Wheeling Algorithm Thoughts - pick 5
William Collin
game. The wheel generates a 3 of 5 match. I will use the following
meanings for this post.
- a draw is 5 numbers selected by the game.
- a pick is 5 numbers that the player buys in hopes of matching 3 or more
numbers from the draw.
- a wheel is a set of picks that guarantees that at least one of picks
will match 3 numbers from the draw.
- a wheel is characterize by the range of numbers it covers,(set size),
and the number of picks it contains.
for any set of size N, there are N*(N-1)*(N-2)*(N-3)*(N-4)/5! possible picks.
for any set of size N, there are N*(N-1)*(N-2)/3! possible "trebles". (3
number covers)
any pick can cover at most 10 trebles (5*4*3/6)
Consider 2 picks that have a union of 2 numbers in common. i.e.
12345
12678
the trebles covered by each are:
123,124,125,134,135,145,234,235,245,345
126,127,128,167,168,178,267,268,278,678
no two of which are the same:
Collin's Wheel Theory
If a player examines a set of picks generated from a set of N and
generates a set of all picks which contain no more than 2 numbers in
common between any 2 picks, then that set of picks is a wheel which
guarantees a 3 match if all 5 numbers of the draw are contained in the set
N.
(If only two numbers match, then the set of all picks was incomplete)
Here is a brief algorithm for this method.
I. set N = 1,2,3,4,5,6,7,8,9,10,11,12
II. Initialize the wheel to 1 pick of 1 2 3 4 5
III. For all possible draws (12*11*10*9*8/120 draws)
a. compare the current draw d to all entries in wheel and
determine a score from 0 to 4, which is the number of
matching digits between d and the wheel entry.
b. If the score is less than 3 for all wheel entries, then
add d to the wheel
IV. done
Examples:
Set size of 12
1 = 1 2 3 4 5
2 = 1 2 6 7 8
3 = 1 2 9 10 11
4 = 1 3 6 9 12
5 = 1 4 7 10 12
6 = 1 5 8 11 12
7 = 2 3 7 11 12
8 = 2 4 8 9 12
9 = 2 5 6 10 12
10 = 3 4 6 8 10
11 = 3 5 7 8 9
12 = 4 5 6 7 11
Set size of 20
1 = 1 2 3 4 5
2 = 1 2 6 7 8
3 = 1 2 9 10 11
4 = 1 2 12 13 14
5 = 1 2 15 16 17
6 = 1 2 18 19 20
7 = 1 3 6 9 12
8 = 1 3 7 10 13
9 = 1 3 8 11 14
10 = 1 4 6 10 14
11 = 1 4 7 9 15
12 = 1 4 8 12 16
13 = 1 4 11 13 17
14 = 1 5 6 11 15
15 = 1 5 7 12 17
16 = 1 5 8 9 13
17 = 1 5 10 16 18
18 = 1 6 13 16 19
19 = 1 7 11 16 20
20 = 1 8 10 15 19
21 = 1 9 14 17 18
22 = 2 3 6 10 15
23 = 2 3 7 9 14
24 = 2 3 8 12 17
25 = 2 3 11 13 16
26 = 2 4 6 9 13
27 = 2 4 7 10 12
28 = 2 4 8 11 15
29 = 2 4 14 16 18
30 = 2 5 6 12 16
31 = 2 5 7 11 18
32 = 2 5 8 10 14
33 = 2 5 9 15 19
34 = 2 5 13 17 20
35 = 2 6 11 14 17
36 = 2 8 9 16 20
37 = 3 4 6 7 11
38 = 3 4 8 9 10
39 = 3 4 12 13 15
40 = 3 4 14 17 19
41 = 3 5 6 8 18
42 = 3 5 7 15 16
43 = 3 5 9 11 17
44 = 3 5 10 12 19
45 = 3 6 13 14 20
46 = 3 7 8 19 20
47 = 3 9 13 18 19
48 = 3 10 11 18 20
49 = 4 5 6 19 20
50 = 4 5 7 13 14
51 = 4 5 9 12 18
52 = 4 5 10 15 17
53 = 4 7 8 17 18
54 = 4 9 11 14 20
55 = 4 10 11 16 19
56 = 5 6 7 9 10
57 = 5 8 11 12 20
58 = 5 8 16 17 19
59 = 5 14 15 18 20
60 = 6 7 12 13 18
61 = 6 7 14 15 19
62 = 6 8 9 11 19
63 = 6 8 10 13 17
64 = 6 9 15 16 18
65 = 6 12 15 17 20
66 = 7 9 12 16 19
67 = 7 10 14 16 17
68 = 8 9 12 14 15
69 = 9 10 12 13 20
70 = 10 11 13 14 15
71 = 11 12 14 18 19
Set size of 34 generates 411 picks.
IMPORTANT. This does not guarantee that all 5 drawn numbers will be in the
wheeled set!!
This does not guarantee the wheel to be optimal!!!
for example the following set for N = 12
Wheel to guarantee 3 if only 3 of the drawn numbers in set of 12
1 = 1 2 3 4 5
2 = 1 2 6 7 8
3 = 1 2 9 10 11
4 = 1 3 6 9 12
5 = 1 4 7 10 12
6 = 1 5 8 11 12
7 = 2 3 7 11 12
8 = 2 4 8 9 12
9 = 2 5 6 10 12
10 = 3 4 6 8 10
11 = 3 5 7 8 9
12 = 4 5 6 7 11
13 = 3 4 9 10 11
14 = 6 7 9 10 11
15 = 1 3 5 7 10
16 = 1 4 5 6 9
17 = 2 3 6 8 11
18 = 5 8 9 10 11
19 = 1 4 7 8 11
20 = 2 5 7 9 12
21 = 1 3 8 10 12
22 = 2 3 4 7 10
23 = 2 4 6 11 12
24 = 1 3 5 6 11
25 = 2 4 5 8 10
26 = 3 6 7 8 12
27 = 1 2 6 8 9
28 = 1 3 4 5 12
29 = 1 9 10 11 12
30 = 1 2 3 7 9
31 = 1 2 5 11 12
32 = 1 5 6 8 10
33 = 4 7 8 9 10
If this is applied to 1 - 12 and 13 - 24 then the resulting
wheel guarantees 3 if 5 of the drawn numbers are in 24 in 66 guesses.
this is less than the number required for set size 20.
Hope someone finds this relevant.
collin
"lottery playing is complex, there is a real part (probability) and a
imaginary part (dreams) "
Normand Veilleux
>From: col...@computek.net (William Collin)
>
>Collin's Wheel Theory
>If a player examines a set of picks generated from a set of N and
>generates a set of all picks which contain no more than 2 numbers in
>common between any 2 picks, then that set of picks is a wheel which
>guarantees a 3 match if all 5 numbers of the draw are contained in the
>set N.
Obviously. They only "problem" with that theory is that there are
wheels which have tickets that match more than 2 numbers for other
tickets in the set and are still a cover. For example, the 174 ticket
cover for the Lotto 6/49. It has 18 tickets that have 5 identical
numbers.
If you could prove that the minimal wheel MUST follow your rule, then
you'd have something. But, that's probably false in general.
>Here is a brief algorithm for this method.
>
>I. set N = 1,2,3,4,5,6,7,8,9,10,11,12
>II. Initialize the wheel to 1 pick of 1 2 3 4 5
>III. For all possible draws (12*11*10*9*8/120 draws)
> a. compare the current draw d to all entries in wheel and
> determine a score from 0 to 4, which is the number of
> matching digits between d and the wheel entry.
> b. If the score is less than 3 for all wheel entries, then
> add d to the wheel
>IV. done
>
>[...]
>for example the following set for N = 12
>Wheel to guarantee 3 if only 3 of the drawn numbers in set of 12
> [33 tickets deleted]
>
>If this is applied to 1 - 12 and 13 - 24 then the resulting
>wheel guarantees 3 if 5 of the drawn numbers are in 24 in 66 guesses.
>this is less than the number required for set size 20.
Actually, for N = 12 you only need 29 tickets to cover all 3 on 3.
And for N = 10 it's possible with 17. So, you can generate a wheel
to cover all 3 on 5 with 2 * 17 = 34 tickets. That is a lot better
than the 74 the algorithm produces.
This algorithm does reasonably well for 3 on 3, but deteriates very
quickly for 3 on 5 or even 3 on 6. I just implemented it for tickets
of 6 numbers and for N = 12 we get:
01 02 03 04 05 06
01 02 07 08 09 10
03 04 07 08 11 12
05 06 09 10 11 12
But, taking these 2 tickets does the job too:
01 02 03 04 05 06
07 08 09 10 11 12
For N = 24, the algorithm generates a set of 74 tickets to guarantee
a 3 on 6, yet it is possible with only 20 tickets.
William Collin
for selected sets. For example the following wheel which guarantees a
2-match on a set of 13 in 4 slots is optimal:
1 2 3 4 3 5 9 13
1 5 6 7 3 6 10 11
1 8 9 10 3 7 8 12
1 11 12 13 4 5 10 12
2 5 8 11 4 6 8 13
2 7 10 13 4 7 9 11
2 6 9 12
It has no more than 1 number in common between and two picks and covers
the 78 doubles in 13 picks which is the minimun.
Before I posted the other optimal wheel which guarantees a
2-match on on a set of 15 in 3 slots:
Here it is:
1 2 3 1 4 5 1 6 7 1 8 9 1 10 11
1 12 13 1 14 15 2 4 6 2 5 7 2 8 10
2 9 11 2 12 14 2 13 15 3 4 7 3 5 6
3 8 11 3 9 10 3 12 15 3 13 14 4 8 12
4 9 13 4 10 14 4 11 15 5 8 13 5 9 12
5 10 15 5 11 14 6 8 14 6 9 15 6 10 12
6 11 13 7 8 15 7 9 14 7 10 13 7 11 12
Additionally the algorithm will duplicate the 3-match on a set of 22 in 6
slots as is contained twice in the 174 ticket
wheel.
The purpose of the algorithm was to generate a covering set that could be
used as well as be a starting point for optimizing operations.
For example, wheels can be represented as matrices. Here is a wheel that
guarentees a 2-match on a set of 8 in 3 slots:
1 1 1 0 0 0 0 0 1 2 3
1 0 0 1 1 0 0 0 1 4 5
1 0 0 0 0 1 1 0 1 6 7
0 1 0 1 0 1 0 0 etc
0 1 0 0 1 0 1 0
0 0 1 1 0 0 1 0
0 0 1 0 1 1 0 0
0 0 1 1 0 0 0 1
0 0 0 0 1 1 0 1
1 0 0 0 0 0 1 1
I know that columns can be swapped with no change in the coverage, but are
there operations on the rows that can lead to optimization???
Some numbers are used 4 times and others only 3, is this a hint to some
optimization operation???
thanks for the other comments.
collin
In article <nveilleu.17...@emr1.emr.ca>, nvei...@emr1.emr.ca
(Normand Veilleux) wrote:
> >From: col...@computek.net (William Collin)
> >
> >Collin's Wheel Theory
> >If a player examines a set of picks generated from a set of N and
> >generates a set of all picks which contain no more than 2 numbers in
> >common between any 2 picks, then that set of picks is a wheel which
> >guarantees a 3 match if all 5 numbers of the draw are contained in the
> >set N.
>
Iliya Bluskov
guarentees a 2-match on a set of 8 in 3 slots:
1 1 1 0 0 0 0 0 1 2 3
1 0 0 1 1 0 0 0 1 4 5
1 0 0 0 0 1 1 0 1 6 7
0 1 0 1 0 1 0 0 etc
0 1 0 0 1 0 1 0
0 0 1 1 0 0 1 0
0 0 1 0 1 1 0 0
0 0 1 1 0 0 0 1
0 0 0 0 1 1 0 1
1 0 0 0 0 0 1 1
How about the pair 2 8 ?