Welcome Guest
You last visited December 5, 2016, 7:43 pm
All times shown are
Eastern Time (GMT-5:00)

# Mega Millions Full Distribution

Topic closed. 81 replies. Last post 2 months ago by wander73.

 Page 1 of 6
Somerset
United Kingdom
Member #9710
December 17, 2004
184 Posts
Offline
 Posted: September 11, 2016, 7:39 pm - IP Logged

I was reading the post here https://www.lotterypost.com/thread/289726 and found it very interesting so I thought this might be of use.

The full distribution for the MegaMillions

I don't know why people would play a pick 5/75 game, OK the rewards are great but the odds are so bad, 18.5 million to 1 to hit 5 and a crazy 259 million to one for the jackpot, why do people play this?

It has been said every software has a bug in it, it has also been said that every program can be reduced by one line of code.

There for any program can be reduced to 1 line that dont work.

Ny
United States
Member #167314
July 5, 2015
1805 Posts
Online
 Posted: September 11, 2016, 7:53 pm - IP Logged

What if I picked numbers like this:

0_ 3_ 4_ 4_ 6_

2_ 3_ 5_ 6_ 7_

2_ 2_ 4_ 4_ 6_

In my experience Mega and Powerball numbers land in between 3-5 groups.. powerball having 1 less group..

But there are 7 groups

0_ 1_ 2_ 3_ 4_ 5_ 6_ 7_

Could choosing the first digit of the sets of groups and leaving the second digit to fate, luck or some calculation give a better chance than showed in these graphs?

Creativity..

" What's more likely to happen will happen.. "

Million dollar operation

Somerset
United Kingdom
Member #9710
December 17, 2004
184 Posts
Offline
 Posted: September 11, 2016, 8:36 pm - IP Logged

The best approach is this.

The draw as with any draw follows a fixed set of rules based upon the rules in this case 5/75, these never change, the graphs show this.

What I would do if I were to play this draw is do the math to find when each number should appear.

So graph 1 show ball 1 appears in 1,150,626 combinations of the 17,259,390 different possible combinations found in 5/75

So 17,259,390 / 1,150,626 = 15

So ball one should appear in ticket position one on average once every 15 draws.

do the same for the others

Ball 2 Ticket position 1 = 15.85 draws

B3 T1 = 16.77 draws

B4 T1 = 17.76 draws

B5 T1 = 18.82 draws

B6 T1 = 19.96 draws

And so on, do this for all 5 graphs

Once you have the data

Position 1 2 3 4 5

Ball 1 : 15 0 0 0 0

Ball 2 : 16 278 0 0 0

Ball 3 : 17 145 6,753 0 0

Ball 4 : 18 101 2,315 243,090 0

Ball 5 : 19 79 1,191 61,641 17,259,390

Ball 6 : 20 66 736 25,014 3,451,878

Ball 7 : 21 57 505 12,691 1,150,626

Ball 8 : 23 51 372 7,360 493,125

Ball 9 : 24 47 287 4,670 246,563

Ball 10 : 25 44 230 3,161 136,979

Ball 11 : 27 41 190 2,247 82,188

Ball 12 : 29 40 161 1,660 52,301

Ball 13 : 31 38 138 1,265 34,867

Ball 14 : 33 37 121 989 24,139

Ball 15 : 35 36 107 790 17,242

Ball 16 : 38 35 96 643 12,644

Ball 17 : 41 35 87 531 9,483

Ball 18 : 44 35 80 445 7,252

Ball 19 : 47 35 73 378 5,640

Ball 20 : 51 35 68 324 4,453

Ball 21 : 55 35 63 280 3,562

Ball 22 : 59 35 60 245 2,884

Ball 23 : 64 35 56 216 2,359

Ball 24 : 69 36 54 191 1,949

Ball 25 : 75 37 51 171 1,624

Ball 26 : 81 37 49 153 1,364

Ball 27 : 89 38 47 138 1,154

Ball 28 : 97 39 45 126 983

Ball 29 : 106 41 44 115 843

Ball 30 : 116 42 43 105 727

Ball 31 : 127 43 42 97 630

Ball 32 : 140 45 41 89 549

Ball 33 : 154 47 40 83 480

Ball 34 : 170 49 40 77 422

Ball 35 : 189 51 39 72 372

Ball 36 : 210 54 39 68 330

Ball 37 : 234 57 39 64 293

Ball 38 : 261 60 39 60 261

Ball 39 : 293 64 39 57 234

Ball 40 : 330 68 39 54 210

Ball 41 : 372 72 39 51 189

Ball 42 : 422 77 40 49 170

Ball 43 : 480 83 40 47 154

Ball 44 : 549 89 41 45 140

Ball 45 : 630 97 42 43 127

Ball 46 : 727 105 43 42 116

Ball 47 : 843 115 44 41 106

Ball 48 : 983 126 45 39 97

Ball 49 : 1,154 138 47 38 89

Ball 50 : 1,364 153 49 37 81

Ball 51 : 1,624 171 51 37 75

Ball 52 : 1,949 191 54 36 69

Ball 53 : 2,359 216 56 35 64

Ball 54 : 2,884 245 60 35 59

Ball 55 : 3,562 280 63 35 55

Ball 56 : 4,453 324 68 35 51

Ball 57 : 5,640 378 73 35 47

Ball 58 : 7,252 445 80 35 44

Ball 59 : 9,483 531 87 35 41

Ball 60 : 12,644 643 96 35 38

Ball 61 : 17,242 790 107 36 35

Ball 62 : 24,139 989 121 37 33

Ball 63 : 34,867 1,265 138 38 31

Ball 64 : 52,301 1,660 161 40 29

Ball 65 : 82,188 2,247 190 41 27

Ball 66 : 136,979 3,161 230 44 25

Ball 67 : 246,563 4,670 287 47 24

Ball 68 : 493,125 7,360 372 51 23

Ball 69 : 1,150,626 12,691 505 57 21

Ball 70 : 3,451,878 25,014 736 66 20

Ball 71 : 17,259,390 61,641 1,191 79 19

Ball 72 : 0 243,090 2,315 101 18

Ball 73 : 0 0 6,753 145 17

Ball 74 : 0 0 0 278 16

Ball 75 : 0 0 0 0 15

Look to the past 100 draws and start applying this math

So if the 3rd draw has the number one in ticket position 1 then set the a counter to 15, expect to see this ball in 15 draws time.

If after 9 draws ball one occurs then it does not reset to 15, remember this math is based on fixed data based upon the rules of the draw, it still remains on average every 15 draws ball one must appear in ticket position 1 so the fact that after 9 + 3 draws its appeared twice means you adjust the average to maintain the rule.

So

Draw 1 No Hit : Odds 15 to 1

Draw 2 No Hit : Odds 14 to 1

Draw 3 Hit : Odds 15 to 1

Draw 4 No Hit : Odds 14 to 1

Draw 5 No Hit : Odds 13 to 1

Draw 6 No Hit : Odds 12 to 1

Draw 7 No Hit : Odds 11 to 1

Draw 8 No Hit : Odds 10 to 1

Draw 9 hit :      Odds 9 to 1 : (15 - 9 = 6 remains)  the rules say on average every 15 draws, we must obey these rules so expand 15 + 6 = 21

Draw 10 No Hit : Odds 21 to 1

Draw 11 No Hit : Odds 20 to 1

and so on

Do this for all balls in all ticket positions and eventually you will see things failing in line, by the math it has to.

Jamie

It has been said every software has a bug in it, it has also been said that every program can be reduced by one line of code.

There for any program can be reduced to 1 line that dont work.

NYC
United States
Member #54483
August 20, 2007
886 Posts
Offline
 Posted: September 11, 2016, 10:07 pm - IP Logged

The best approach is this.

The draw as with any draw follows a fixed set of rules based upon the rules in this case 5/75, these never change, the graphs show this.

What I would do if I were to play this draw is do the math to find when each number should appear.

So graph 1 show ball 1 appears in 1,150,626 combinations of the 17,259,390 different possible combinations found in 5/75

So 17,259,390 / 1,150,626 = 15

So ball one should appear in ticket position one on average once every 15 draws.

do the same for the others

Ball 2 Ticket position 1 = 15.85 draws

B3 T1 = 16.77 draws

B4 T1 = 17.76 draws

B5 T1 = 18.82 draws

B6 T1 = 19.96 draws

And so on, do this for all 5 graphs

Once you have the data

Position 1 2 3 4 5

Ball 1 : 15 0 0 0 0

Ball 2 : 16 278 0 0 0

Ball 3 : 17 145 6,753 0 0

Ball 4 : 18 101 2,315 243,090 0

Ball 5 : 19 79 1,191 61,641 17,259,390

Ball 6 : 20 66 736 25,014 3,451,878

Ball 7 : 21 57 505 12,691 1,150,626

Ball 8 : 23 51 372 7,360 493,125

Ball 9 : 24 47 287 4,670 246,563

Ball 10 : 25 44 230 3,161 136,979

Ball 11 : 27 41 190 2,247 82,188

Ball 12 : 29 40 161 1,660 52,301

Ball 13 : 31 38 138 1,265 34,867

Ball 14 : 33 37 121 989 24,139

Ball 15 : 35 36 107 790 17,242

Ball 16 : 38 35 96 643 12,644

Ball 17 : 41 35 87 531 9,483

Ball 18 : 44 35 80 445 7,252

Ball 19 : 47 35 73 378 5,640

Ball 20 : 51 35 68 324 4,453

Ball 21 : 55 35 63 280 3,562

Ball 22 : 59 35 60 245 2,884

Ball 23 : 64 35 56 216 2,359

Ball 24 : 69 36 54 191 1,949

Ball 25 : 75 37 51 171 1,624

Ball 26 : 81 37 49 153 1,364

Ball 27 : 89 38 47 138 1,154

Ball 28 : 97 39 45 126 983

Ball 29 : 106 41 44 115 843

Ball 30 : 116 42 43 105 727

Ball 31 : 127 43 42 97 630

Ball 32 : 140 45 41 89 549

Ball 33 : 154 47 40 83 480

Ball 34 : 170 49 40 77 422

Ball 35 : 189 51 39 72 372

Ball 36 : 210 54 39 68 330

Ball 37 : 234 57 39 64 293

Ball 38 : 261 60 39 60 261

Ball 39 : 293 64 39 57 234

Ball 40 : 330 68 39 54 210

Ball 41 : 372 72 39 51 189

Ball 42 : 422 77 40 49 170

Ball 43 : 480 83 40 47 154

Ball 44 : 549 89 41 45 140

Ball 45 : 630 97 42 43 127

Ball 46 : 727 105 43 42 116

Ball 47 : 843 115 44 41 106

Ball 48 : 983 126 45 39 97

Ball 49 : 1,154 138 47 38 89

Ball 50 : 1,364 153 49 37 81

Ball 51 : 1,624 171 51 37 75

Ball 52 : 1,949 191 54 36 69

Ball 53 : 2,359 216 56 35 64

Ball 54 : 2,884 245 60 35 59

Ball 55 : 3,562 280 63 35 55

Ball 56 : 4,453 324 68 35 51

Ball 57 : 5,640 378 73 35 47

Ball 58 : 7,252 445 80 35 44

Ball 59 : 9,483 531 87 35 41

Ball 60 : 12,644 643 96 35 38

Ball 61 : 17,242 790 107 36 35

Ball 62 : 24,139 989 121 37 33

Ball 63 : 34,867 1,265 138 38 31

Ball 64 : 52,301 1,660 161 40 29

Ball 65 : 82,188 2,247 190 41 27

Ball 66 : 136,979 3,161 230 44 25

Ball 67 : 246,563 4,670 287 47 24

Ball 68 : 493,125 7,360 372 51 23

Ball 69 : 1,150,626 12,691 505 57 21

Ball 70 : 3,451,878 25,014 736 66 20

Ball 71 : 17,259,390 61,641 1,191 79 19

Ball 72 : 0 243,090 2,315 101 18

Ball 73 : 0 0 6,753 145 17

Ball 74 : 0 0 0 278 16

Ball 75 : 0 0 0 0 15

Look to the past 100 draws and start applying this math

So if the 3rd draw has the number one in ticket position 1 then set the a counter to 15, expect to see this ball in 15 draws time.

If after 9 draws ball one occurs then it does not reset to 15, remember this math is based on fixed data based upon the rules of the draw, it still remains on average every 15 draws ball one must appear in ticket position 1 so the fact that after 9 + 3 draws its appeared twice means you adjust the average to maintain the rule.

So

Draw 1 No Hit : Odds 15 to 1

Draw 2 No Hit : Odds 14 to 1

Draw 3 Hit : Odds 15 to 1

Draw 4 No Hit : Odds 14 to 1

Draw 5 No Hit : Odds 13 to 1

Draw 6 No Hit : Odds 12 to 1

Draw 7 No Hit : Odds 11 to 1

Draw 8 No Hit : Odds 10 to 1

Draw 9 hit :      Odds 9 to 1 : (15 - 9 = 6 remains)  the rules say on average every 15 draws, we must obey these rules so expand 15 + 6 = 21

Draw 10 No Hit : Odds 21 to 1

Draw 11 No Hit : Odds 20 to 1

and so on

Do this for all balls in all ticket positions and eventually you will see things failing in line, by the math it has to.

Jamie

Hi, Jamie:

I like your LSA.

But could you tell us when do you finish your V7 of LSA?

Best regards,

lb

• Saying Lotto #s Can be Predicted means that Lotto's ODD can be reduced down  to an economical level by a system.
• Saying a Lotto System Works means that we can win constantly (not each draw)  and economically (get a real profit) by using the System.
• Practice is the only criterion for testing truth.
Somerset
United Kingdom
Member #9710
December 17, 2004
184 Posts
Offline
 Posted: September 12, 2016, 5:10 am - IP Logged

Thanks

LSAv7, I have put so many hours to this yet it still won’t be available until next year, I code LSA myself and simply cannot justify putting other developers on to this as they work on ‘bread and butter’ projects for me.

LSA is a labour of love, over the years I have logged user suggestions and v7 is to implement many of them making a lot of work for myself.

2017 some time.....

It has been said every software has a bug in it, it has also been said that every program can be reduced by one line of code.

There for any program can be reduced to 1 line that dont work.

NYC
United States
Member #54483
August 20, 2007
886 Posts
Offline
 Posted: September 12, 2016, 9:51 am - IP Logged

Thanks

LSAv7, I have put so many hours to this yet it still won’t be available until next year, I code LSA myself and simply cannot justify putting other developers on to this as they work on ‘bread and butter’ projects for me.

LSA is a labour of love, over the years I have logged user suggestions and v7 is to implement many of them making a lot of work for myself.

2017 some time.....

Hi, Jamie:

Could you give us some smart strategies for using LSA V6.08?

And the best way is to explain them by using a living example such as Pick-5 (5/39).

If so, I believe firmly that not only we can develop our winning chance but also you can get

more useful suggestions for your V7 since there are more users  than your site. Isn't it?

Best regards.

lb

• Saying Lotto #s Can be Predicted means that Lotto's ODD can be reduced down  to an economical level by a system.
• Saying a Lotto System Works means that we can win constantly (not each draw)  and economically (get a real profit) by using the System.
• Practice is the only criterion for testing truth.
Zaperlopopotam
Belgium
Member #173932
March 26, 2016
952 Posts
Offline
 Posted: September 14, 2016, 9:47 am - IP Logged

Ball 1 = Lowest 1st
Ball 2 = lowest 2nd
...

Of course 1 2 3 4 5 has the same chance as 1 18 35 43 69

Waffles, beer, fries, choc'lat's, cigarettes ... some more mayonaise?
Sauna, massage ... ?
Halal koscher mouton?
Professeur Mamadou vous prédit votre future!
Drugs straight from Colombia?
Coke with sugar and caffeine?
Vodka Red Bull?
Bet on the winning numbers and win the jackpot!
Chi chi's, Mac Do, Starbucks, Apple ...

.
Somerset
United Kingdom
Member #9710
December 17, 2004
184 Posts
Offline
 Posted: September 16, 2016, 7:56 pm - IP Logged

Yes, correct.

Ball placement probability among other things can provide a user with the information to locate their chosen numbers in the best possible positions on tickets.

Looking at the chart above, the combination 13, 23, 29, 36, 37 would be a bad choice whereas the combination 1, 19, 38, 57, 75 has the highest hit probability you can choose for this draw.

For those that don’t understand the importance of distributions take a look to the last distribution graph I posted above titled “05 75: Normal Distribution” and you will see this draw has 1,150,626 lines with ball 75 as the highest number, this fact is very significant when targeting smaller win combinations and very useful when building wheels.

Some wheels look great but if you compare the numbers in each ticket line against these charts, it can be shocking how bad the probability of intercepting some combinations actually are.

The distributions are an excellent way of determining you placements on tickets.

It has been said every software has a bug in it, it has also been said that every program can be reduced by one line of code.

There for any program can be reduced to 1 line that dont work.

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3675 Posts
Online
 Posted: September 17, 2016, 8:52 am - IP Logged

The formula for finding any Combinatorial Distribution is as follows:

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3675 Posts
Online
 Posted: September 17, 2016, 9:00 am - IP Logged

You can find the Average Hit Frequency for any given draw using the following:

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3675 Posts
Online
 Posted: September 17, 2016, 9:06 am - IP Logged

The Average Rate at which a number will recur in a given column is a follows:

Please note, this is on a draw to draw recurrence.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

Zaperlopopotam
Belgium
Member #173932
March 26, 2016
952 Posts
Offline
 Posted: September 17, 2016, 10:35 am - IP Logged

C(x,y) = V(x,y)/y!

.
The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3675 Posts
Online
 Posted: September 17, 2016, 11:06 am - IP Logged

Combinatorial Distribution

Factorial - n! = n * (n -1) * (n - 2) * ... * 3 * 2 * 1 , and 0! = 1

Permutation - P(n,r) = n! / (n - r)!

Combination - C(n,r) = P(n,r) / r!

Combinatorial Distribution - D(n,r,c,z) = C(z - 1, c - 1) * C(n - z, r - c)

n - total number of items
r - number of items in a combinatorial or permutational set
c - column number of the distribution
z - item number of the distribution

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

Zaperlopopotam
Belgium
Member #173932
March 26, 2016
952 Posts
Offline
 Posted: September 17, 2016, 11:09 am - IP Logged

C(70,20) = 393849377343759797528386895216640000 / 2432902008176640000 = 161884603662657876

.
Zaperlopopotam
Belgium
Member #173932
March 26, 2016
952 Posts
Offline
 Posted: September 17, 2016, 11:12 am - IP Logged

p(n, r) = M(i=0;i<r): (n-i)/(r-i)

n = h - l + 1

.

 Page 1 of 6