Hello JADELottery
Mega million 14 15 37 42 67 22 1A1B2C1E
There are 85 combinaations from the 53 k wheel numbers you gave me
Of which 21 combinations are 2a1b1c1d 2a1b1d1e 1a2b1c1e 1a2c1d1e and offcourse 1a1b2c1e drawn on the 5th
The rest of the 85 combinations are 2a2b1c 2a2c1d 2b2c1e 2a2c1d 4a1b 4b1c 3a1b1c 1a3b1d etc total 85 minus 21=64
21 being the 2a1b1c1d group of combinations and this group of combinations represent 36% of the total draw during two year period
so I choosed this group for mega draw on the fifth and the draw was 1a1b2c1e
If I choose all the 85 combinations it is going to be 53 thousands so by choosing one of the group the total combinationsis much less
in this case is 21/85 x 53thousand=13091 combination
I think by fixing the odd and even number per my post such as 3ODD 1 3 4 or 2 4 5 etc will further reduce the total combinations
So I was right on the group of 1a1b2c1e and I was right on the 3odd but wrong on the choice of 2 4 5 and 1 4 5 because the odd was 2 3 5
A 14 B15 C37 42 E67 bonus B 22
I do not have 14 on A but I have 12
I do not have 15 on B but I have 17
I do not have 37 and 41 on C but i have 38 and 42 and if I can have 2odd numbers in this two column then its either 37 or 39 and 41 only
because 43 belong to D
E67 and I only have 65 and 69 on E
As I said my english is not so good and please forgive me if I say anything wrong
But I do have bonus 22 included in but my first batch number and second batch number
If you look at superlotto drawn on the sixth 01 02 07 12 24 23 4a1b B 2odd 1 3
I have 07 12 24 23 on the first batch number
i do not have 02 but have 03
i have 2 odd and 1 3 as drawn
Unfortunately I didnt sent the third batch number for superlotto
Thank you very much for your time
H.Y.