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Probability IA.docx
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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

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