Hello Everyone,

My name is Francesc, or Flynnt on the forums, and Dan has invited me to generate content for this site. I am actually pretty happy to contribute!

I decided to begin my collaboration with a post on Guildball dice math. In this post, I will give you three tables that show how many successes you should expect when performing a TN test. This is useful when performing an attack, using a play or kicking the ball. At the end of the article, I will also explain how the numbers on the tables are generated , but you can ignore it and take the tables to your next game!

I added three tables, depending on the probability of the outcome. The first table has the number of successes that should happen at least 80% of the time (or 4 out of 5 times), the next one is 75% (or 3 out of 4 times) and the last one is 66% (or 2 out of 3 times).

Now, how can you use this information?

It can be use to measure your odds on a **goal**, for instance, let’s say that you are playing with Morticians (my home team!) and that Obulus has the ball and is ready to score. Since his dice pool is 2, we will be using the first row on every table. If you want your play to have at least 75% chance of success you are good, since for a TN of 4+ and 2 dice, the 75% table says that at least you will get one success. However, if you want your odds to be at least of 80% you will have to double time it as with 2 die you have 0 successes on a TN 4+ but with 3 die you do have 1 success on a TN 4+.

Alternatively, this could also be used to see how much **damage** you could make to someone. Therefore, only for fun, let’s see what a fully loaded Cosett can do with a 75% of confidence! I’m going to assume that Dirge went before her and Singled Out [+2 TAC] a model, let’s say Boar [def 2+], and now she goes Crazy [+3 TAC] and charges with a double time [1 MP]. Her first attack uses 14 die, which should generate at least 11 hits allowing 4 + 3 damage (from the playbook). Those numbers should be 5+4 since she should be close to Dirge and generate 1 MP. Then, she buys another attack, this one she double times with the MP generated on her previous one, so she has a pool of 10 die, causing 8 successes which means 4 + 1 damage, which again is 5+2. If she can double time the next one, she should cause another 4+1 (or 5+2) which should kill Boar!

However, that strategy falls apart when checking to a confidence of 80%, since the extra two attacks would only generate 7 hits which could only cause 4 each.

TN 2+ | TN 3+ | TN 4+ | TN 5+ | TN 6+ | |
---|---|---|---|---|---|

2 die | 1 | 1 | 0 | 0 | 0 |

3 die | 2 | 1 | 1 | 0 | 0 |

4 die | 3 | 2 | 1 | 1 | 0 |

5 die | 4 | 2 | 2 | 1 | 0 |

6 die | 4 | 3 | 2 | 1 | 0 |

7 die | 5 | 4 | 2 | 1 | 0 |

8 die | 6 | 4 | 3 | 2 | 0 |

9 die | 7 | 5 | 3 | 2 | 1 |

10 die | 7 | 5 | 4 | 2 | 1 |

11 die | 8 | 6 | 4 | 2 | 1 |

12 die | 9 | 7 | 5 | 3 | 1 |

13 die | 10 | 7 | 5 | 3 | 1 |

14 die | 11 | 8 | 5 | 3 | 1 |

15 die | 11 | 8 | 6 | 3 | 1 |

TN 2+ | TN 3+ | TN 4+ | TN 5+ | TN 6+ | |
---|---|---|---|---|---|

2 die | 1 | 1 | 1 | 0 | 0 |

3 die | 2 | 1 | 1 | 0 | 0 |

4 die | 3 | 2 | 1 | 1 | 0 |

5 die | 4 | 3 | 2 | 1 | 0 |

6 die | 4 | 3 | 2 | 1 | 0 |

7 die | 5 | 4 | 3 | 1 | 0 |

8 die | 6 | 4 | 3 | 2 | 1 |

9 die | 7 | 5 | 3 | 2 | 1 |

10 die | 8 | 6 | 4 | 2 | 1 |

11 die | 8 | 6 | 4 | 3 | 1 |

12 die | 9 | 7 | 5 | 3 | 1 |

13 die | 10 | 8 | 5 | 3 | 1 |

14 die | 11 | 8 | 6 | 3 | 1 |

15 die | 12 | 9 | 6 | 4 | 1 |

TN 2+ | TN 3+ | TN 4+ | TN 5+ | TN 6+ | |
---|---|---|---|---|---|

2 die | 2 | 1 | 1 | 0 | 0 |

3 die | 2 | 2 | 1 | 1 | 0 |

4 die | 3 | 2 | 2 | 1 | 0 |

5 die | 4 | 3 | 2 | 1 | 0 |

6 die | 5 | 4 | 2 | 1 | 0 |

7 die | 6 | 4 | 3 | 2 | 1 |

8 die | 6 | 5 | 3 | 2 | 1 |

9 die | 7 | 5 | 4 | 2 | 1 |

10 die | 8 | 6 | 4 | 3 | 1 |

11 die | 9 | 7 | 5 | 3 | 1 |

12 die | 10 | 7 | 5 | 3 | 1 |

13 die | 10 | 8 | 6 | 4 | 1 |

14 die | 11 | 9 | 6 | 4 | 2 |

15 die | 12 | 9 | 7 | 4 | 2 |

Now to explain the actual math behind the tables.

There is a formula that lets you measure what are the odds of, for instance, rolling three 5’s when using a given number of die. Assuming you are rolling **n** die, the odds of rolling at least three 5’s is:

because the odds of rolling 5’s on a D6 is 2/6.

Therefore, the **probability of rolling exactly k of a number in n die** is:

where *p* is the probability of rolling the number we are looking for (so 1/6 for a 6, 2/6 for a 5 and so on).

The **probability of rolling at least k in n die** is a summation of the previous odds:

Therefore, each entry on the table has the highest k in which the probability is still the number set (so 0.8, 0.75 or 0.66).