I've been to a couple small tournaments and league events that have done this. I was told this is commonplace at many larger events. Is this true?
The idea is: when choosing random doubles partners, everyone would flip their own disc at once. Heads people would go to one side, and tails would go to the other. Then begins a second round of flipping. Each group splits again based on heads/tails. These divisions continue until two people remain and they become partners. If an odd person remains, they become Cali.
I've thought about this system a little bit and it doesn't make sense to me in instances of multiple odd-men-out. I've tried to communicate this problem before, but I alone doubt the method. Am I crazy?
Here's a example of where the method breaks down:
Start flipping with 10 golfers: Half get heads and half get tails.
Now, the 5 heads flip; there is exactly 1 tails.
The 4 remaining heads flip; there is exactly 1 tails.
The 3 remaining heads flip; there is exactly 1 tails.
The 2 remaining people become partners.
Meanwhile, on the 5-Tails side: the exact same thing happens.
Thus, we're left with two pairings completed and 6 remaining ungrouped people who were an odd-man-out at the end of their flips.
In rows, it could be represented like this:
10
5 & 5
4 / 1 & 4 / 1
3 / 1 / 1 & 3 / 1 / 1
[2] / 1 / 1 / 1 & [2] / 1 / 1 / 1
Matched twosomes are removed to yield:
1 / 1 / 1 & 1 / 1 / 1
Lets rename these players as:
A / B / C & X / Y / Z
Here, there are many ways to group the remaining players.
We could say those who only flipped twice (or three times) are paired together, we get: AX, BY, CZ
Or, we could say the first two odd men out on a side get paired together, we get: AB, XY, CZ
In short, there are many logical ways to make this determination. But, in reality, I've never heard such rule intricacies explained beforehand (or even during). For larger numbers of players, I doubt a simple rule could suffice for all possible outcomes. What usually ends up happening is the odd-men-out just wander around until they find another odd-man-out and they group up.
Has anybody else thought about this?
The idea is: when choosing random doubles partners, everyone would flip their own disc at once. Heads people would go to one side, and tails would go to the other. Then begins a second round of flipping. Each group splits again based on heads/tails. These divisions continue until two people remain and they become partners. If an odd person remains, they become Cali.
I've thought about this system a little bit and it doesn't make sense to me in instances of multiple odd-men-out. I've tried to communicate this problem before, but I alone doubt the method. Am I crazy?
Here's a example of where the method breaks down:
Start flipping with 10 golfers: Half get heads and half get tails.
Now, the 5 heads flip; there is exactly 1 tails.
The 4 remaining heads flip; there is exactly 1 tails.
The 3 remaining heads flip; there is exactly 1 tails.
The 2 remaining people become partners.
Meanwhile, on the 5-Tails side: the exact same thing happens.
Thus, we're left with two pairings completed and 6 remaining ungrouped people who were an odd-man-out at the end of their flips.
In rows, it could be represented like this:
10
5 & 5
4 / 1 & 4 / 1
3 / 1 / 1 & 3 / 1 / 1
[2] / 1 / 1 / 1 & [2] / 1 / 1 / 1
Matched twosomes are removed to yield:
1 / 1 / 1 & 1 / 1 / 1
Lets rename these players as:
A / B / C & X / Y / Z
Here, there are many ways to group the remaining players.
We could say those who only flipped twice (or three times) are paired together, we get: AX, BY, CZ
Or, we could say the first two odd men out on a side get paired together, we get: AB, XY, CZ
In short, there are many logical ways to make this determination. But, in reality, I've never heard such rule intricacies explained beforehand (or even during). For larger numbers of players, I doubt a simple rule could suffice for all possible outcomes. What usually ends up happening is the odd-men-out just wander around until they find another odd-man-out and they group up.
Has anybody else thought about this?