magusprovernum
Member
Was looking at the numbers 1-49.
36.7% are prime numbers.
33% are divisible by 3.
18.4% are divisible by 5.
In six numbers drawn, two should be prime; two divisible by 3 and one divisible by five. Look at the following set:
Canada 649, May 05, 2007
01 23 26 28 33 40
Primes: 01, 23
Div3 : 26, 33
Div5 : 40
I am playing the following 14 random numbers in tonight's Ontario49 (May 16):
2 5 9 10 11 24 28 30 35 38 41 45 47 48
When filtered by my decade-one rules, sums 125-175 and the above new rules we get the following 10 sets:
2 9 11 35 38 48
2 9 28 35 41 48
2 9 28 35 47 48
2 9 28 38 41 45
2 9 28 38 45 47
5 9 24 28 38 41
5 9 24 28 38 47
9 10 11 24 28 41
9 10 11 28 47 48
9 10 11 38 47 48
You will notice that the number 30 got filtered out.
Just a new idea to toy with. How many numbers do you play? I have found that you should never play more than 14, because after that the number of new sets you have to play are greater than the difference in odds.
36.7% are prime numbers.
33% are divisible by 3.
18.4% are divisible by 5.
In six numbers drawn, two should be prime; two divisible by 3 and one divisible by five. Look at the following set:
Canada 649, May 05, 2007
01 23 26 28 33 40
Primes: 01, 23
Div3 : 26, 33
Div5 : 40
I am playing the following 14 random numbers in tonight's Ontario49 (May 16):
2 5 9 10 11 24 28 30 35 38 41 45 47 48
When filtered by my decade-one rules, sums 125-175 and the above new rules we get the following 10 sets:
2 9 11 35 38 48
2 9 28 35 41 48
2 9 28 35 47 48
2 9 28 38 41 45
2 9 28 38 45 47
5 9 24 28 38 41
5 9 24 28 38 47
9 10 11 24 28 41
9 10 11 28 47 48
9 10 11 38 47 48
You will notice that the number 30 got filtered out.
Just a new idea to toy with. How many numbers do you play? I have found that you should never play more than 14, because after that the number of new sets you have to play are greater than the difference in odds.