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.

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?

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.

 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.