We have been determining flight order for each round of stunt at the Nats by random draw. This might not be the fairest way to do it. Stunt lore has it that flying late in a round is advantageous. A purely random draw can give one guy a bunch of early flights over the three or four rounds of a phase of the contest, and give another guy a bunch of late flights. I talked to Dennis Adamisin about this yesterday. Long ago his mother came up with a mapping scheme that would determine subsequent-round flight orders from a randomly drawn first round. I proposed putting that or something like that into the Nats tabulation program. He suggested that it might be worth doing for the Open finals, but he was reluctant to have me further complicate the rest of the contest. I set out to concoct a mapping.
First I think we should come up with a figure of merit: something measureable to shoot for. I propose adding the flight orders for each guy and trying to make that sum come out the same for everybody. Here’s an example randomly drawn Nats Open finals (five guys, three flights each):
guy flight 1 flight 2 flight 3
John 1 2 5
Paul 2 4 4
George 3 5 1
Ringo 4 1 2
Howard 5 3 3
Sums of flight orders in this case are 8 for John, 10 for Paul, 9 for Ringo, and so on. The attached PDF shows the probabilities for each possible sum for randomly drawn flights. It goes from 3 (all 1’s) to 15 (all 5’s).
A more uniform way to assign flight orders is to randomly draw the first round, then move everybody down two notches for each of the other flights. Here’s what that looks like. Behold in all these cases I’m sorting by the first-flight order.
guy flight 1 flight 2 flight 3
Howard 1 4 2
Paul 2 5 3
Ringo 3 1 4
George 4 2 5
John 5 3 1
This yield sums of 7, 10, 8, 11, and 9: a little more uniform and the same every time. It has the property of having each guy follow the same person in most cases for each round, which is probably not good.
Better, I think, would be something like this:
guy flight 1 flight 2 flight 3
Paul 1 4 4
George 2 5 2
Ringo 3 1 5
Howard 4 2 3
John 5 3 1
The numbers in the table would stay fixed. The sum of flight orders in each case is 9. We would randomly assign guys to the five rows. What do you think?
Topic: Nats Stunt Flight Order Scheme (Read 3334 times)