Here are a few thoughts on what MIGHT come to pass for tonight's draw.
We are getting close to a draw where the first number is a double digit -- maybe tonight. After 10 single digit draws in a row, the probability is only 3% for another, i.e. 0.97 probability that the first decade will miss -- i.e. the first number drawn will be 10 or higher.
Looking at the columns, the 1st column (1-7) has hit 7 times in a row and has a 0.98 probability of a miss. The second column (8-14) has missed twice in a row and now has a 0.94 probability of a hit -- so I would guess that the first digit in tonight's draw might fall in the 10-14 range. Column E (29-35) also has a 0.95 probability of a hit.
As discussed in the previous thread, Row 6 (6,13,20,27,34,41,48) has missed for 6 draws in a row and has a 0.99 probability of a hit. Based on the above, we could eliminate the "6". Also, Row 4 (4,11,18,25,32,39,46) has hit 5 times in a row and has a 0.93 probability of a miss.
The group of numbers with skips in the 5-10 range (5,6,8,9,14,15,16,17,27,27,28,32,35,36,38,39,40) has missed for the last two games (quite unusual) and is due for a hit. I would drop 5,6,8 and 9 and we could eliminate a couple from Row 4, above. We might see 1-3 of the numbers coming from this group.
I would not expect the 43 to repeat again after 2 times in a row, in fact, looking at the Vtracs, the 18/43 combination will likely not repeat as Vtrac 18 has hit in the last 4 draws. Vtrac group 4 (13/38, 14/39, 15/40, 16,41) hasn't hit for 6 games in a row and is due for a hit while Vtrac group 2 (5/30, 6/31, 7/32, 8/33) hasn't hit for 4 games and is also due -- obviously with more emphasis on the 30's than on the single digits.
One or two of tonight's numbers should come from the last two draws (1,2,4,18,20,21,22,23,25,37,41,42,43). I would eliminate 1,2,4,18 and 43.
Numbers with a root sum of 8 have been missing for 7 games in a row and may hit: 8,17,26,35,44.
Hope this is helpful and GOOD LUCK!
We are getting close to a draw where the first number is a double digit -- maybe tonight. After 10 single digit draws in a row, the probability is only 3% for another, i.e. 0.97 probability that the first decade will miss -- i.e. the first number drawn will be 10 or higher.
Looking at the columns, the 1st column (1-7) has hit 7 times in a row and has a 0.98 probability of a miss. The second column (8-14) has missed twice in a row and now has a 0.94 probability of a hit -- so I would guess that the first digit in tonight's draw might fall in the 10-14 range. Column E (29-35) also has a 0.95 probability of a hit.
As discussed in the previous thread, Row 6 (6,13,20,27,34,41,48) has missed for 6 draws in a row and has a 0.99 probability of a hit. Based on the above, we could eliminate the "6". Also, Row 4 (4,11,18,25,32,39,46) has hit 5 times in a row and has a 0.93 probability of a miss.
The group of numbers with skips in the 5-10 range (5,6,8,9,14,15,16,17,27,27,28,32,35,36,38,39,40) has missed for the last two games (quite unusual) and is due for a hit. I would drop 5,6,8 and 9 and we could eliminate a couple from Row 4, above. We might see 1-3 of the numbers coming from this group.
I would not expect the 43 to repeat again after 2 times in a row, in fact, looking at the Vtracs, the 18/43 combination will likely not repeat as Vtrac 18 has hit in the last 4 draws. Vtrac group 4 (13/38, 14/39, 15/40, 16,41) hasn't hit for 6 games in a row and is due for a hit while Vtrac group 2 (5/30, 6/31, 7/32, 8/33) hasn't hit for 4 games and is also due -- obviously with more emphasis on the 30's than on the single digits.
One or two of tonight's numbers should come from the last two draws (1,2,4,18,20,21,22,23,25,37,41,42,43). I would eliminate 1,2,4,18 and 43.
Numbers with a root sum of 8 have been missing for 7 games in a row and may hit: 8,17,26,35,44.
Hope this is helpful and GOOD LUCK!