Groups format

[JV24601]

I'm on a trip of 7 people playing 6 consecutive days as a 4ball and a 3ball.

Any ideas how I can compose the groups each day so they're fairly levelled out without playing with the same people all the time and without always being in the 4ball or 3ball group?
I imagine there will be an online creator somewhere but I can't find one.

Any help appreciated cheers!

 

[mikejohnchapman]

You could play a bowmaker across both groups and emply a "ghost" player to make the 3 into a 4. Either draw or pick one of the players in the 4 to have his / her score added to the 3 to calculate the team score. You would need to run this as an individual stableford to get the fairest result but it would be good fun to debate the final result.

Could pick the ghost by trying to equalise the total team handicaps - take the total for the team of 4 and use the handicap to add to the team of 3to try to get them to the same total - or you could just draw it from the 4.

 

[full_throttle]

try this,

1234 567
2345 671
3456 712
4567 123
5671 234
6712 345

 

[duncan mackie]

try this,

1234 567
2345 671
3456 712
4567 123
5671 234
6712 345

The flaw in this is highlighted when you see that 7 and 3 never play in the same game!

 

[JV24601]

The flaw in this is highlighted when you see that 7 and 3 never play in the same game!

So they don’t. That’s a shame.
I’ve been trying all weekend and can’t come up with the perfect draw.

Need a mathematician on the case!

 

[JV24601]

1234 567
2376 541
3456 712
4517 623
4671 235
3512 674

1 – 4 x 4 balls / 2 x 3 balls
2 – 3 x 4 balls / 3 x 3 balls
3 – 4 x 4 balls / 2 x 3 balls
4 – 4 x 4 balls / 2 x 3 balls
5 – 3 x 4 balls / 3 x 3 balls
6 – 3 x 4 balls / 3 x 3 balls
7 – 3 x 4 balls / 3 x 3 balls

1 & 2 – 3 times
1 & 3 – 2 times
1 & 4 – 4 times
1 & 5 – 3 times
1 & 6 – 1 time
1 & 7 – 3 times
2 & 3 – 5 times
2 & 4 – 1 time
2 & 5 – 2 times
2 & 6 – 2 times
2 & 7 – 2 times
3 & 4 – 2 times
3 & 5 – 3 times
3 & 6 – 3 times
3 & 7 – 1 time
4 & 5 – 3 times
4 & 6 – 3 times
4 & 7 – 3 times
5 & 6 – 2 times
5 & 7 – 2 times
6 & 7 – 4 times

I’d be interested to see if anybody can beat this and make it so there’s nobody who only plays once with another??
Problematic groups highlighted.

 

[richart]

The flaw in this is highlighted when you see that 7 and 3 never play in the same game!

Just give numbers 3 and 7 to two golfers that can't stand each other. Problem solved.:thup: