Dumb question?

[janjan]

Hi,
I have a simple question(or dumb question?)
Let's take the wheel 106464
1 2 3 4 5 6
1 2 7 8 9 10
3 4 5 6 7 8
With 10 numbers you have 210 cobin with 4 numbers COMBIN(10,4)=210, but 3XCOMBIN(6.4)=3x15=45
So, how this 45 combinations cover all 210 combinations and please can somebody tell me how wheels

are build?

Thanks.

 

[Brad]

Let's see if I got it right this time

What you posted is a (10,6,4,6)=3 wheel which will guarantee a 4 hit minimum if you correctly match any 6 from a pool of 10 numbers.

You’re correct, Janjan, there are a total of 210 possible 4 number combinations from 10, and that the wheel’s 3 lines only cover 45 of those.

But you don't need to cover all the 210 combos of 4. All you need to do, to satisfy the minimum guarantee, is to cover only one 4 no. combo from each possible 6 number line that can be made from a 10 number pool.

So theoretically, since there are 42 six (6) number lines, your wheel must cover 42 different four (4) number combos for a full guarantee.

Try this - write out all the 45 combos from the wheel you presented, ie:

1.2.3.4
1.2.3.5
1.2.3.6
...
...
7.8.9.10

and you'll see that some (3?) of them repeat, probably leaving you with 42 unique combinations from the 45 total (I'm guessing a bit here cuz I have not actually written all out).

Note also there are overlaps (repeats), ie: for lines 1.2.3.4.5.6 and 1.2.3.4.5.7 there are five (5) 4 number combos that are the same, but only one is needed to satisfy the guarantee for both 6 number lines ... the same could be the case for other lines, so there maybe more repeated combinations in the wheel.

It's not difficult to write all the 4 number combos from your wheel alongside the possible 6 number lines (from 10) and then visually confirm the matches.

Doing that will go a long way to show you how wheels are made ... or use CoverMaster or LottoDesigner (or other programs) to build them.

If I read your question correctly this should help a bit, others have provided more input below. Some believe in playing with wheels and some don’t.


Cheers

PS I edited this post cuz the way I presented it initially did not seem clear (not because I thought your Q was dumb).

 

[Dennis Bassboss]

Hi,
I have a simple question(or dumb question?)
Let's take the wheel 106464
1 2 3 4 5 6
1 2 7 8 9 10
3 4 5 6 7 8
With 10 numbers you have 210 cobin with 4 numbers COMBIN(10,4)=210, but 3XCOMBIN(6.4)=3x15=45
So, how this 45 combinations cover all 210 combinations and please can somebody tell me how wheels

are build?

Thanks.

Garantee and number of appearances in the wheel is the bottom line here....

 

[CMF]

Re: Primer question?

Janjan

We all started off knowing nothing. A willingness to learn should never be deprecated.

A wheel is simply the desired enumerated combinations from a reduced set of integers. Various dubious filters are used to arrive at the set of integers and subsequently to reduce the number of combinations.

A Cover is where say in a 649 Lotto the sixes in your Cover will guarantee a low level prize provided you get a certain number of hits from the winning six in the pool of numbers you have enumerated your combinations from. For example, if you wheeled 8 integers to give 28 combinations and you were using a cover of four sixes which guaranteed a three if you got three right in your 8 then at least one of the threes will be there (not all of them).

No one yet has produced a cover which when tested by me gives any significant improvement over using random selections. Very few people join my website but as someone learning the ins and outs of Lotto numbers you should find it useful to get to know the jargon and to get an opposing point of view. Most of the Lotto sites follow a toady line with wheels and covers apart from mine, Ion Saliu's and one or two others.

Regards
Colin

 

[hot4]

Hi,
I have a simple question(or dumb question?)
Let's take the wheel 106464
1 2 3 4 5 6
1 2 7 8 9 10
3 4 5 6 7 8
With 10 numbers you have 210 cobin with 4 numbers COMBIN(10,4)=210, but 3XCOMBIN(6.4)=3x15=45
So, how this 45 combinations cover all 210 combinations and please can somebody tell me how wheels

are build?

Thanks.

Those 3 lines cover all 210 lines *if* 6 numbers are drawn from them,e.g. 1,3,5,7,8,9.

If you want to check by hand you'll see that this line covers 115 lines with four of these 6 numbers.

What you are asking it's not possible with 3 lines; you are asking for how many lines give 4 numbers if 4 numbers of the 10 are drawn, isn't it?

To cover every tetras is different from covering 4numbers if 6 are drawn.

 

[gsobier]

...for 6/49, all you need is 2 DNs to hit and the right quad... ...finding the correct quads to go with the DNs is required if you are to do well... ...any random quads are NOT going to help you...

 

[Dennis Bassboss]

...for 6/49, all you need is 2 DNs to hit and the right quad... ...finding the correct quad to go with the DNs is also part of the key...

 

[gsobier]

...Dennis and and a few others here know what we are talking about...

 

[tomtom]

...for 6/49, all you need is 2 DNs to hit and the right quad... ...finding the correct quads to go with the DNs is required if you are to do well... ...any random quads are NOT going to help you...

2DNs...the missing point here is a hint how to get near to that correct quad..

 

[gsobier]

...here is your hint, look in the history... ...quads which repeat are in the target zone... ...its a hit and miss process...

2DNs...the missing point here is a hint how to get near to that correct quad..

 

[tomtom]

...here is your hint, look in the history... ...quads which repeat are in the target zone... ...its a hit and miss process...

Well, can't get the hint...

Tickets are very expensive these days..and in my opinion any high prices shouldn't be a part of a game that pretends to be both fair and interesting. Btw, from that reason presently I'm not interested in that 649 game at all...however, for those lottery fans who deserve a good win from any reason here is a possible beginning procedure in looking for those 2 DNs, that might help in saving some money : check in history for two repeats out of 6 ... that means once in (X?) draws you might have the first 2 numbers in one of 15 lines containing 2 numbers only...

If combination you expected to have 2 repeats looked like 1,2,3,4,5,6
the 15 lines containing 2 possible DNs in one of lines would look
1,2
1,3
1,4
1,5
1,6
2,3
2,4
2,5
2,6
3,4
3,5
3,6
4,5
4,6
5,6


This might be quite Ok in a case of just one or even 3 repeats also....with this strategy it seems there are still some chances for 5/6 or 5+B/6 in about 40-45 % of draws

 

[janjan]

Thanks all for replys,hot4 now i understand.
But please tell me how lower bounds are calculated (Quantity of Combinations required for 100% cover)
and why these are rarely reached.

 

[hot4]

Thanks all for replys,hot4 now i understand.
But please tell me how lower bounds are calculated (Quantity of Combinations required for 100% cover)
and why these are rarely reached.

Why they usually are not reached is the rule!! You calculate the minimum theoretically cover and at last you get the minimum you or someone else (or a software) can get.
Exception are wheels like 22_6_3_3=77 or 26_6_3_3= 130; these two wheels are posted somewhere and equals the theoric minimum!

To calculate the minimum amount of lines that wins 3 hits *IF* 3of22 hits, you can make: (22*21*20)/(6*5*4)=77

To calculate 22,6,4IF4 just: (22*21*20*19)/(6*5*4*3).

So, you calculate how many combinations of 3 (or 4) there are in 22 numbers and you divide by the amount of 3 (triplets) (or 4) there are in 6 numbers.

To calculate how many lines are covered by one line (1,2,3,4,5,6) regarding 3 numbers (fifth prize) you have to know how many combinations of 3 numbers there are in 6 numbers (6*5*4/3*2*1= 20) and multiply by the combinations of 3 numbers existing in the remaining 43 numbers (43*42*41/3*2*1=12341)

20*12,341 = 246,820 lines covered (3if6)

Now you divide the total possible lines by this number and you know the minimum possible lines to get a 3if6 of 49 numbers: 13,983,816 / 246,820 = 56,65 lines. [you may know that till now, the smaller wheel that *guarantees* a 3if6 in every possible draw has 163 lines, which is +-3* bigger than 57!; don't worry because if you play 57 well balanced lines every draw, you may expect a media of 3-hits per draw, while not every draw].

Hoping this helps. Feel free to ask if my explanation didn't fullfill your aims. I'll try again

 

[BushHappy]

hot4,

I am trying to use your formulas for a 22,6,4IF6 wheel, but I cannot get it right. Please can you help me?

Cheers,
BushHappy

 

[hot4]

hot4,

I am trying to use your formulas for a 22,6,4IF6 wheel, but I cannot get it right. Please can you help me?

Cheers,
BushHappy

Formulas are not mine

let's see: C(22,6)/[C(6,4)*C(16,2)]

OR

C((TOTAL),6)/[C(6,4)*C((TOTAL-6),2)]=>

=> 74,613/((15)*(16*15/2))= +- 41,5 [parenthesis are used, only trying to show factors correspondence