Surprisingly, winning lotto numbers are not repeated. This is true based on probability tests studied for lottery systems 6/42, 6/45, 6/49, 6/55 and 6/58. As of 2008, only 1 combination had won twice out of 5805 results, a probability of 0.03% (related article). I don't know for now if such has remained true since I have lost track of all results since then. At least, for lottery 6/58, such hypothetical statement has remained true that no jackpot numbers have won more than once; every combination is unique.

Assuming, such argument remains true this day, how would you know if the combination you are petting had won before if you don't have access to a database?

Here's the answer: use the drop technique. Is it 100% absolute? No, but at least, you'd be more confident.

## The Drop Technique To Make Your Combination Unique

The purpose of the Drop Technique is to come up with a combination, in absence of access to actual database, that may have not won yet previously. This is useful, if from then on, you would like to pet such formed combination.

### Simple Method

Without considering some probability factors, you can easily form a combination that may probably has not won before. Here's how.

- Pick any jackpot combination from the most recent draws. Arrange the numbers from the lowest to the highest number.
- Pick any 3 numbers from the combination and drop the other 3. By default, keep the lowest number and pick other two of your own choice.
- Replace the numbers that you dropped with any number. Make sure that the numbers you are substituting with are between the number on its left and right.

On 6 Dec 2019, the winning numbers were 10 • 11 • 17 • 41 • 55 • 58 arranged from the lowest to the highest number. Let's form a combination base on these winning numbers.

- Retain the lowest number, which is 10. We're targeting results that begin with 10.
- Pick 2 other numbers, say, your favourite numbers are 41 • 58. So, the numbers that you are keeping are 10 • 41 • 58.
- The numbers that you will drop are 11 • 17 • 55. Replace these numbers with your other favourite numbers. Make sure that the number you are substituting with is between the numbers on its left and its right.
- Let's substitute. Replace 11 • 17 • 55 with say, 13 • 23 • 51.
- Your new combination is now 10 • 13 • 23 • 41 • 51 • 58

Let's peek inside the database and see if your new combination has not won previously.

Sorted numerically, following is a snapshot of lotto 6/58 results that all begin with number 10. From this snapshot, you will find that the actual Dec 6 results in the second row; while your new combination does not exist yet, which means it has not won before. If your new combination would win, it would be placed between the 4th and 5th row.

24 Jun 18 | 10 | 11 | 17 | 24 | 25 | 58 |

12/6/2019 | 10 | 11 | 17 | 41 | 55 | 58 |

14 Nov 17 | 10 | 11 | 18 | 23 | 34 | 41 |

15 Feb 19 | 10 | 12 | 42 | 48 | 55 | 56 |

21 Aug 16 | 10 | 13 | 24 | 25 | 30 | 42 |

11 Oct 15 | 10 | 13 | 25 | 38 | 42 | 46 |

10 Nov 17 | 10 | 14 | 35 | 40 | 43 | 47 |

15 May 18 | 10 | 15 | 18 | 22 | 23 | 32 |

17 Sep 19 | 10 | 19 | 21 | 26 | 36 | 38 |

9 Apr 19 | 10 | 19 | 23 | 39 | 40 | 51 |

7 May 19 | 10 | 22 | 26 | 30 | 36 | 46 |

21 Nov 17 | 10 | 22 | 27 | 51 | 54 | 58 |

24 Jul 16 | 10 | 22 | 28 | 30 | 34 | 50 |

10 Sep 17 | 10 | 23 | 26 | 34 | 37 | 39 |

#### Additional Tips

- If you are replacing the 2nd number (2nd lowest), don't deviate much from the original number. A minus 1 or 2 or plus 1 or 2 is good enough.
- Your options will be more flexible if you will also keep the 2nd lowest number.
- If the number you are replacing is an odd number, replace it with an odd number. If even, replace it with an even number. Just limit your odd or even numbers to 4 maximum.

### Complex Method

If you want to be more strategic, use the Complex Drop Technique. With the complex method, you will be needing a list of the last 21 draws.

Following is a snapshot of 21 consecutive draws from Ultra Lotto 6/58.

Following is a snapshot of 21 consecutive draws from Ultra Lotto 6/58.

21 | 14 | 25 | 35 | 36 | 45 | 54 | ||||||||||||

20 | 13 | 29 | 30 | 34 | 44 | 54 | 1 | 1 | 29 | 30 | ||||||||

19 | 3 | 6 | 34 | 38 | 39 | 43 | 2 | 1 | 34 | 38 | ||||||||

18 | 2 | 10 | 22 | 26 | 37 | 50 | ||||||||||||

17 | 3 | 10 | 14 | 18 | 23 | 54 | 1 | 1 | 18 | 23 | ||||||||

16 | 13 | 16 | 22 | 28 | 31 | 37 | 2 | 37 | ||||||||||

15 | 8 | 11 | 32 | 45 | 49 | 52 | ||||||||||||

14 | 8 | 22 | 31 | 32 | 52 | 54 | ||||||||||||

13 | 7 | 13 | 15 | 22 | 33 | 54 | 1 | 3 | 4 | 1 | 7 | 13 | 22 | 33 | ||||

12 | 11 | 15 | 28 | 43 | 49 | 53 | 2 | 2 | 1 | 43 | 49 | 53 | ||||||

11 | 19 | 21 | 32 | 39 | 41 | 56 | 1 | 56 | ||||||||||

10 | 2 | 12 | 14 | 15 | 35 | 55 | 1 | 2 | 12 | 35 | ||||||||

9 | 1 | 6 | 15 | 17 | 20 | 55 | 2 | 1 | 6 | 20 | ||||||||

8 | 1 | 2 | 25 | 31 | 32 | 50 | 2 | 3 | 2 | 1 | 31 | 50 | ||||||

7 | 14 | 27 | 28 | 41 | 44 | 57 | 4 | 3 | 1 | 14 | 28 | 57 | ||||||

6 | 25 | 45 | 46 | 52 | 54 | 58 | 3 | 25 | ||||||||||

5 | 8 | 19 | 21 | 26 | 27 | 36 | 3 | 2 | 2 | 2 | 8 | 21 | 26 | 27 | ||||

4 | 5 | 15 | 16 | 44 | 54 | 55 | 1 | 5 | 2 | 7 | 5 | 15 | 16 | 54 | ||||

3 | 2 | 39 | 44 | 46 | 52 | 55 | 4 | 3 | 4 | 2 | 4 | 2 | 39 | 44 | 46 | 52 | ||

2 | 10 | 11 | 17 | 41 | 55 | 58 | 3 | 3 | 2 | 3 | 2 | 10 | 11 | 17 | 41 | 58 | ||

1 | 3 | 19 | 32 | 36 | 45 | 55 | 3 | 3 | 5 | 3 | 4 | 6 | 3 | 19 | 32 | 36 | 45 | 55 |

#### Selecting the Ideal Combination

Follow this procedure in forming your combination to hypothetically ensure that it has not won yet.- List the results of the last 21 draws. Arrange each combination from lowest to highest number.
- Cross out the duplicate numbers, leaving only 1 of the many. In the snapshot above, the numbers on the left are the winning numbers or jackpot combination (brightly colored numbers). After removing the duplicates, the result are the winning numbers on the right (pastel-coloured area).
- This is the trickiest part. Your basis of picking the perfect combination are the winning numbers on the right; the ones with duplicates removed.
- Pick a combination with its lowest number still intact plus 3 other numbers. In the snapshot above, three jackpot results are available: 4th, 5th and 13th draws.
- Why pick a combination with only 4 numbers intact? The reason is that winning numbers usually come from different past draws. If you will pick the very latest draw, you will need to drop 5 numbers because these 5 numbers are unlikely to win again in the next draw. If you will pick the 2nd draw combination, you will need to drop 4 numbers for the same reason. Likewise, if you will pick the last 3rd draw, you will have to drop the other 4 numbers for the same reason; or in other words, only 1 of them can likely win again in the next draw. So, the last 4th draw is most ideal.
- The combination from the 4th draw is the most ideal choice. 05 • 15 • 16 • 44 • 54 • 55. Now, which ones do we drop and which ones do we retain?
- By default, we keep the lowest number, which is 05.
- If you will look at the section where the duplicates were removed, what's left with the combination are: 05 • 15 • 16 • 54. These are the numbers (except 5) that you are going to drop for the reason that they are unlikely to win again all together in the next draw.
- Thus, the numbers that you keep are: 05 • 44 • 55. Are they likely to win again together? Yes. Remember that winning numbers come from different draws of the past. In our example, 05 will be coming from the 4th draw; 44 will be coming from the 3rd draw; and 55 will be coming from the first draw. Examine the snapshot above (duplicates removed section): where do you see the numbers 5, 44 and 55? They do not appear on the same row.

#### Substituting the Numbers You Dropped

- The combination we picked is: 05 • 15 • 16 • 44 • 54 • 55.
- By dropping 15 • 16 • 54, we are left with: 05 • ? • ? • 44 • ? • 55. So, we are looking for two numbers between 5 and 44; and one number between 44 and 55.
- Observe these probability rules: 2 to 4 winning numbers usually come from the last 6 draws (pink area in the snapshot); 1 number can from the other areas: yellow, blue and white areas in the snapshot.
- 5, 44, and 55 are located in the pink area. Therefore, we will be looking for 1 number each in the yellow, blue and white area.
- Let's look for a number between 44 and 55. Your choices are: 50 in the yellow area; 49, 53 or 56 in the blue area; and none in the white area.
- Let's look for the 2nd lowest number. Originally, the 2nd lowest number was 15. We won't deviate much from this figure; so we will be looking for numbers around 15 such as 13, 14, or 17. The number 14 is in the yellow area together with #50. In the blue area, your choices are 12 and 13. In the white are, your only choice is 18.
- Assuming you have decided with 18 in the white area, and 50 from the yellow area, your partial combination now is: 05 • 18 • ? • 44 • 50 • 55.
- Next is to look for a number between 18 and 44. You have used the yellow and white area, so, you can only get a number from the blue area. And your choices are: 35, 43, 22, 33.
- Assuming you picked 35. Your new combination is now: 05 • 18 • 35 • 44 • 50 • 55
- Odd:even ratio. Examine further your new combination. How many odd and even numbers? The common odd:even ratio are 4:2, 2:4 or 3:3.
- Low:high number ratio. How many lower numbers (1 - 29) and how many high numbers (30 - 58). Ideal ratio are: 4:2, 2:4, 3:3.
- Distances. Calculate the difference between every 2 adjacent numbers. Ideally, avoid repeating numbers. In our example, distances are: 13, 17, 9, 6, 5.

How good is your new number? I peeked into my database; there were only 2 results that start with 5 • 18. One of which had 55 but that is okay. Your new combination is good.

## Theory of a Single Number To Win Repeatedly

Definitely, a single winning number can win again anytime for the reason that there are only, depending on lottery type, 42 to 58 numbers to draw from randomly. Think of the numbers 42 or 58 as one cycle that for each cycle, a number can only win once. So, if there were 156 draws in a year, that's equivalent to 22 cycles for a system 6/42 calculated as 156 times 6^{1}divided by 42. Therefore, a single number can probably win, more or less, 22 times a year.

^{1}A cycle is composed of instances of being drawn. In a lottery system six, 6 balls are drawn out. Each time a ball is drawn, that's equivalent to 1 instance. This means, that for every draw, each number has 6 chances of being drawn. Thus, 156 draws times 6 instances is equal to 936 instances. Divide 936 by 42, you'd get 22 cycles.

As the number of combinations increase, the chance of a combination to win again becomes rare. For example, if you would combine 2 numbers from 1 to 42, you would be forming 903 combinations. For every 6-number jackpot numbers, Therefore, a cycle would consist of 903 draws.

To be continued....

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