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Topic closed. 3 replies. Last post 7 years ago by Ramijami.

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Ramijami's avatar - wired shell.jpg
Cape Town
South Africa
Member #66343
October 23, 2008
115 Posts
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Posted: April 16, 2010, 5:02 am - IP Logged

Hi All,

In BobP's book he gives the odds of having the winning numbers amongst a reduced field. So for example in a 6/49 game the odds of having all the numbers in a reduced field of 18 is 1 in 753.28 combinations, for 21 it is 257.70 and for 24 it is 103.89. Does anybody have these combinations, is there a website where I can get it, or is there any software that can generate these? I've tried to use LottoDesignerXL but it doesnt' generate 24 number combinations, and with the lower numbers it gives an error message when I try and generate more than 200 sets.

Any help would be much appreciated :-)

Ramijami

    Winlotta's avatar - Lottery-001.jpg

    United States
    Member #60014
    April 3, 2008
    735 Posts
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    Posted: April 16, 2010, 7:24 am - IP Logged

    Send BobP a personal message and ask him.

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      NY
      United States
      Member #23835
      October 16, 2005
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      Posted: April 16, 2010, 6:39 pm - IP Logged

      He's only describing odds, which in the case of 18 numbers is 1 in 753.28, period.  There are no "combinations" to have.

      For a 6/49 game there are 13,983,816 possible combinations to choose from using all 49 numbers. If you "pre-choose" 18 numbers and make all of your choices from those 18 numbers there will be 18,564 combinations you can  choose from.  Those 18,564 combinations are 1/753.28 of the 13,983,816 combinations  possible with all 49 numbers. The chances that all 6 winnning numbers will be among the 18 you started with are 1 in 753.28.

      If that's not clear, imagine a simpler game, where you just pick 1 number from a field of 100. Since there are 100 numbers your chances of picking the winning number are 1 in 100. Now suppose instead of choosing your number from all 100 you decide to choose from a group of 5. Those 5 numbers cover 5 of the 100 possible numbers, or 1 in 20.

      I'm not sure why anyone bothers to point out those particular odds, other than as an exercise in math. Your question presumably comes from the concept of "wheeling" where you choose all of the combinations you play from a subset of the whole field of numbers. Unless the numbers in that subset are more (or less) likely to be drawn than the other numbers you've done absolutely nothing to change your odds of winning. The 1 in 753.28 chance that all 6 winning numbers will be among your 18 times the 1 in 18,564 chance that any one combination of those 18 numbers would be the winning combination works out to the same chance of winning as using all 49 numbers: 1 in 753.28 * 1 in18,564 = 1 in13,983,816. If the 18 numbers are more (or less) likely to be drawn as winning numbers then the true chance of all 6 winning numbers being in those 18 would be more "or less" than 1 in 753.28.

        Ramijami's avatar - wired shell.jpg
        Cape Town
        South Africa
        Member #66343
        October 23, 2008
        115 Posts
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        Posted: April 22, 2010, 5:22 am - IP Logged

        Thanks winlotta and KY Floyd for responding.

        KY Floyd, I am familiar with the odds as BobP is pretty clear that those are just odds and that ultimately it all comes down to the 13 million odd combinations. I am looking to play around with the sets of numbers to see if and how I can select the set (of for eg. 24 numbers) containing the 6 winning numbers. The way I figure is that it should be easier to select 1 from 103+ (in the case of 24 number sets/combinations) than it is to select 1 from 13+ million............not sure if anybody on the forum has approached it like this before.