I would like to know if this method is as good as random for generating a number for Pick 3, because I think it results in a skewed distribution and some numbers being left out. Like for instance I don't think you can ever get the number 349 out of it. I mentioned this to the person who swears by this method, but they said prove it. Well I think it requires someone with a computer background so I couldn't prove my hunch. The method is : multiply together the last two results, then multiply that product by 7, then use just the last 3 digits. Any super mathy people can confirm my intuition?

Dec 16, 2019, 6:10 am

my math skills are not up to answering your question, so I will ask a question. If I understand this method, it rejects 999 combinations and leaves 1. over the past 600 games would it have worked? yes, then you at least break even. does it then matter that 1 or more combinations are not possible?

i did a quick check of 4000 games and it appears that approximately half of the possible combinations are never used. could there be an error in my testing? yes

scrolling through the test results i saw some interesting results that will bear further checking, would a number other than 7 give better results?

you gave me something to look at this morning so thanks for that.

In response to db101

**Quote:** Originally posted by db101 on Dec 16, 2019

I would like to know if this method is as good as random for generating a number for Pick 3, because I think it results in a skewed distribution and some numbers being left out. Like for instance I don't think you can ever get the number 349 out of it. I mentioned this to the person who swears by this method, but they said prove it. Well I think it requires someone with a computer background so I couldn't prove my hunch. The method is : multiply together the last two results, then multiply that product by 7, then use just the last 3 digits. Any super mathy people can confirm my intuition?

This is a really cool math problem and it turns out you're partially right. All the numbers from 000 to 999 actually can be generated with your friend's method, but as you intuited, they will not be generated uniformly if you consider all 1 million possible ways of choosing the first seed and second seed values. Here is the distribution I found when I ran a script to enumerate the outputs

400 numbers: 0.04% chance of occurring each

280 numbers: 0.08% chance of occurring each

116 numbers: 0.12% chance of occurring each

40 numbers: 0.16% chance of occurring each

4 numbers: 0.17% chance of occurring each

100 numbers: 0.2% chance of occurring each

28 numbers: 0.24% chance of occurring each

2 numbers: 0.34% chance of occurring each

4 numbers: 0.36% chance of occurring each

20 numbers: 0.4% chance of occurring each

1 number: 0.51% chance of occurring

4 numbers: 0.6% chance of occurring each

1 number: 0.85% chance of occurring

Total: 1000 numbers

The script I wrote considers the seed pair (A, B) as distinct from (B, A) even though the work out to the same output.

The four hundred numbers that occur at a rate of 0.04% each in the output space are all odd, so this method greatly favors the production of even numbers, i.e., numbers with the last digit 0, 2, 4, 6, or 8. Going back to your number 349, you can generated it many ways, for example starting with initial values 521 and 067. When you multiply them and multiply the product by 7 you get 244**349**.

Another interesting thing I found is that the seed value 143 and any other three digit number XYZ always gives you back XYZ when you apply the method. Ex: 705*143*7 = 705705. It's not too difficult to see why. 143*7 = 1001, and when you multiply any three digit number by 1001, the last three digits will be the same you started with.

I can PM you the script if you want. Thanks for the brain food, and good luck to your friend!

Dec 16, 2019, 7:34 pm

In response to db101

**Quote:** Originally posted by db101 on Dec 16, 2019

I would like to know if this method is as good as random for generating a number for Pick 3, because I think it results in a skewed distribution and some numbers being left out. Like for instance I don't think you can ever get the number 349 out of it. I mentioned this to the person who swears by this method, but they said prove it. Well I think it requires someone with a computer background so I couldn't prove my hunch. The method is : multiply together the last two results, then multiply that product by 7, then use just the last 3 digits. Any super mathy people can confirm my intuition?

Using 1000 OH actual combined draws with no altering

371 numbers appeared 0 times

376 numbers appeared 1 time

168 numbers appeared 2 times

59 numbers appeared 3 times

19 numbers appeared 4 times

7 numbers appeared 5 times

0 numbers appeared 6 times

0 numbers appeared 7 times

Using 1000 OH combined draws with last 3 digits from (num 1 x num 2) x 7

412 numbers appeared 0 times

320 numbers appeared 1 time

168 numbers appeared 2 times

69 numbers appeared 3 times

20 numbers appeared 4 times

10 numbers appeared 5 times

0 numbers appeared 6 times

1 numbers appeared 7 times

Using 1000 OH Eve only draws with last 3 digits from (num 1 x num 2) x 7

377 numbers appeared 0 times

358 numbers appeared 1 time

185 numbers appeared 2 times

58 numbers appeared 3 times

14 numbers appeared 4 times

6 numbers appeared 5 times

2 numbers appeared 6 times

0 numbers appeared 7 times

Using 1000 RANDBETWEEN in Open Office

383 numbers appeared 0 times

355 numbers appeared 1 time

171 numbers appeared 2 times

66 numbers appeared 3 times

20 numbers appeared 4 times

5 numbers appeared 5 times

0 numbers appeared 6 times

0 numbers appeared 7 times

Using 1000 PC based Random function

360 numbers appeared 0 times

380 numbers appeared 1 time

182 numbers appeared 2 times

61 numbers appeared 3 times

14 numbers appeared 4 times

2 numbers appeared 5 times

0 numbers appeared 6 times

1 numbers appeared 7 times

In response to cottoneyedjoe

**Quote:** Originally posted by cottoneyedjoe on Dec 16, 2019

This is a really cool math problem and it turns out you're partially right. All the numbers from 000 to 999 actually can be generated with your friend's method, but as you intuited, they will not be generated uniformly if you consider all 1 million possible ways of choosing the first seed and second seed values. Here is the distribution I found when I ran a script to enumerate the outputs

400 numbers: 0.04% chance of occurring each

280 numbers: 0.08% chance of occurring each

116 numbers: 0.12% chance of occurring each

40 numbers: 0.16% chance of occurring each

4 numbers: 0.17% chance of occurring each

100 numbers: 0.2% chance of occurring each

28 numbers: 0.24% chance of occurring each

2 numbers: 0.34% chance of occurring each

4 numbers: 0.36% chance of occurring each

20 numbers: 0.4% chance of occurring each

1 number: 0.51% chance of occurring

4 numbers: 0.6% chance of occurring each

1 number: 0.85% chance of occurring

Total: 1000 numbers

The script I wrote considers the seed pair (A, B) as distinct from (B, A) even though the work out to the same output.

The four hundred numbers that occur at a rate of 0.04% each in the output space are all odd, so this method greatly favors the production of even numbers, i.e., numbers with the last digit 0, 2, 4, 6, or 8. Going back to your number 349, you can generated it many ways, for example starting with initial values 521 and 067. When you multiply them and multiply the product by 7 you get 244**349**.

Another interesting thing I found is that the seed value 143 and any other three digit number XYZ always gives you back XYZ when you apply the method. Ex: 705*143*7 = 705705. It's not too difficult to see why. 143*7 = 1001, and when you multiply any three digit number by 1001, the last three digits will be the same you started with.

I can PM you the script if you want. Thanks for the brain food, and good luck to your friend!

Thank you cottoneyedjoe! This is **exactly** what I was looking for. I will dust off my coding skills and try to work it out myself before asking you for the code you used. I am a little miffed that I was wrong about not being able to generate all numbers but I'm glad my intuition was right about the non-uniform distribution of the outputs. I think my friend will be very interested in what you discovered about the number 143 and the evens vs odds issue.

@mr B 216: thanks for also taking a look. Although it's not what I was looking for it is interesting to see the comparison to Obio results and I appreciate the effort you put into making those lists.

Dec 27, 2019, 9:52 am

In response to db101I would like to know if this method is as good as random for generating a number for Pick 3, because I think it results in a skewed distribution and some numbers being left out. Like for instance I don't think you can ever get the number 349 out of it. I mentioned this to the person who swears by this method, but they said prove it. Well I think it requires someone with a computer background so I couldn't prove my hunch. The method is : multiply together the last two results, then multiply that product by 7, then use just the last 3 digits. Any super mathy people can confirm my intuition?

**Quote:** Originally posted by db101 on Dec 16, 2019

If you want 3 pseudo generated random numbers, why not throw out the guesswork of that method and just generate 3 random numbers?

| A | B | C | D | E | |

1 | =RAND() | =RAND() | =RAND() | =RAND() | =A1+B1+C1+D1 | =RANK(E1,$E$1:$E$10)-1 |

2 | =RAND() | =RAND() | =RAND() | =RAND() | ||

3 | =RAND() | =RAND() | =RAND() | =RAND() | ||

4 | =RAND() | =RAND() | =RAND() | =RAND() | ||

5 | =RAND() | =RAND() | =RAND() | =RAND() | ||

6 | =RAND() | =RAND() | =RAND() | =RAND() | ||

7 | =RAND() | =RAND() | =RAND() | =RAND() | ||

8 | =RAND() | =RAND() | =RAND() | =RAND() | ||

9 | =RAND() | =RAND() | =RAND() | =RAND() | ||

10 | =RAND() | =RAND() | =RAND() | =RAND() |

Column A to D (or as many as you want): a bunch of pseudo random numbers, hopefully seeded, um, randomly

Column E: Drag E1 down for all 10 cells in that column. SUM or add up all RAND() columns. Or use MDAS without the division because 0.000 is a possible and that will error. This column proved to be sufficiently shuffled. Running a simulation for an hour (over one billion tests) only one had a duplicate. Over one billion 0 to 9 with no duplicates in there.

Column F: Drag F1 down for all 10 cells in that column. The -1 at the end is to shift the ranking from 1-10 to 0-9. Use 3 digits here.

Much easier, less guesswork, and no back-testing required.

Or just quick picks? Seems like this doesn't do anything if you prefer hot or cold, or any other method. Maybe it's just to satisfy knowing your picks are really random if your lottery terminal P3 quick picks seem weird? Thoughts?

Edit: weird table behavior and cell shifting due to empty table cells. Where did column F header go? I don't know.

In response to GoogilyMoogily

**Quote:** Originally posted by GoogilyMoogily on Dec 27, 2019

If you want 3 pseudo generated random numbers, why not throw out the guesswork of that method and just generate 3 random numbers?

| A | B | C | D | E | |

1 | =RAND() | =RAND() | =RAND() | =RAND() | =A1+B1+C1+D1 | =RANK(E1,$E$1:$E$10)-1 |

2 | =RAND() | =RAND() | =RAND() | =RAND() | ||

3 | =RAND() | =RAND() | =RAND() | =RAND() | ||

4 | =RAND() | =RAND() | =RAND() | =RAND() | ||

5 | =RAND() | =RAND() | =RAND() | =RAND() | ||

6 | =RAND() | =RAND() | =RAND() | =RAND() | ||

7 | =RAND() | =RAND() | =RAND() | =RAND() | ||

8 | =RAND() | =RAND() | =RAND() | =RAND() | ||

9 | =RAND() | =RAND() | =RAND() | =RAND() | ||

10 | =RAND() | =RAND() | =RAND() | =RAND() |

Column A to D (or as many as you want): a bunch of pseudo random numbers, hopefully seeded, um, randomly

Column E: Drag E1 down for all 10 cells in that column. SUM or add up all RAND() columns. Or use MDAS without the division because 0.000 is a possible and that will error. This column proved to be sufficiently shuffled. Running a simulation for an hour (over one billion tests) only one had a duplicate. Over one billion 0 to 9 with no duplicates in there.

Column F: Drag F1 down for all 10 cells in that column. The -1 at the end is to shift the ranking from 1-10 to 0-9. Use 3 digits here.

Much easier, less guesswork, and no back-testing required.

Or just quick picks? Seems like this doesn't do anything if you prefer hot or cold, or any other method. Maybe it's just to satisfy knowing your picks are really random if your lottery terminal P3 quick picks seem weird? Thoughts?

Edit: weird table behavior and cell shifting due to empty table cells. Where did column F header go? I don't know.

I completely agree with you and it's not my system. Excel's pseudorandom number generator is known to be pretty nonrandom so I would use something else if I was hardcore about randomness. Some polyhedral dice maybe.

Jan 8, 2020, 7:09 pm

It would be interesting to back test the strategy to see how many times the system hits. There is only a 63.23% chance of winning off a straight combo in 1000 consecutive games.

In response to db101

**Quote:** Originally posted by db101 on Dec 27, 2019

I completely agree with you and it's not my system. Excel's pseudorandom number generator is known to be pretty nonrandom so I would use something else if I was hardcore about randomness. Some polyhedral dice maybe.

Actually the quality of Excel's RAND() function is pretty good.

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