Data table of cost/win amounts for a 3 of 5 method in a pick 5 game
For California's Fantasy 5 it happens about once every 15 draws that the computer picks 3 numbers from the same previous draw from anywhere from 1-10 last draws, like it did four times in the first table, shown at the right column.
Date | Draw1 | Draw2 | Draw3 | Draw4 | Draw5 | Dr1 in last 6 | Dr2 in last 6 | Dr3 in last 6 | Dr4 in last 6 | Dr5 in last 6 | DblCt | Draws Back | MultDbl | Actual Numbers | TrplCt | 3 from X draws
Back |
3 Repeat #s |
4/22/2011 | 7 | 10 | 13 | 26 | 31 | 1 | 1 | 1 | 1 | 1 | 1 | 4,6,10 | 1 | 7,13,
10,26, 10,13 |
1 | 9 | 7,26,31 |
4/21/2011 | 2 | 6 | 19 | 27 | 31 | 1 | 1 | 1 | 8 | 19,31 | |||||||
4/20/2011 | 9 | 10 | 37 | 38 | 39 | 1 | 1 | 1 | 1 | 1 | 1 | 6,8 | 1 | 37,39,
10,38 |
1 | 3 | 9,37,38 |
4/19/2011 | 5 | 6 | 28 | 36 | 39 | 1 | 1 | 1 | 1 | 5 | 28,39 | 1 | 8 | 6,28,36 | |||
4/18/2011 | 3 | 7 | 13 | 14 | 28 | 1 | 1 | 1 | 1 | 7 | 13,28 | ||||||
4/17/2011 | 6 | 9 | 34 | 37 | 38 | 1 | 1 | 1 | 1 | 10 | 34,37 | ||||||
4/16/2011 | 10 | 21 | 26 | 29 | 30 | 1 | 1 | 1 | |||||||||
4/15/2011 | 1 | 8 | 16 | 19 | 32 | 1 | 1 | ||||||||||
4/14/2011 | 17 | 25 | 28 | 37 | 39 | 1 | 1 | 1 | 1 | 1 | 6 | 17,39 | 1 | 4 | 17,28,37 | ||
4/13/2011 | 7 | 19 | 20 | 26 | 31 | 1 | 1 | 1 | 1 | 1 | 4 | 19,26 | |||||
4/12/2011 | 10 | 11 | 13 | 22 | 38 | 1 | 1 | 1 | |||||||||
4/11/2011 | 6 | 13 | 16 | 28 | 36 | 1 | 1 | ||||||||||
4/10/2011 | 17 | 20 | 28 | 30 | 37 | 1 | 1 | 1 | 1 | 1 | 3,7 | 1 | 30,37,
28,37 |
||||
4/9/2011 | 15 | 19 | 26 | 28 | 34 | 1 | 1 | 1 | 1 | 6,9,10 | 1 | 28,34,
19,34, 15,19 |
|||||
4/8/2011 | 7 | 10 | 17 | 24 | 39 | 1 | 1 | 1 |
Assuming you have a strong hunch that 3 numbers are going to repeat from the same previous draw of anywhere in the last 1-10 last draws, how many times and different prior draws covered would provide an interesting return? On 4/19 I re-played all 5 numbers from 4/11 5 times, a jpeg should show below, for 5 3 of 5 winners and I picked up a 3 or 5 in 3 other ways for a total of 8 3 of 5 winners on the LP prediction board and in real life for $136. I made a "data table" that forms a matrix of how many times played multiplied by how many different previous draws are covered multiplied by an assumed $15 payout for 3/5 winner less how much you played.
For example, where the 4 meets the 3 in the matrix (in bold): you won four times by repicking all 5 numbers from three different prior draws four times, and let's say you won so $60 payout less (3*4) = $48. Has anyone else tried this method or any variation thereof?
X | T | i | m | e | s | w | o | n | |||
$15 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
D | 1 | $14 | $28 | $42 | $56 | $70 | $84 | $98 | $112 | $126 | $140 |
a | 2 | $13 | $26 | $39 | $52 | $65 | $78 | $91 | $104 | $117 | $130 |
y | 3 | $12 | $24 | $36 | $48 | $60 | $72 | $84 | $96 | $108 | $120 |
s | 4 | $11 | $22 | $33 | $44 | $55 | $66 | $77 | $88 | $99 | $110 |
C | 5 | $10 | $20 | $30 | $40 | $50 | $60 | $70 | $80 | $90 | $100 |
o | 6 | $9 | $18 | $27 | $36 | $45 | $54 | $63 | $72 | $81 | $90 |
ve | 7 | $8 | $16 | $24 | $32 | $40 | $48 | $56 | $64 | $72 | $80 |
red | 8 | $7 | $14 | $21 | $28 | $35 | $42 | $49 | $56 | $63 | $70 |
9 | $6 | $12 | $18 | $24 | $30 | $36 | $42 | $48 | $54 | $60 | |
10 | $5 | $10 | $15 | $20 | $25 | $30 | $35 | $40 | $45 | $50 |
You might also lose so I'll add a "data table" that shows how losses can add up if you try the strategy across an array x-y combos played.
X | t | i | m | e | s | w | o | n | |||
$0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
D | 1 | -$1 | -$2 | -$3 | -$4 | -$5 | -$6 | -$7 | -$8 | -$9 | -$10 |
a | 2 | -$2 | -$4 | -$6 | -$8 | -$10 | -$12 | -$14 | -$16 | -$18 | -$20 |
y | 3 | -$3 | -$6 | -$9 | -$12 | -$15 | -$18 | -$21 | -$24 | -$27 | -$30 |
s | 4 | -$4 | -$8 | -$12 | -$16 | -$20 | -$24 | -$28 | -$32 | -$36 | -$40 |
C | 5 | -$5 | -$10 | -$15 | -$20 | -$25 | -$30 | -$35 | -$40 | -$45 | -$50 |
o | 6 | -$6 | -$12 | -$18 | -$24 | -$30 | -$36 | -$42 | -$48 | -$54 | -$60 |
ve | 7 | -$7 | -$14 | -$21 | -$28 | -$35 | -$42 | -$49 | -$56 | -$63 | -$70 |
red | 8 | -$8 | -$16 | -$24 | -$32 | -$40 | -$48 | -$56 | -$64 | -$72 | -$80 |
9 | -$9 | -$18 | -$27 | -$36 | -$45 | -$54 | -$63 | -$72 | -$81 | -$90 | |
10 | -$10 | -$20 | -$30 | -$40 | -$50 | -$60 | -$70 | -$80 | -$90 | -$100 |
Here's the one pic of the 8 x 3 of 5 and another unusual souvenir: the machine kicked out two replays with the exact 5 numbers when two different tickets with the same numbers were put into the machine as 2 of 5 "replays".
The winning numbers for jpeg below was 5,6, 28, 36, 39 and I re-played all from 4/11 which was 6,13,16,28,39. Every so often CA F5 will have 4 numbers repeat from the same previous draw and did have all 5 numbers repeat from 10 days back in 2000. Of course the trick is in figuring which prior draw.