Trying to Solve Math Problem for Real World Use - Combinatorics

Question

I'm trying to solve a math problem that hasn't been solved - to anyone's knowledge - in the community it's being used in. I am sure it is not difficult, but I am not smart enough to figure it out.

In England, when on a country shoot (part of Britain's heritage) there are 8 "pegs" (shooting position in a straight line numbered 1-8) and shoot four "drives" (45 minute period of shooting). People draw pegs blind and then there are several ways that people change pegs across the 4 drives. Move two up: 1 goes to 3 goes to 5 goes to 7. Move up three: 1 goes to 4 goes to 7 goes to 2. Odds up 3, evens down 3, etc. 4 and 5 are considered the best "pegs" and 1 and 8 are considered the worst.

The questions is this: How would you solve this problem trying to solve for two different parameters: 1) Everyone get an equal distribution of being at 4/5 and 1/8 or at least close to them such that no one is advantaged over the course of the four "drives" and everyone is equally in the center or on the ends. 2) People get to stand next to different people across the course of the day and not always next to the same people (the reason odds up and evens down was invented).

No one particularly likes the current numbering system and many are looking for an alternative where you draw a number sequence as opposed to a number. (IE, you draw a card that has the "peg" order pre-determined for the 8 people - eg 3,1,5,7)

Thanks for your help! :)

Rand

PS Someone tried to solve this problem previously and could only make it work with 9 "pegs" and not 8. See link - https://www.gunsonpegs.com/articles/shooting-talk/alternatives-to-moving-up-2-the-durnford-wheel

Answer 1

How about this schedule? Everyone gets one drive on the end and one in the middle, and also one drive which is one from the middle and one which is one from the end. No two people stand together twice.

Letters are people, rows are drives, columns are pegs.

A B H C G D F E
B C A D H E G F
C D B E A F H G
D E C F B G A H

Answer 2

If you want a wheel-like system, by my (corrected) calculation there are 4 possibilities (8 with reversals).

• $1 \to 2 \to 4 \to 6 \to 8 \to 7 \to 5 \to 3 \to 1$ (almost evens-up odds-down)
• $1 \to 2 \to 5 \to 3 \to 8 \to 7 \to 4 \to 6 \to 1$
• $1 \to 3 \to 5 \to 7 \to 8 \to 6 \to 4 \to 2 \to 1$ (almost odds-up evens-down)
• $1 \to 3 \to 4 \to 2 \to 8 \to 6 \to 5 \to 7 \to 1$
• $1 \to 7 \to 4 \to 3 \to 8 \to 2 \to 5 \to 6 \to 1$
• $1 \to 7 \to 5 \to 6 \to 8 \to 2 \to 4 \to 3 \to 1$
• $1 \to 6 \to 5 \to 2 \to 8 \to 3 \to 4 \to 7 \to 1$
• $1 \to 6 \to 4 \to 7 \to 8 \to 3 \to 5 \to 2 \to 1$

The first or third seem like they might be easiest to sell from a cultural standpoint.