Found this piece on X
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The chance of winning a 3 x 3 x 3 x 3 crisscross wager is 12.96% (1 out of 7.7 plays) and the cost of the $2 ticket would be $162.00. To achieve a profit one would have to have a winning ticket that pays over $1,100.00 once in every 7.7 attempts (7.7 x 162). I would like to share with everyone a system of playing the pick 4. I believe that it is superior to anything I have seen to date. The goal is to minimize the number of combinations without minimizing the number of horses under consideration to win the race. Let’s say we wanted to play a pick 4 with 3, 3, 4, and 5 horses respectively in each of the 4 legs. That would be a total of 180 combinations or $360.00. Using, what I call a “Conditional Pick-4”, the total number of combinations can be reduced to 172, 132, 62, or as low as 8, depending on the set conditions. The number of horses remains the same. I will give several examples. Please, DO NOT be intimidated by what follows. It is very, very simple; perhaps difficult to explain, but simple nevertheless.
Example 1:
Pick - 4 with 3, 3, 4, and 5 horses in consideration. Total combinations = 180 ($2.00 ticket playing all horses would cost $360.00). We like some horses more than others and those horses will be labeled as the Key Horses. In the first leg there is one key horse, second and third legs have 2 key horses, while the fourth leg has three key horses. Now, the condition must be set. The condition applies to the Key horses only. Whatever condition is set, it must be fulfilled. We have to ask ourselves: How many of the key horse legs can we pick correctly? 3 out of 4, 2 out of 4, or 1 out of 4. In this first example, the condition set is: 3 out of 4! (Applies to Key Horses column) Starting in the top left box, write in all combinations of 3 using only the KEY horses selected as depicted below.
1a
What do we do with the empty boxes? Well, that depends on where the empty boxes are. If the empty boxes are to the right of the LAST Key entry (yellow), then the horses in the All Horses Considered column are inserted. In the above example only leg 1 has an empty box to the right of the last Key horse A. Thus, horses A, B, and C are entered in the last box of leg 1. In all other empty boxes, that is, all empty boxes to the left of the LAST Key entry (green), write in all horses in that particular leg that are not our Key plays. Therefore, in leg 2 - horse C, leg 3 - horses C and D, and in leg 4 - horses D and E. See completed chart below. The total number of combinations has been reduced from 180 to 62 (8+12+6+36 = 62) !!! As long as ANY 3 out of the 4 conditions are met, a winning ticket is guaranteed (assuming the winners are amongst all horses considered, of course)
1b
Example 2:
Let’s now assume we want only 2 conditions. The procedure is the same as above. First, insert all combinations of 2 using only Key Horses in the matrix. See below.
2a
Now in leg 1, there are 3 empty boxes to the right of the last Key entry (yellow), thus, A, B, and C are inserted in all three boxes. In leg two, there is one empty box to the right of the last Key entry, thus A, B, and C are inserted and in the two empty boxes to the left, only C is inserted. In legs 3 and 4 there are no empty boxes to the right, thus C and D for leg 3 and D and E for leg 4 are entered in the empty boxes to the left. There are a total of 132 combinations (8+4+6+24+36+54 = 132) !!! See below:
2b
Example 3:
Finally, for completeness sake, 1 condition. The procedure is the same.
3a
The condition, 1 out of 4 is the easiest to fulfill and will reduce the number of combinations the least. The total number of combinations is 172 (4+24+36+108 = 172). See below:
3b
The above system can obviously be applied to the Pick – 3 as well as the Pick – 6. I believe this is a superior way of playing because it can be adjusted to meet one’s own criteria on a race to race and day to day basis. One can use any number of key horses, consider as many horses as contenders, and minimize the number of combinations.
