Original Blog Entry: Data table of cost/win amounts for a 3 of 5 method in a pick 5 game

LottoMining — Sun, 24 Apr 2011 02:49:40 GMT

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.

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