A few weeks ago I posted a system regarding the infamous 8 combos that guarantee at least two numbers as long as there are no doubles or triples. They are below.
308-719-843-175-283-659-024-716
I just saw something that was interesting. A good portion of the time in Florida, the winning third elusive number will be in an adjacent combo, either to the right or left.
In addition, since a good portion of the time at least one number from the previous draw shows up in the next draw, it would make sense to use those numbers when making the selection preocess for the next draw.
Of course double will ruin your day with this system, so you'll have to be determined to wait patiently for a non-double to show.
Example, Florida,
Dec. 22 eve-995
Dec. 23 mid-946
Using the 9 and the 5 in 995, we look at the eight combos.Any combo that has a 9 and a 5 will be used. From left to right we have 719, which has a 9 in it. Now we make combos from that 9 as pairs with the combos to the left and right of it. Eliminate the dupes.
308-719-843-175-283-659-024-716
Left:
917-918-910-913
978-970-973
Right: (918- dupe, so omit) 914-974
Now continuing to the right we see 659, which has both 9 and 5 in the combo.
Left:
956-952-958-953
962-968-963
Right:
950-952-954
960-962-964
Total combos: 22
Dec. 23 mid-946-boxed
Straights can show too, so play them str/bx.
<308-719-843-175-283-659-024-716>
So for tonight's Florida, I would use today's mid-946
Start with 9 in 719.
917-918-910-913-978-970-973-914
Then the 9 in 659.
956-953-958-952-963-968-962-950-959-960-969
Then the 4 in 843.
483-489-481-487-439-431-437-485-435
The the 4 in 024.
420-429-425-426-409-405-406-427-421-407-401
Then the 6 in 659.
653-658-652-650-654-690-694
Then the 6 in 716
617-614-612-610-674-672-670
Total, 53 combos, ouch, so filter by using only sums between 11-19 and eliminate dupes.
917-918-913-970-973-914-953-952-963-950-960-483-481-487-439-431-437-485-435-420-429-425-409-405-406-427-421-426-407-401-653-658-652-692-650-654-694-617-614-612-610-674-672-670
Final combos: 44