Easy method for counting rounds
from UDE's Player Management Forums
At a recent event, one of my floor judges asked how many rounds we were playing. When I counted them out 8 rounds for the 224 players on just the fingers of my left hand, he said, “Woah, you have got to teach me that.” So, if it helps, I thought others could benefit from this method. The math will be below:
Each finger is double that which precedes it, starting at 2. So, 1st finger is 2, 2nd is 4, 3rd is 8, 4th is 16, 5th is 32, etc. Each finger represents rounds played and the sum equals the maximum players for each round. Keep in mind that 2 players is not enough and 4 players is a minimum, not a maximum and you have your start of 3 rounds for (4-)8 max. 4 Rounds for (9-)16 max, etc. The break down is as follows (it’s easier to demonstrate in person):
1 finger, 2 people =/= tournament rounds
2 fingers, 4 people = minimum, keep counting.
3 fingers, 8 people max.= 3 rounds (4-8)
4 fingers, 16 people max. = 4 rounds (9-16)
5 fingers, 32 people max. = 5 rounds (17-32)
From here, if you use American Sign Language counting you can go to ten on one hand. Touch thumb to pinky for 6, ring finger for 7, middle finger for 8, index for 9 and thumb up for ten.
6 ->64 max = 6 rounds
7->128 max = 7 rounds
8->256max = 8 rounds
9->512max = 9 rounds
10->1024max = 10 rounds.
The Math:
Basically you begin with 2 (makes sense, you can’t even play the game without 2 people). Each finger, then, represents the exponent (“to the power of”) of 2. So, 2 ^1 =2, 2 ^2 (squared) =4, 2 ^3 (cubed) = 8, 2 ^4 =16, etc. where exponent = rounds and sum = maximum people per round.
This formula is a modified version of a binary counting (on/off switches) method I developed (although I am sure others have thought of it before me) utilizing each finer to represent a number that added together would produce twice the previous finger number (-1). This allows me to count to 1023 on one hand and 1,048,575 on both hands. Index finger =1, middle =2, index+middle =3, ring =4, ring+index =5, ring+middle =6, ring+middle+index =7, pinky =8, etc.
Let me know if this is something you think could help other judges. Or, if you have your own easy way (besides pestering the Scorekeeper, or memorizing the chart), please share.
Zaphenathpaneah.
http://entertainment.upperdeck.com/COMMUNITY/forums/thread/1053402.aspx
from UDE's Player Management Forums
At a recent event, one of my floor judges asked how many rounds we were playing. When I counted them out 8 rounds for the 224 players on just the fingers of my left hand, he said, “Woah, you have got to teach me that.” So, if it helps, I thought others could benefit from this method. The math will be below:
Each finger is double that which precedes it, starting at 2. So, 1st finger is 2, 2nd is 4, 3rd is 8, 4th is 16, 5th is 32, etc. Each finger represents rounds played and the sum equals the maximum players for each round. Keep in mind that 2 players is not enough and 4 players is a minimum, not a maximum and you have your start of 3 rounds for (4-)8 max. 4 Rounds for (9-)16 max, etc. The break down is as follows (it’s easier to demonstrate in person):
1 finger, 2 people =/= tournament rounds
2 fingers, 4 people = minimum, keep counting.
3 fingers, 8 people max.= 3 rounds (4-8)
4 fingers, 16 people max. = 4 rounds (9-16)
5 fingers, 32 people max. = 5 rounds (17-32)
From here, if you use American Sign Language counting you can go to ten on one hand. Touch thumb to pinky for 6, ring finger for 7, middle finger for 8, index for 9 and thumb up for ten.
6 ->64 max = 6 rounds
7->128 max = 7 rounds
8->256max = 8 rounds
9->512max = 9 rounds
10->1024max = 10 rounds.
The Math:
Basically you begin with 2 (makes sense, you can’t even play the game without 2 people). Each finger, then, represents the exponent (“to the power of”) of 2. So, 2 ^1 =2, 2 ^2 (squared) =4, 2 ^3 (cubed) = 8, 2 ^4 =16, etc. where exponent = rounds and sum = maximum people per round.
This formula is a modified version of a binary counting (on/off switches) method I developed (although I am sure others have thought of it before me) utilizing each finer to represent a number that added together would produce twice the previous finger number (-1). This allows me to count to 1023 on one hand and 1,048,575 on both hands. Index finger =1, middle =2, index+middle =3, ring =4, ring+index =5, ring+middle =6, ring+middle+index =7, pinky =8, etc.
Let me know if this is something you think could help other judges. Or, if you have your own easy way (besides pestering the Scorekeeper, or memorizing the chart), please share.
Zaphenathpaneah.
http://entertainment.upperdeck.com/COMMUNITY/forums/thread/1053402.aspx