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.