United States Member #59354 March 13, 2008 4091 Posts Offline

Posted: February 28, 2013, 10:18 am - IP Logged

Quote: Originally posted by mathhead on February 19, 2013

I am interested in learning about lottery wheel algorithms, specifically for abbreviated wheels.

And I am interested in techniques for proving that the resulting wheel is "mathematically minimal" (and "balanced"?)

(What makes a wheel "well-balanced"? Is it just some subjective concept of the "variety" or "scope" of numbers; for example, 1 2 3 4 5 might be "poorly balanced", whereas 1 12 23 34 45 might be considered "well-balanced"? Or is there a more-rigorous definition of "balanced"?)

Please note that I am interested in algorithms, not the wheels themselves.

So pointing me to the Lottery Wheels link is not sufficient, unless I overlooked a link to algorithms per se (where?). Also, since I am not a gold or platinum member, the links there to particular wheels are not useful to me (other than as good examples).

However, there is a reference there to a book by Iliya Bluskov. Does that book describe wheeling algorithms in sufficient detail to implement in a computer language? Or does the book simply provide the wheels themselves?

(Or perhaps someone here can offer an algorithm in sufficient detail.)

A Google search has turned up some implementations. I am still looking through the links. However, the ones that I have seen so far are implemented in C++. Although I have a lot of experience with C, I struggle to understand C++ implementations. An implementation in C or VBA would be helpful.

In any case, I'm not sure I will know how to prove that the results are "good", much less "mathematically minimal".

For example, the wiki lottery wheel page (as well as a link in this forum's Lottery Wheel Pick-6 page) claims that an abbreviated wheel for 4-if-4-of-8 can be covered with as few as 7 tickets for 6-draw lottery. I can easily do that in 15 tickets. But I have not yet been able to pare that down to 7, even manually, much less algorithmically. (Moreover, my algorithm is not extensible to more 8 favored numbers.)

I would be grateful if someone could show me a 7-ticket wheel for 4-if-4-of-8 for a 6-draw lottery. Then at least I know what I am shooting for.

MH

Most likely all you will find are general examples used to explain the theory. I am sure most

would not be interested in sharing the actual code. A good wheel has commerical value so it

would not be in the best interest of the programmer to share it. Google "lottery wheel theory"

and go from there. Testing is fairly easy as you can compare against known wheels to see how

your code stacks up and to test coverage just build a pair, tray, quad, etc... generator and run

the results against the wheeled sets. There use to be a website that listed the best of the

best for lines/coverage. There is a free program called covermaste that I think can still be

downloaded that would be great for testing against your code. I have never been able to build

Condition - If 4 Drawn Numbers are in the Set of 9 Numbers, - Then at least 1 Combination has 3 Winning Numbers. ____________________________________________ 05 06 07 08 09 01 03 04 06 09 02 03 04 08 09 01 02 06 07 08 01 02 03 04 05 ____________________________________________

When reading the condition you should understand the second part '- Then at least 1 Combination has 3 Winning Numbers.' to mean that a line will have At Least that many or more winning numbers.

This means if the condition is '3 Winning Numbers', then there could be a winning line with 3 or 4 winning numbers, because the other part of the condition is '4 Drawn Numbers are in the Set of 9 Numbers,".

Also, these mean the following:

Set - The wheel pool of numbers you are playing; typically less than the lottery pool of numbers being drawn. Pick - The number of balls being drawn for that lottery pool of numbers. Match - The quantity of drawn numbers that match your wheel pool numbers. Win - Guaranteed to have at least 1 line with that many matching numbers in the wheeled combination or more.

If there is a line with 4 winning numbers, it definitely has any combination of 3 winning numbers, and thus qualifies as having 3 Winning numbers; even if there are no lines with just 3 winning numbers.

We'd prefer the 4 winning numbers line than just 3.

JADELottery wrote: ``The wheel you've generated should have the additional information as follows: [....] 3.968254% Coverage of 126 Combinations``.

Thanks for adding your expert and practical knowledge to this discussion. Your step-by-step description of the behavior of your algorithm in the aforementioned discussion was very clear and helpful to me. (Thanks to Ramijami for pointing me to it.)

Could you explain the derivation of that percentage figure?

That particular wheel covers 25 of the possible 126 4-tuples, which is about 19.84%.

It also covers 46 of the possible 84 3-tuples, which is about 54.76%.

-----

Also, it does not seem like your algorithm makes any effort to derive an "optimal" wheel -- the minimum number of tickets to play.

How would we compute the theoretical minimum wheel for the t<m requirement?

The computation is straight-forward for the t=m requirement like my original (8,6,4,4)=b example. Using Excel nomenclature, it is ROUNDUP(COMBIN(8,4)/COMBIN(6,4),0) = 5.

(Of course, I know we might not -- probably cannot -- achieve that theoretical minimum, as "lotteryarchitect" noted. In fact, for (8,6,4,4)=b, I can demonstrate that b=7 is indeed the minimum (optimum); and there are 280 such 7-ticket wheels.)

Likewise, for (9,5,4,4)=b, it would be ROUNDUP(COMBIN(9,4)/COMBIN(5,4),0) = 26.

Obviously, the theoretical minimum is much smaller for (9,5,3,4)=b. Your algorithm generated b=5 in your example.

But I cannot predict that or a smaller theoretical minimum, if any.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2136 Posts Offline

Posted: February 28, 2013, 12:55 pm - IP Logged

Hello, mathhead, one could also think of a new model of wheels, ie generate Wheels with delta values ??at each position, so instead of choosing numbers At each position delta values ??which can be coordinated vaores of deltas Value of the horizontal (the same bet delta position by position) and also Delta vertical line pair another hit = condition has saddled the delta of horizontal and vertical delta (the latter for the penultimate draw) the pivot is the last draw, this type of wheel has not been created in the world, wants to know how to create algorithm Get well,

West Concord, MN United States Member #21 December 7, 2001 3684 Posts Offline

Posted: February 28, 2013, 5:24 pm - IP Logged

Quote: Originally posted by mathhead on February 28, 2013

JADELottery wrote: ``The wheel you've generated should have the additional information as follows: [....] 3.968254% Coverage of 126 Combinations``.

Thanks for adding your expert and practical knowledge to this discussion. Your step-by-step description of the behavior of your algorithm in the aforementioned discussion was very clear and helpful to me. (Thanks to Ramijami for pointing me to it.)

Could you explain the derivation of that percentage figure?

That particular wheel covers 25 of the possible 126 4-tuples, which is about 19.84%.

It also covers 46 of the possible 84 3-tuples, which is about 54.76%.

-----

Also, it does not seem like your algorithm makes any effort to derive an "optimal" wheel -- the minimum number of tickets to play.

How would we compute the theoretical minimum wheel for the t<m requirement?

The computation is straight-forward for the t=m requirement like my original (8,6,4,4)=b example. Using Excel nomenclature, it is ROUNDUP(COMBIN(8,4)/COMBIN(6,4),0) = 5.

(Of course, I know we might not -- probably cannot -- achieve that theoretical minimum, as "lotteryarchitect" noted. In fact, for (8,6,4,4)=b, I can demonstrate that b=7 is indeed the minimum (optimum); and there are 280 such 7-ticket wheels.)

Likewise, for (9,5,4,4)=b, it would be ROUNDUP(COMBIN(9,4)/COMBIN(5,4),0) = 26.

Obviously, the theoretical minimum is much smaller for (9,5,3,4)=b. Your algorithm generated b=5 in your example.

But I cannot predict that or a smaller theoretical minimum, if any.

Thanks for adding your expert and practical knowledge to this discussion. Your step-by-step description of the behavior of your algorithm in the aforementioned discussion was very clear and helpful to me. (Thanks to Ramijami for pointing me to it.)

Could you explain the derivation of that percentage figure?

That particular wheel covers 25 of the possible 126 4-tuples, which is about 19.84%.

It also covers 46 of the possible 84 3-tuples, which is about 54.76%.

5 wheel combos / 126 total possible combos of 9 numbers taken 5 at a time = 5 / 126 ≈ .03968254 or 3.968254%.

Those other percentages you've derived are likely accurate.

-----

Also, it does not seem like your algorithm makes any effort to derive an "optimal" wheel -- the minimum number of tickets to play.

Actually it does.

However, it's based on a different definition of 'optimal'.

'Quantumly Optimal' is the better description for what the algorithm is doing.

You'll notice that each time you build a new wheel, it usually has a different set of combinations depending on the wheel generator settings.

This is because the algorithm is randomly working toward a minimal set of combinations or combinatoric ground state for those wheel settings.

For larger wheel pool sets, trying to find a minimum combination set gets very time consuming.

We're talking geometrically, exponentially far beyond the simple total set of all possible wheel combinations.

The algorithm works near the combinatoric ground state, giving you a wheel that is very near what you would consider 'optimal'.

How would we compute the theoretical minimum wheel for the t<m requirement?

There might be one out there, but we haven't come across any or thought of deriving any.

We look at this as kind of like data compression.

You want to get as much crammed in to as small a space as possible.

That's why most compression is algorithmic, with sprinkles of mathematical computation to make that batch of goodies come out just right... hmmm, goooddiees... ssshhhuppp...

Anyway, if there is one, it will be a while before it becomes known, or sooner than we think.

This is the Quantum approach to the problem.

The computation is straight-forward for the t=m requirement like my original (8,6,4,4)=b example. Using Excel nomenclature, it is ROUNDUP(COMBIN(8,4)/COMBIN(6,4),0) = 5.

(Of course, I know we might not -- probably cannot -- achieve that theoretical minimum, as "lotteryarchitect" noted. In fact, for (8,6,4,4)=b, I can demonstrate that b=7 is indeed the minimum (optimum); and there are 280 such 7-ticket wheels.)

Likewise, for (9,5,4,4)=b, it would be ROUNDUP(COMBIN(9,4)/COMBIN(5,4),0) = 26.

Obviously, the theoretical minimum is much smaller for (9,5,3,4)=b. Your algorithm generated b=5 in your example.

But I cannot predict that or a smaller theoretical minimum, if any.

If 5 is better than 7, hey, let's go with that.

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Greece Member #2815 November 18, 2003 502 Posts Offline

Posted: March 1, 2013, 4:51 am - IP Logged

How would we compute the theoretical minimum wheel for the t<m requirement?

Check my forums at the "knowledge base" section. I have that equation for the general case posted there. It is the absolute minimum lowest bound however which assumes the wheel is a packed design which means a combination is covered by one and only one block in our wheel.

Dump Water Florida United States Member #380 June 5, 2002 3114 Posts Offline

Posted: March 1, 2013, 5:07 am - IP Logged

Pool=v=the total balls we want to wheel Pick=k=the ticket size (e.g. a 6 ball game has k=6) Match=t=the prize division we want to guarantee a win Hit=m=the condition that has to be met, in order to guarantee the t prize division win; m defines the least number of balls from our v set that must be correct. Tickets=b=the total tickets required to play. Assumes wheel is 100% less show percent= 66%(49,6,3.6)=46 may be 100% for lower tier prizes. Match=t on some software can be set for multiple prizes x3 etc.

New Mexico United States Member #86099 January 29, 2010 11167 Posts Offline

Posted: March 1, 2013, 12:23 pm - IP Logged

Quote: Originally posted by mathhead on February 28, 2013

That's an ironic comment, considering the "systems" that you suggest in other discussions (e.g, click here).

Humor aside, I want to reiterate that I am not interested in using wheels. I am interested in understanding the algorithms for generating wheels. It is an academic exercise for my own edification.

So comments like yours are neither helpful nor wanted in this discussion. In fact, they are never constructive, IMHO.

It was not meant to funny, but some people would take it as such. It is very helpful in the fact that the big games can be won without wheels. Wheels are fine but an element of luck and timing are part of the winning equation. So, endless babbling about balanced wheels, and which wheel is bestserves no purpose in the pursuit of winning.

Dump Water Florida United States Member #380 June 5, 2002 3114 Posts Offline

Posted: March 1, 2013, 2:07 pm - IP Logged

Quote: Originally posted by lakerben on March 1, 2013

It was not meant to funny, but some people would take it as such. It is very helpful in the fact that the big games can be won without wheels. Wheels are fine but an element of luck and timing are part of the winning equation. So, endless babbling about balanced wheels, and which wheel is bestserves no purpose in the pursuit of winning.

In general I agree. I am always amazed when people talk about testing wheels against a draw or history to see which would be the better wheel for future draws.

However, there are better wheels to use, those with multiple prize tier guarantees or better percentage of coverage for the prize desired. Some wheels just plain blow others away.

Like this one with an 88% (12,6,5,6)=22 where if you have all six winning numbers among the 12 and don't win the 5# prize you are guaranteed to win 9 4# prizes plus 4 3# prizes instead.

Condition - If 4 Drawn Numbers are in the Set of 9 Numbers, - Then at least 1 Combination has 3 Winning Numbers. ____________________________________________ 05 06 07 08 09 01 03 04 06 09 02 03 04 08 09 01 02 06 07 08 01 02 03 04 05 ____________________________________________

When reading the condition you should understand the second part '- Then at least 1 Combination has 3 Winning Numbers.' to mean that a line will have At Least that many or more winning numbers.

This means if the condition is '3 Winning Numbers', then there could be a winning line with 3 or 4 winning numbers, because the other part of the condition is '4 Drawn Numbers are in the Set of 9 Numbers,".

Also, these mean the following:

Set - The wheel pool of numbers you are playing; typically less than the lottery pool of numbers being drawn. Pick - The number of balls being drawn for that lottery pool of numbers. Match - The quantity of drawn numbers that match your wheel pool numbers. Win - Guaranteed to have at least 1 line with that many matching numbers in the wheeled combination or more.

If there is a line with 4 winning numbers, it definitely has any combination of 3 winning numbers, and thus qualifies as having 3 Winning numbers; even if there are no lines with just 3 winning numbers.

We'd prefer the 4 winning numbers line than just 3.

I am having a look at this wheel for keno. So for 5 euro costs I get with 4/9:

Payout €

P()

2

0,26

4

0,42

5

0,10

6

0,12

7

0,08

9

0,02

I have to play by two tickets, so I must play 6 euro. The expectation for the payout is 4.16 € with 4/9.

For 5/9: E(payout)=14.68

Payout €

P()

6

0,09

7

0,10

8

0,20

9

0,33

10

0,03

11

0,14

12

0,06

14

0,02

15

0,01

152

0,02

154

0,02

That is guaranteed break even, on the other side 5/9 as pick 9 pays 2 for 1.