Last data update: 2024-06-15

Bonus numbers are excluded for calculation

Bonus numbers are excluded for calculation

Consider:
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What will be the next? ■ as it appeared the more? Or would it be ■ this time?

We find out by finding winning counts of each number.

Number | Win Count |
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34 | 172 |

18 | 165 |

12 | 164 |

45 | 163 |

13 | 162 |

Number | Win Count |
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9 | 118 |

32 | 133 |

22 | 134 |

28 | 135 |

23 | 136 |

Consider:
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What will be the next? Will it be ■ since it was drawn recently? Or will it see ■ as it has not appeared for a long time?

We calculate the number of days since a number's last draw.

Number | Draws Ago | Last Drawn |
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5 | 73 | 2023-01-28 |

27 | 39 | 2023-09-23 |

18 | 33 | 2023-11-04 |

28 | 24 | 2024-01-06 |

29 | 19 | 2024-02-10 |

Number | Draws Ago | Last Drawn |
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3 | 1 | 2024-06-15 |

8 | 1 | 2024-06-15 |

17 | 1 | 2024-06-15 |

30 | 1 | 2024-06-15 |

33 | 1 | 2024-06-15 |

34 | 1 | 2024-06-15 |

13 | 2 | 2024-06-08 |

19 | 2 | 2024-06-08 |

21 | 2 | 2024-06-08 |

24 | 2 | 2024-06-08 |

35 | 2 | 2024-06-08 |

6 | 3 | 2024-06-01 |

31 | 4 | 2024-05-25 |

32 | 4 | 2024-05-25 |

38 | 4 | 2024-05-25 |

44 | 4 | 2024-05-25 |

2 | 5 | 2024-05-18 |

26 | 5 | 2024-05-18 |

41 | 5 | 2024-05-18 |

Consider:
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What will be the next? Won't it be ■ since it was periodicially appearing with interval 1 but it didn't for a while?

This is quite similar to the number of days since the last win, but this one is a bit more formal. Assuming that each number usually appears at regular interval, we can estimate how unlikely a ball's inapperance is as well as if now is the avg time for a number to appear.

More formally, we assume that each ball’s occurrence interval follows a Gaussian distribution (a bell curve). We then calculate the p-value of the time since the ball’s last appearance. If it's close to 0.0, a ball didn't win weirdly long time, e.g., as ■ did. If the p-value is around 0.5, a number is supposed to be drawn now if it's appearing at the usual interval.

See the diagram to understand the meaning of the p-value:

Number | p-value |
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5 | 0.000 |

27 | 0.000 |

18 | 0.000 |

28 | 0.016 |

37 | 0.065 |

Number | p-value |
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4 | 0.469 |

14 | 0.497 |

11 | 0.523 |

16 | 0.527 |

15 | 0.532 |

This is similar to Periodicity of each number: The days since the last win, but this time we assume that all numbers share a single distribution in their average time between wins. After that, it's determined if a number isn't winning for a long time or if a number is supposed to be drawn now if it's appearing at the usual interval (of all numbers).

This diagram shows the case of using separate distributions for each number:

This shows the case of using a single distribution for all numbers:

Number | p-value |
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5 | 0.000 |

27 | 0.000 |

18 | 0.000 |

28 | 0.009 |

29 | 0.049 |

Number | p-value |
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4 | 0.468 |

11 | 0.525 |

14 | 0.525 |

15 | 0.525 |

16 | 0.525 |

Consider:
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What will be the next? We see winning interval shift of ■ as it started appear frequently, i.e., ■ ■ ■ ■ became ■ ■ ■. To the contrary, ■ took longer time to appear in the last wins.

As in the previous section, we assume that each number has a fixed interval between wins. From the distribution, we detect if the most recent winning interval of a number has shown any shift from the usual by computing the p-value.

If the last interval was shorter than usual, it has larger p-value, while longer interval has smaller.

Number | p-value |
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31 | 0.843 |

27 | 0.839 |

21 | 0.829 |

26 | 0.828 |

18 | 0.811 |

Number | p-value |
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45 | 0.000 |

15 | 0.000 |

34 | 0.012 |

22 | 0.025 |

10 | 0.054 |

["This is similar to the previous very recent interval change, but this time we assume that all numbers share a single distribution in their average time until win. After that, p-value of the each number's last winning interval is calculated."]

Number | p-value |
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5 | 0.821 |

18 | 0.821 |

21 | 0.821 |

26 | 0.821 |

27 | 0.821 |

31 | 0.821 |

39 | 0.821 |

2 | 0.781 |

3 | 0.781 |

6 | 0.781 |

14 | 0.781 |

16 | 0.781 |

36 | 0.781 |

42 | 0.781 |

38 | 0.737 |

40 | 0.737 |

12 | 0.688 |

28 | 0.688 |

29 | 0.688 |

7 | 0.636 |

9 | 0.636 |

19 | 0.636 |

25 | 0.636 |

41 | 0.636 |

43 | 0.636 |

Number | p-value |
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15 | 0.000 |

45 | 0.000 |

22 | 0.006 |

34 | 0.026 |

10 | 0.049 |

Consider:
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What will be the next? Conditional probability estimates how likely is an event given the previous. In the example, ■ appeared 2 times after ■, while ■ appeared 3 times. Thus, given the last ■, ■ is more likely to appear.

More formally, we calculate conditional probabililty and then estimate prob(next|prev) where prev is the numbers that appeared in the last draw. To avoid overfit, we adopt add 1 smoothing.

Number | Prob. |
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13 | 0.614 |

34 | 0.609 |

4 | 0.596 |

14 | 0.585 |

18 | 0.581 |

Consider:
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What will be the next? Do you see 3 different probability patterns in it? Let's split the sequence into 3 parts of ■■■■ / ■■■ / ■■■■, and name first and the last as state A, and the middle as state B. State A has 50% probability of emiting ■ and ■, while state B has 100% probability of ■.

Hidden Markov Model (HMM) assumes that the state of the system is hidden and only the output is observed. To apply it to the lottery, we need to 1) determine the # of state, 2) determine when a state starts and ends just given the winning numbers, and 3) estimate the probability of winning numbers in each state.

When using HMM, state corresponds to the lottery machine configuration and we're assuming that the machine is switching to different states over time.

Below is the output of HMM algorithm. It shows numbers and their winning probability (Prob.) from the predicted machine state.

Number | Prob. |
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22 | 0.061 |

31 | 0.061 |

27 | 0.060 |

42 | 0.057 |

10 | 0.054 |