I have posted a little bit about this process previously in the Mathematics Forum under the Topic “Predictions from a Sequence”.
Divide and Conquer is the General idea behind the approach.
Playing Daily 4 only requires 4 choices, one for each number.
Decision 1 (Pick First Number) has 10 possible options as do each succeeding Decision. I replace the first decision with an option that has only 5 options. The 5 options are the number of hits expected from 1 of 2 groups of numbers. Generally the Number Groups could be set in several ways, (0-4 & 5-9; Odd & Even; Alternate Numbers; etc.) My groups are formed by taking the numbers from the 4 Strings Broken on the previous Draw, placing those in the “R” Group with the remaining numbers placed in the “N” Group. For each draw there will be from 0 to 4 numbers selected from the “R” Group. The number of digits in each group varies from 6 to 8 in the “R” Group based on the number of broken strings. An all Single Draw breaks 15 Strings and places 6 digits in the “R” Group. A Double Draw breaks 35 Strings and places 7 Digits in the “R” Group. A Double-Double Draw breaks 70 Strings and places 8 digits in the “R” Group. A Triple breaks 70 Strings and places 8 Digits in the “R” Group. A Quadruple draw breaks 126 Strings and places 9 digits in the “R” Group (Note his has happened only 3 times in the 2743 games in the data). The following is a summary of the 5 options for Ca. Daily 4 as of Game 2777 (Dec 25, 2015).
Games in Data 2743 Last Game 2777
"R" Hits Hits Frequency Games Out GO/F
1 305 9 21 234%
4 544 5 7 139%
3 1050 2.6 3 115%
2 802 3.4 1 29%
0 49 56 2 4%
Looking at the options, 0 can be eliminated for a few draws. I should noted that there are not that many combinations created from this option so it can be played for less than the other 4 options. Monitoring the Games out Divided by the Frequency shows which options are due or overdue.
With this decision there is a 1:5 Chance of being right (1:4 if 0 is excluded)
Decision 2 is what positions will the R and N Groups occupy in the Draw. The number of options is determined by the first decision as follows: 0=NNNN; 1= NNNR, NNRN, NRNN, RNNN; 2= NNRR, NRNR, NRRN, RNNR, RNRN, RRNN; 3= NRRR, RNRR, RRNR, RRRN; 4= RRRR. The odds of getting this right (assuming decision 1 is correct) vary from 1:1 for 1:6.
Looking at Game 2778 there were 7 numbers in the “R” Group. (Draw 2777 was a Double) The “R” Numbers 0135689; The “N” Numbers 247. There were 2 numbers from the “R” Group drawn The position of the group Numbers was RRNN.
At this point My Wheel generates all of the Single, Double, Double-Double, Triple and Quadruple Combinations from the First 2 Decisions. There are 146 Single Combinations, 168 Double Combinations, and 12 Double-Double Combinations.
At this point there are 255 Combinations that could be played from the First 2 decisions. Playing all combinations as a Straight would have produced 1 winning combination. There was 1 winning Straight ticket that paid $5925 so dividing by 2 Win $2962.50. Playing all combinations as Boxes would have Won $826.00. (there were 28 Box winning tickets at $236.00 each=$6608.00 total prize. Divided by 32=$206.5 each). There were 4 Box Combinations in the Wheeled combinations. The Cost to Win would be $255.00
Making a 3rd Decision and Selecting only the Single Game Combinations would have won the same at a cost of $146.00
I am going to forward test this to catch up to the current game. I will let you know how I do.
I am curious about some basic math here. 10x10x10x10 is the odds of picking the right Straight Combination. If you correctly pick the first 2 numbers there are 100 combinations (of those numbers and 2 more) left. There is a 1:100 Chance of being right. After 2 decisions with this workout there are 255 Combinations left. I have a 1:20 chance of being right. The odds of winning are 1:5100 at a cost of $255.00. I am drawing a blank on the General odds of winning when playing 255 tickets.
Waiting for thoughts from the LP Class