# Question for Wheel Design Gurus

I Read Book called "How Not To Be Wrong" by Jordan Ellenberg.

There is a chapter where he discuss the MIT guys the took advantage of the Massachusetts Cash Winfall roll down lottery. He was trying to figure out whether the MIT guys utilized some sort of algorithm.

He mentioned a combinatorial design paper by R.H.F Denniston "Some New 5-Designs" and give an example of the first few combination in the design. Here is a brief excerpt from the book:

"Denniston wrote down a list of 285,384 six-number combinations from a choice of forty-eight numbers.

The list starts like this:

1,2,48,3,4,8
2,3,48,4,5,9
1,2,48,3,6,32...

The first two tickets have four numbers in common: 2,3,4 and 48. But---and here is the miracle of the Denniston system---you will never find any two of those 285,384 tickets that have five numbers in common."

I was playing around with CoverMaster and couldn't see how to design such a wheel.

This is for you Wheel gurus. Is such a design even possible? and If so, how would I configure CoverMaster?

It's a special type of combinatorial design called a steiner system. Explation https://www.dmgordon.org/steiner/ and cover https://ljcr.dmgordon.org/cover/LARGE/C_48_6_5.html

The second link is very large, so don't click it if your device can't handle 285,384 lines of text on a webpage.

If you are trying to make a Key/King/Monarch number wheel with one number the same in every line using CoverMaster try this.

For a six number wheel with one key number, setup to make a five number wheel one number less then the size of the wheel, then using the drop down menu expand the grid from five numbers to six.  CM will ask you if you want to specify a specific number or random, a specific number will put the same number on every line.

For two key numbers, setup to make a four number wheel and expand the grid.  For a five number key number wheel, setup to make a four number wheel and expand the grid, etc.

Best of luck.  BobP

code for a wheel for 1..40 of 5 numbers  change \$num to yourown number eg  34,3,2,33   \$num = 34,3,2,33,1,22,6

\$num = 1..40

\$try1 = \$num
\$try2 = \$num
\$try3 = \$num
\$try4 = \$num
\$try5 = \$num
\$try6 = \$num

foreach (\$line in \$try1)
{
foreach (\$line1 in \$try2)
{
foreach (\$line2 in \$try3)
{
foreach (\$line3 in \$try4)
{
foreach (\$line4 in \$try5)
{

\$tline = (\$line,\$line1,\$line2,\$line3,\$line4) -join(",")
\$e = \$tline.Split(",")| select –unique
\$num = \$e.Count

if (\$num -lt 5 ) {

}

else {
\$e=(\$tline.Split(",")| foreach {[int]\$_ } | sort | select –unique) -join (",")
write-host \$e

}
}
}
}
}
}

example output :

1,2,3,4,5
1,2,3,4,6
1,2,3,4,7
1,2,3,4,8
1,2,3,4,9
1,2,3,4,10
1,2,3,4,11
1,2,3,4,12
1,2,3,4,13
1,2,3,4,14
1,2,3,4,15
1,2,3,4,16
1,2,3,4,17
1,2,3,4,18
1,2,3,4,19
1,2,3,4,20
1,2,3,4,21
1,2,3,4,22
1,2,3,4,23
1,2,3,4,24
1,2,3,4,25
1,2,3,4,26
1,2,3,4,27
1,2,3,4,28
1,2,3,4,29
1,2,3,4,30
1,2,3,4,31
1,2,3,4,32
1,2,3,4,33
1,2,3,4,34
1,2,3,4,35
1,2,3,4,36
1,2,3,4,37
1,2,3,4,38
1,2,3,4,39
1,2,3,4,40
1,2,3,4,5
1,2,3,5,6
1,2,3,5,7
1,2,3,5,8
1,2,3,5,9
1,2,3,5,10
1,2,3,5,11
1,2,3,5,12
1,2,3,5,13
1,2,3,5,14
1,2,3,5,15
1,2,3,5,16
1,2,3,5,17
1,2,3,5,18
1,2,3,5,19
1,2,3,5,20
1,2,3,5,21
1,2,3,5,22
1,2,3,5,23
1,2,3,5,24
1,2,3,5,25
1,2,3,5,26
1,2,3,5,27
1,2,3,5,28
1,2,3,5,29
1,2,3,5,30
1,2,3,5,31
1,2,3,5,32
1,2,3,5,33
1,2,3,5,34
1,2,3,5,35
1,2,3,5,36
1,2,3,5,37
1,2,3,5,38
1,2,3,5,39
1,2,3,5,40
1,2,3,4,6
1,2,3,5,6
1,2,3,6,7
1,2,3,6,8
1,2,3,6,9
1,2,3,6,10
1,2,3,6,11
1,2,3,6,12
1,2,3,6,13
1,2,3,6,14
1,2,3,6,15
1,2,3,6,16
1,2,3,6,17
1,2,3,6,18
1,2,3,6,19
1,2,3,6,20
1,2,3,6,21
1,2,3,6,22
1,2,3,6,23
1,2,3,6,24
1,2,3,6,25
1,2,3,6,26
1,2,3,6,27
1,2,3,6,28
1,2,3,6,29
1,2,3,6,30

Thanks guys!

Yea, that's what you have to do..reduce the amount of tickets, but still offer the guarantee.. it's hard to do, tedious, but Combinatorical Math says it's possible..

take for instance a 10n 5-pick full wheel, requires 210 combinations in offering a 5 of 5 guarantee,... I designed a Combinatorical wheel that will do the same in 84 combinations...I call it a "Black Hole" wheel..!! .. can't get away!.. Like an Anaconda, squeezes til ya lights go out...

ANA-CONDA plan on using it on that illusive prey.. the Lottery...because it "runs away"..!!

It's my drawing board answer to the " ole 1 digit away disease"...

Imo squeeze that Jackpot, til they won't have enough money left to pay their light bill...LOL

It's a numerical jungle out here...that's why there's ME..!!

-Stat\$talker

A full wheel 5 if 5 of 10 for a pick 5 game has 252 tickets. See: https://www.lotterypost.com/wheels/pick5

Not sure where you're getting 210. The full wheels can't be reduced while maintaining the guarantee because they are designed to match the full set, not a subset. A full wheel that has 210 tickets is a 6 if 6 of 10 for a pick 6 game, as seen here: https://www.lotterypost.com/wheels/pick6

You must be referring to some abbreviated wheel with a size of 210 that you were able to reduce.

Rade Belic's covering bounds page gives the smallest sizes of loads of (v, k, t, m) coverings, aka wheels. It's a good place to challenge yourself to create the most efficient coverings. For instance, the 3 if 5 of 39 wheel for a pick 5 game can theoretically be as small as 240 tickets according to the page. (Scroll 2/3 of the way down the page to the entry that reads "39,5,3,5") I managed to create one that's 241 tickets.

Sorry, my mistake... I didn't check... I recalled seeing 210, but now that you mention it.. Yea , it is 252.. But, the wheel, still guarantees 5 of 5...in 84 combinations... I've tested it..that's what Combinatorical arrangements are about,.. reducing the combinations, but still guaranteeing the 5 of 5...

-Stat\$talker

I'm not asking you to reveal your wheel here, but can you explain to me your understanding of what it means to guarantee a 10 number, 5 of 5? I'm thinking we are not on the same page here.

My understanding is this:

I choose 10 numbers in my 5/39 game.

I wheel those 10 numbers and end up with 252 lines to play, with a guarantee 5 of 5 win if any 5 of my 10 numbers are drawn.

Whaatsup B 216?

I figured you'd surface on this one..

Combinatorical Wheels reduces the amount of tickets, but still offer the guarantee..by causing the chosen numbers to work together in the minimal amount of combinations that still guarantees the 5 of 5 .,or whatever tier guarantee you're trying to achieve...

It's a Mathematical "shortcut"..it's possible..but one must "think outside the box"...that's as much as I can say..

It's my drawing board answer to the " ole 1 digit away disease"...We Gamblers often come down with it .. because we don't consume enough vitamin "J"... for JaCkPoTs...!!  LOL

Ole Cottoneyedjoe caught my typo mistake there on my earlier post.. I stand corrected,.. but my wheel still gives me the 5 of 5 guarantee in 84 combinations.

I've tested it, by putting in any 5 winning numbers, then randomly choosing the remaining 5 to complete the 10n... and it's always there, so it works..

What Cottoneyedjoe & I were agreeing on , is that a full wheel for 10n pick-5 is 252 combinations.. So, yes, your understanding is correct..

-Stat\$talker

I've tested it, by putting in any 5 winning numbers, then randomly choosing the remaining 5 to complete the 10n

Now it sounds like you're describing a pick 10 game, like 10-spot keno or NY Pick 10.

Note there is a difference between the terms "5 of 5" and "5 if 5." I don't know if you're making a typo all the time or you don't understand the language used to describe wheels (also called combinatorial designs, covering designs, combinatorial covers, or combinatorial covering designs -- all equivalent terms). A covering design C(v, k, t, m) is a collection of k-element subsets of the set {1, 2, ..., v} such that every m-element subset of {1, 2, ..., v} intersects at least one member of C(v, k, t, m) in at least t elements. In lottery terms, its a "t if m of v guarantee for a pick k game."

For example, C(v=30, k=5, t=2, m=4) is a 2 if 4 of 30 wheel for a pick 5 game.

The La Jolla Covering Repository hosted on D. Gordon's site (see my first post in this thread) has minimal covers for t=m, and if you have a Platinum LP membership you can download efficient coverings for when t<m on the wheels page.

It's not clear what your v, k, t, and m values are from your description of your wheel.

"I've tested it, by putting in any 5 winning numbers, then randomly choosing the remaining 5 to complete the 10n... and it's always there, so it works.."

Putting in 5 winning numbers tells me you are testing AFTER the drawing. Try testing BEFORE the drawing with 10 random numbers of your choice and get back to us. It appears you are not comparing apples to apples here.

Yea, that I am,.. but I said at least that..Not trying to hide anything here...

Those results are just a test, before I expand it to higher wheel number count..!

I'm thinking outside the box... If it's consistent with 10n 5 if 5.. then it will be as consistent when I expand the same Mathematical properties to higher count wheels... get it?

I figure, by the time I get it up to 30n 5 if 5.. it'll still perform comparatively as the 10n wheel .. and at around 18-20,000 less combinations..

I already know how to pick Statistically viable numbers, that's not my drawing board problem.. It's getting all the winning numbers on 1 ticket, or at least enough of the winning numbers to make a decent profit...maybe buy some XO and a nice Cuban Cigar..!!!... while I sit back and count my money..!!

-Stat\$talker

"There is a chapter where he discuss the MIT guys the took advantage of the Massachusetts Cash Winfall roll down lottery. He was trying to figure out whether the MIT guys utilized some sort of algorithm."

He sounds like he might be too stupid to write a book that's worth reading. There was no need for an algorithm, because a rolldown resulted in a statistical probability of a payout that was more than \$1 for every \$1 you bet, so all you had to do was bet enough to have results that are statistically normal. You could calculate a fairly precise value for the minimum safe bet but if you know you can win you should bet as much money as you can, and that's exactly what they did.

It would be comparable to pick 3 offering a payout of more than \$1,000 for a \$1 bet. If you bet \$999 there's still a 1 in 1000 chance that you'll lose, but if you bet \$1 million the chances of not turning a profit are zero as a practical matter.  As a rough guess the Winfall game probably had overall odds of something like 1 in 35, so buying just a few thousand bucks worth of tickets would have given them an almost certain chance of turning a (small) profit.

It's a virtual certainty that people from MIT wouldn't have been stupid enough to waste time trying to predict numbers and eliminate any from play.  For a modest number of bets eliminating  some numbers is no different than concentrating on others, but actually hitting the jackpot would have been better than just winning the statistically normal number of smaller prizes. If they could have managed to play all possible combinations they would have, but if they had avoided particular numbers they would have run the risk of deliberately avoiding winning numbers.

What the people who promote wheels don't usually mention is that unless you wheel at least half of the numbers they're more likely to guarantee that you'll lose.

"reduce the amount of tickets, but still offer the guarantee"

That's exactly what I do. for MM and PB I eliminate almost all of the numbers and play wheels that only require 1 ticket. If the numbers I play win I'm guaranteed to win the jackpot. The wheel also guarantees that I won't lose anything over \$2.

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