of

In the case

consider a table of possible outcomes:

(1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |

(2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |

(3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |

(4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |

(5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |

(6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |

Next, we can consider the sums associated with these outcomes:

2 | 3 | 4 | 5 | 6 | 7 |

3 | 4 | 5 | 6 | 7 | 8 |

4 | 5 | 6 | 7 | 8 | 9 |

5 | 6 | 7 | 8 | 9 | 10 |

6 | 7 | 8 | 9 | 10 | 11 |

7 | 8 | 9 | 10 | 11 | 12 |

Since there are 36 outcomes, each equally likely, we see that the probabilities are:

X |
count | P()
X |
---|---|---|

2 | 1 | 1/36 |

3 | 2 | 2/36 |

4 | 3 | 3/36 |

5 | 4 | 4/36 |

6 | 5 | 5/36 |

7 | 6 | 6/36 |

8 | 5 | 5/36 |

9 | 4 | 4/36 |

10 | 3 | 3/36 |

11 | 2 | 2/36 |

12 | 1 | 1/36 |

If there are more than two dice, it is difficult to make tables such as these. The number

of outcomes will be 6

number of times each sum occurs among the 6

by considering the generating function for the number of times each sum appears:

f(x) = (x + x

(x + x

The coefficient of x