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A Different type of Wheel, Call it: Groups wheel or Sets wheel...

Topic closed. 3 replies. Last post 6 years ago by dr san.

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Honduras
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August 29, 2005
4715 Posts
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Posted: February 23, 2011, 11:55 pm - IP Logged

lets use Pick6, let say you want to wheel, 30 numbers for a 5/6....It might costs you like, $4,000 i dont know i am just saying....

let´s say the first 30 numbers in the Pick6/49 were going to play...

you use 2 wheels 1 from 1-15 pick3 abbreviated wheel the other from: 16-30 abbreviated wheel...

you could get a: 2/3 abbreviated wheel from: 1-15 and a 2/3 or 3/3 abbreviated wheel from: 16-30...

lets say the, 1-15 came out to: 20 tickets and the 16-30 came out to, 30 tickets...YOU THEN MULTIPLY THE 20 TICKETS PAIR THEM UP WITH THE 30 TICKETS IT SHOULD BE: 20 X 30 = $600 less than the: $4,000 you was going to pay for the same....

The Forex trades: 1.6 Trillion dollars EVERY day, that´s more than the GDP of the Carribbean Central America, COMBINED. Enough to feed every crook out there for centuries...To all Geniuses & Powers Countries of the World the Planet needs breakthroughs in all Medicine, Veterinary, Biology related fields, Psychology, Population Psychology/Sociology..They need to genetically ingeneer new plants species/types to give more variety of plants and thus have more resources for combating diseases¨


 


 


 

 


    Boney526's avatar - NjlpLogo
    New Jersey
    United States
    Member #99032
    October 18, 2010
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    Posted: February 24, 2011, 12:37 am - IP Logged

    I'm sorry you're going to have to explain this more clearly if you want anybody to understand it.  There's literally no possible way to have the same wheel that would regularly cost 4000 dollars for 600 dollars.

    But if I understand what you are sayingis that you'd split the field in half, and then play 2 seperate  wheels, one 2 if 3 wheel and one 3 if 3 with it,  I'm not a big fan of splittin the field of numbers you chose in half, but if you meant you were just playing two seperate wheels, the I guess it's all the same.  Because then you're just playing 2 seperate wheels.  It wouldn't necessarily achieve the same results as a 5/6 wheel - but those wheels are expensive in general, so I wouldn't consider playing any of them.

      BobP's avatar - bobp avatar.png
      Dump Water Florida
      United States
      Member #380
      June 5, 2002
      3114 Posts
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      Posted: February 24, 2011, 4:31 am - IP Logged

      lets use Pick6, let say you want to wheel, 30 numbers for a 5/6....It might costs you like, $4,000 i dont know i am just saying....

      let´s say the first 30 numbers in the Pick6/49 were going to play...

      you use 2 wheels 1 from 1-15 pick3 abbreviated wheel the other from: 16-30 abbreviated wheel...

      you could get a: 2/3 abbreviated wheel from: 1-15 and a 2/3 or 3/3 abbreviated wheel from: 16-30...

      lets say the, 1-15 came out to: 20 tickets and the 16-30 came out to, 30 tickets...YOU THEN MULTIPLY THE 20 TICKETS PAIR THEM UP WITH THE 30 TICKETS IT SHOULD BE: 20 X 30 = $600 less than the: $4,000 you was going to pay for the same....

      Let's make it a little easier to discuss by working with 20 numbers for that 5/6 guarantee.  That wheel would require 1,154 lines.  (20,6,5,6)=1154

      Divided into two groups of 10

      For the 3/3 that takes (10,3,3,3)=120

      For the 2/3 that takes (10,3,2,3)=8

      To make our Pick-6 combinations we'd have to pair up each of the 120 lines with each of the 8 for a total of 8 x 120 = 960 which is darn close to our 1154 lines, but not exactly and I'll tell you why. 

      Our new wheel is not a 100% (20,6,5,6)=960, it is conditional on 3 of the winning numbers falling on each half, no more, no less on a side.  In short, the savings is a result of the greater risk, hence no real savings at all.

      Good idea though.  BobP

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        bgonçalves
        Brasil
        Member #92564
        June 9, 2010
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        Posted: February 24, 2011, 11:32 am - IP Logged

        pumpi 76=

        Note that  C(50,5,3,5)=C(26,5,3,3)+C(24,5,3,3)
        because if you divide 50 numbers into 2 groups, then at least 3 of 5
        numbers must be either in one or in another group.  You can find
        C(26,5,3,3) =260 and C(24,5,3,3)<=231  Taking 2*C(25,5,3,3)<=2*256=512 would be bigger, because 26 has a nice
        symmetry.

        So, obviously 17296 is far too much. But let’s take a look at your
        idea to take one number from [2..18], one from [6..36] and one from
        [34..49]. This gives you 17*31*16=8432 VARIATIONS. The number of
        COMBINATIONS is of course not 6 times smaller, because the three
        intervals don’t coincide. But since at least two and two partly
        overlap, it should be somewhere between this number and half less. Rough
        guess about 6000. Even if you avoid overlapping, say [2..17],[18..33]
        and [34..49], you still have 16*16*16 = 4096 games, which is over 8
        times more expensive than 491 above.
         
        Even so, it’s not exactly 100% guarantee. But with such redundancy,
        it obviously can’t be much less either. You are right about that. But
        the same is true for any random choice of 4096 or so games. Ordering
        makes no difference here, all balls are equal.

        But even with only 491, I doubt you can make a profit. I’m sure that
        the prize for 3 in 5 is small enough that the Lottery, not you, makes
        the profit. On average of course, not when you are lucky.