Instead of a 1 n 10 chance.......Do ya want to find a way to have a 1 n 3 chance at getting thefirst digit correct digit in Pick 3? How about a 1n3 chance at at the right Sum instead of 1 n 10 ? Increase our chances ...... yes. Very much.
Find the 10 short sums chart and then find an instance where 3 of the 10 are all missing at the same time. Important ( All 3 must be missing for at least 15 draws) Together.
Example: Short sums 0,,,,,,1,,,,,,, and 2 are all missing at the same time. The 0 has been missing for 43 days. The digit 1 for 23 days. The digit 0 for 15 days. All in first position only.
Now ..... since these are "Short Sums" we must divide each by 3 sooooo......
43/3= 14.33 23/3=7.6 and last digit 0 for 15/3=5
Now instead of 43 days ...its 14 draws out. Instead of 23 its 7 days out ...... and its not 15 days it's 5 days out. Divide by three ...and its equal to or the same as all 3 digits missing in all positions for that length of time.
As long as the shortest missing digit /sum has been out for at least 15 days or 15/3 = 5 normal days. It will work.
Almost the same thing is involved for a 1 n 3 chance at first position odds . Beats 1 n 10 right?
Ex. Digits 4,,,,,5,,,,and 6 are all missing at the same time from position one. The digit 4 has been missing for 52 days .... the digit 5 for 24 days ....and the digit 6 for 16 days.
Since this is for a single position and not ALL POSITIONS missing digits then we must divide by 3 AGAIN. So 52/3= 17 24/3=8 and the last digit missing 16/3= 5.3
ODDS are very high that the next draw will come from one of these 3 missing digits once all three have crossed over the line. When the least of the 3 digits has been missing for at least 15 draws.
" They can hide one of something forever and 2 of something almost as long....but they just can't hide 3 of something very long at all"
The only real failure .....is the failure to try.
Luck is a very rare thing....... Odds not so much.
Odds never change .....but probability does.