15 spot raffle for 08 Sugartree custom

[gmcole]

This is what I have so far, I see I have some Pm's. In order to keep everything on the up and up,I think you must post your wish to be in and pick your #. thanks-jeff

List of current AZer's in raffle. In order to keep raffle legit-ONLY az members can be in raffle
...Handle.................raffle #
1. Subsonic2u ................1
2. Remy..........................6
3. Buckeyejim..................7
4. Gluton1......................13
5. retail1lo......................8
6.windowsrefund..............?
7.soldtonight...................5
8.
9.
10.
11.
12.
13.
14.

 

[windowsrefund]

I will pick #12

 

[brd]

I'm in let me have #3 let me know where to send payment.

 

[windowsrefund]

$100 sent via paypal

 

[gmcole]

I'm in let me have #3 let me know where to send payment.

Paypal is
gmcole@aol.com
good luck-jeff

 

[gmcole]

I'm going to bed.
PLEASE post your intent via quick reply and pick your number.
that way there is no confusion on who had what number etc.

The current avalible numbers are, if I deducted properly.
0,2,4,9,10,11,and 13

 

[gmcole]

If I read the powerball right, there is a drawing tonight and the next one is Saturday. We have 8 in now, if we have 13 by Friday(paid by Saturday), we can do this Saturday. Tonights powerball means nothing,as far as this raffle is concerned
thanks-jeff

 

[barefoot3zk5]

I'm in, my number of preference is "4". Thanx,Howard

 

[gmcole]

updated list
List of current AZer's in raffle. In order to keep raffle legit-ONLY az members can be in raffle
...Handle.................raffle #
1. Subsonic2u ................1
2. Remy..........................6
3. Buckeyejim..................7
4. Gluton1......................13
5. retail1lo......................8
6. windowsrefund.............12
7. soldtonight..................5
8. Brd.............................3
9. Barefoot3zk5................4
10.
11.
12.
13.
14.
15.

 

[custmqs]

I'm in... I'll take number 10.

 

[gmcole]

List of current AZer's in raffle. In order to keep raffle legit-ONLY az members can be in raffle
...Handle.................raffle #
1. Subsonic2u ................1
2. Remy..........................6
3. Buckeyejim..................7
4. Gluton1......................13
5. retail1lo......................8
6. windowsrefund.............12
7. soldtonight..................5
8. Brd.............................3
9. Barefoot3zk5................4
10.custmqs.....................10
11.
12.
13.
14.
15.

 

[gmcole]

The following #'s are avalible
0,2,9,11,14

 

[custmqs]

Winning number calculations

In reading through this again I'm not sure I understand how the winner will be determined. Under the "remainder" method wouldn't 1 and 10 be the same. How will the rounding work?

Help please...

EDIT: If there are 42 possible PB numbers shouldn't the "spots" be limited to 14 with each person getting 3 balls?

1 gets: 1, 15, 29
2 gets: 2, 16, 30
3 gets: 3, 17, 31
...
...
14 gets: 14, 38, 42

 

[gmcole]

I have been ask to clarify the math on the powerball.

The red #,or powerball, in the drawing is from 0-42(I believe)
We will divide that number by 15 then we will be left with a remainder, the remainder is the number you are trying to match.
example #1
powerball # is 40, 40 divided by 15 =2 with a remainder of 10, in that case whoever had 10 would win.
example#2
powerball # is 7, 15 goes into 7 0 times with a remainder of 7, in that case, 7 would be the winner.
Any questions,PLEASE, let's get them cleared up before the drawing.
As soon as we get to 13, looks like real soon, I will ask for payment. (paypal...gmcole@aol.com). I will post who is paid etc.
thanks jeff

 

[gmcole]

Please read above post.
I hope that clears it up. if not please ask.
thanks-jeff

 

[sokaiba]

Question

The people with 13,14,15 get the shaft.
1=1 16=1 31=1
2=2 17=2 32=2
3=3 18=3 33=3
4=4 19=4 34=4
5=5 20=5 35=5
6=6 21=6 36=6
7=7 22=7 37=7
8=8 23=8 38=8
9=9 24=9 39=9
10=10 25=10 40=10
11=11 26=11 41=11
12=12 27=12 42=12
13=13 28=13
14=14 29=14
15=0 - 15? 30=0 - 15?
Too iffy of a method, maybe use the last to numbers in the NYSE instead, imo.

 

[custmqs]

I think there is a flaw in the math... who wins if the PB is 15 or 30?

As I see it, 13, 14 and 15 only get two chances to win... 12 and below get 3 chances.

PB Remainder
1... 1
2... 2
3... 3
4... 4
5... 5
6... 6
7... 7
8... 8
9... 9
10... 10
11... 11
12... 12
13... 13
14... 14
15... 0
16... 1
17... 2
18... 3
19... 4
20... 5
21... 6
22... 7
23... 8
24... 9
25... 10
26... 11
27... 12
28... 13
29... 14
30... 0
31... 1
32... 2
33... 3
34... 4
35... 5
36... 6
37... 7
38... 8
39... 9
40... 10
41... 11
42... 12

 

[gmcole]

In reading through this again I'm not sure I understand how the winner will be determined. Under the "remainder" method wouldn't 1 and 10 be the same. How will the rounding work?

Help please...

EDIT: If there are 42 possible PB numbers shouldn't the "spots" be limited to 14 with each person getting 3 balls?

1 gets: 1, 15, 29
2 gets: 2, 16, 30
3 gets: 3, 17, 31
...
...
14 gets: 14, 38, 42

Here's the breakdowm
Pb# Winning raffle #
0,15,30....................0
1,16,31....................1
2,17,32....................2
3,18,33....................3
4,19,34....................4
5,20,35....................5
6,21,36....................6
7,22,37....................7
8,23,38....................8
9,24,39....................9
10,25,40..................10
11,26,41..................11
12,27,42..................12
13,28......................13
14,29......................14

Any questions

 

[gmcole]

The person with #15 gets the shaft. 15/15=1 with a remainder of zero. It looks like 29 is the only number 15 can win with while the others have 2 or 3 chances. Too iffy of a method, maybe use the last to numbers in the NYSE instead, imo.

You don't understand the calculation, read table above.
thanks-jeff

 

[custmqs]

yeah... why would I want to be 13 or 14? And we don't have a 0.