Finally, if you've made this far then you may be thinking at what point does the number of competitors change the Starting Round from Semi-Final to Quarter-Final and Quarter-Final to Pre-Quarter-Final?

\( C \) | = | Number of Competitors |

\( a \) | = | Attrition Rate |

\( s \) | = | Success Rate = \( (1 - a) \) |

\( r \) | = | Number of Rounds excluding the Final |

\( F \) | = | Number of Finalists |

A fraction \(s\) of competitors succeed in going through to the next round, so the number of Finalists \(F\) remaining from an original pool of competitors \(C\) going through a number of rounds \(r\) with a success rate of \(s\) is given by:

\[ F = C \times a^r \]

The number of competitors at the start of a competition is:

\[ C = {F \over a^r} \]

For many Modern Jive competitions \( F = 6 \) and our preferred success rate \( s = 0.5 \) giving,

\[ C = {6 \over 0.5^r} = 6 \times 2^r \]

For each value of \(r\) corresponding to our competition rounds we get,

\( r \) | 3 (PQF) | 2 (QF) | 1 (SF) | 0 (F) |

\( C \) | 48 | 24 | 12 | 6 |

To get the maximum number of competitors for each round we can see what the number would be for half round values. This is a better indicator than just taking the average because there is a power relationship between the number of competitors in each round.

\( r \) | 3.5 | 2.5 | 1.5 | 0.5 |

\( C \) | 68 | 34 | 17 | 8 |

This also tells us that we shouldn't have more than 8 entries in a final.

#### Attrition Rate

We can determine the success rate \(s\) from the number of rounds \(r\), competitors \(C\) and finalists \(F\):

\[ s = \left( F \over C \right) ^ {1 \over r} \]

Giving the attrition rate \(a = 1 - s \),

\[ a = 1 - \left(F \over C\right) ^ {1 \over r} \]

#### Calculating the Points Starting Round

Right from the start we wanted to know what the starting round \( r \) should be for a given number of competitors, finalists and rate of attrition / success.

\[ r = { \ln \left( F \over C \right) \over \ln \left( s \right) } \]

\[ r = { \ln \left( F \over C \right) \over \ln \left( s \right) } = { { \ln \left( C \over F \right) } \over { \ln \left( 1 \over s \right) } } \]

If \( F = 6 \) and \( s = 0.5 \) then this can be written as:

\[ r = { { \ln \left( C \over 6 \right) } \over { \ln \left( 1 \over 0.5 \right) } } \]

\[ r = { { \ln \left( C \over 6 \right) } \over { \ln \left( 2 \right) } } \text{, where } C \geq 6 \]

If \( C < 6 \) then \( r = 0 \) as we can't have a negative number of rounds (well we can in theory just like we have fractional rounds but we wouldn't or *couldn't* have them in practice.

This may seem overly complicated at first but hopefully what it does achieve is a sense of finality in that can this be analysed or taken further? By providing a nice mathematical approach (which even includes logs - and you can't argue with logs) then it shows that this whole problem / challenge has been thought out thoroughly as our dancers have come to expect of us.