Announcement

Collapse
No announcement yet.

Any Good Mathematicians Here???

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Any Good Mathematicians Here???

    Im working on a lottery system for my board... Just for its birthday :-)

    We are having 5 prizes...

    I need someone to work out how many balls out of a selecion of ??? balls would be needed to have a good probability that this will be the likely outcome... :-

    1st Prize) 1 winner
    2nd Prize) 3 winners
    3rd Prize) 8 winners
    4th prize) 10 winners
    5th prize) 17 winners

    The number of winners can be slightly changed... We have 1104 members and there are usually 20 members online 30 max but I think there will be alot more on board for the boards birthday

    He he I was never any good at maths

    Cloud

  • #2
    How many do you expect to enter?

    Comment


    • #3
      Originally posted by okrogius
      How many do you expect to enter?
      I dont know... maybe about 300 -400 ... But im not sure

      Comment


      • #4
        How do you determine who wins first prize, second prize, etc.?
        http://www.arsegaming.com/dice/index.php/stuff.jpg

        Comment


        • #5
          umm who ever choses the correct balls or something like that :-)

          Comment


          • #6
            Here's the formula and you can figure it out yourself

            C(n,k) = n! / (k! * (n-k)!)

            where:
            n is the number of total balls you're picking from
            k is the number of balls you actually choose

            For example, to determine the odds of picking all 6 numbers correctly in a 6 number drawing of 50 balls:

            C(n,k) => C(50,6) = (50!) / (6! * (50-6)!) = 15,890,700 (1 in 15,890,700 that is)

            To determine if you got a subset of the 6 numbers correctly from the 50 total balls, it's a little more complicated:

            C(n,k) / [ C(k,m) * C(n-k,k-m) ]

            where:
            n = number of balls (50 in this case)
            k = number of balls to choose (6 in this case)
            m = number of balls correctly picked (1,2,3,4,5, or 6)

            For example, to determine the odds of picking 5 of the 6 numbers correctly in a 6 number drawing of 50 balls:

            C(50,6) / [ C(6,5) * C(50-6,6-5) ] = 15,890,700 / [ 6 * 44] = 60,192 (1 in 60,192)

            Yes I'm bored.

            Comment


            • #7
              Ha ha ha...
              Thats a nice formula... Yes I was looking for something like that cheers *Takes out his dusty steam driven calculator* only joking it can hadle factorials



              Cloud

              Comment


              • #8
                Hey treszoks, wanna write my discreet math exam? Ive been doig Perms and Combs all day....

                Comment


                • #9
                  Originally posted by 0ptima
                  Hey treszoks, wanna write my discreet math exam? Ive been doig Perms and Combs all day....
                  As tempting as that sounds, no

                  Comment

                  widgetinstance 262 (Related Topics) skipped due to lack of content & hide_module_if_empty option.
                  Working...
                  X