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# Data table of cost/win amounts for a 3 of 5 method in a pick 5 game

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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.

Entry #13