in reply to Re: Picking (potentially) winning lottery numbers
in thread Picking (potentially) winning lottery numbers

g r i n d e r ,

Thanks. All of what you say is largely true, although one thing that was picking at the back of my brain was what if the balls do have some history. For instance, what if the paint used to put numbers on the balls resulted in some balls being (slightly) heavier than others? Presumably then, with enough tests, it would become apparent which balls were more likely to come out. (I don't know how they are actually picked from the machine but presumibly gravity is involved somehow.) Anyway, I don't really buy that arguement, and I think there would have to be too many trials to determine if it were true.

But it is amuzing that you could make equally rational sounding aurguments for picking most common and least common balls; that's why I display both :-)

Scott

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Re (tilly) 3: Picking (potentially) winning lottery numbers
by tilly (Archbishop) on Jul 25, 2001 at 00:00 UTC
    You can make a better argument for not putting money into the lottery.

    But that said, if you flip a US penny, there is a slight bias towards coming up heads. Not enough of one in practice to be useful though. (It is easiest to see if you balance the pennies on end and then whack the table.)

[reply]
Re: Re: Re: Picking (potentially) winning lottery numbers
by RatArsed (Monk) on Jul 24, 2001 at 19:51 UTC
    In the UK, at least, there are multiple sets of balls, and multiple machines to pick them and balls are replaced frequently. This all stops trends developing, and making the results as random as ever.

    I'd be interested in it being able to calculate my chances of winning anything, and doing some manipulation with that (like what if I choose two sets of numbers for a week.)

    --
    RatArsed

[reply]
      RatArsed,

      Calculating the chances of winning is fairly trivial. Assume the drawing has 50 balls from which 6 balls are drawn without replacement. The chance of picking:

      • one right is 6/50
      • the second one right is 5/49
      • third 4/48
      • forth 3/47
      • fifth 2/46
      • sixth 1/45
      To get the chances of picking them all right, multiply all those probablities together: 6.29299e-8, or about 1 in 15.9 million.

      Scott

[reply]
        That's the simple case for matching all six balls

        Now add other prizes (UK: match 3, match 4, match 5, match 5 + bonus ball, match 6)

        And work it out for multiple lines. The data is dependent on the other line; for example if I have "1, 2, 3, 4, 5, 6" as one choice, and "2, 3, 4, 5, 6, 7" as my other choice, I haven't greatly increased my overall chance of winning something, although my chance of the jackpot has doubled. Had the second line not included any of the numbers from the first line, I would have doubled my chance.

        The answer involves a lot of permutations and combinations work (nPr and nCr), which in turn is a lot of dealing with factorials.

        the bit I don't know how to do is to discount combinations that appear using both sets (2,3,4 is valid, and duplicated, but 1,3,7 is not valid)

        --
        RatArsed

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