the denominator is due to being a constant since probability of getting any number on the dice for him is and he only gets to throw a dice once. The reason the exponent is is that is the number of throws to be done by Ann and is for the which is constant.
In order to prove whether the formula works the values are going to be substituted into the formula. I am going to use the values for Ann losing when she throws her die twice and Bob once only:
Furthermore for the case when Bob and Ann throw a die one time each, the formula was tested:
Since the results of probabilities for Bob to win calculated using the values in the table and the formula: were found to be identical with using the general formula I came up with I can say that the general formula for Bob winning for the infinite number of throws by Ann however only for the case when Bob throws the dice once is correct.
Now I am going to 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. I am going to do that from Bob’s perspective (if he wins and Ann loses).
Number on the face of the dice |
Bob’s combinations from 2 throws to win |
Number of combinations for Bob |
Ann’s combinations from two throws to lose |
Number of combinations Ann |
1 |
(1,1) |
1 |
(1,1) |
1 |
2 |
(1,2),(2,1),(2,2) |
3 |
(1,1),(1,2),(2,1),(2,2) |
4 |
3 |
(3,1),(3,2),(3,3),(1,3),(2,3) |
5 |
(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3),(1,3),(3,1) |
9 |
4 |
(4,1),(4,2),(4,3),(4,4),(1,4),(2,4),(3,4) |
7 |
(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3),(1,3),(3,1),(4,1),(4,2),(4,3),(4,4),(1,4),(2,4),(3,4) |
16 |
5 |
(5,1),(5,2),(5,3),(5,4),(5,5),(1.5),(2,5),(3,5),(4,5) |
9 |
(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3),(1,3),(3,1),(4,1),(4,2),(4,3),(4,4),(1,4),(2,4),(3,4),(5,1),(5,2),(5,3),(5,4),(5,5),(1.5),(2,5),(3,5),(4,5) |
25 |
6 |
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),(1,6),(2,6),(3,6),(4,6),(5,6) |
11 |
(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3),(1,3),(3,1),(4,1),(4,2),(4,3),(4,4),(1,4),(2,4),(3,4),(5,1),(5,2),(5,3),(5,4),(5,5),(1.5),(2,5),(3,5),(4,5),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),(1,6),(2,6),(3,6),(4,6),(5,6) |
36 |