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Odds of lottery

  1. Feb 23, 2005 #1
    if there is lottery where there are six numbers of 49 drawn and each number is discarded one drawn what are the total combinations.
    I think it would be 49 choose 6 but that only gives 14 million and I figured it would be higher. can someone verify this?
    thanks :smile:
     
  2. jcsd
  3. Feb 23, 2005 #2

    mathman

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    I get 13,983,816. It looks like you are right.
     
  4. Feb 23, 2005 #3

    chroot

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    No, it's not.

    It's 49 * 48 * 47 * 46 * 45 * 44 = 10,068,347,520 = 10 billion.

    - Warren
     
  5. Feb 23, 2005 #4

    jcsd

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    The order doesn't matter, so mathman is correct (it's the UK National Lottery I guess).
     
    Last edited: Feb 23, 2005
  6. Feb 23, 2005 #5

    chroot

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    Ah, okay. Silly me.

    - Warren
     
  7. Feb 23, 2005 #6
    Thats [tex] \left(\begin{array}{cc}49\\6\end{array}\right) [/tex], right?
     
  8. Feb 23, 2005 #7

    jcsd

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    Yes, that's correct
     
  9. Feb 23, 2005 #8
    ok thanks. its a candian lottery Jcsd.
    I've got another question not really the same but has to do with choosing so i'll put it here. It was a question I had on a quiz today. Hopefully I got it right
    There is a committee to consist of 5 people if there are 15 students and 18 teachers to choose from and the committee needs atleast one student and one teacher how many different committees can be formed.
    I believe i did 15c1 * 18c1 * 31c3
    is this right?
     
  10. Feb 28, 2005 #9

    shmoe

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    You're overcounting. If the students are labelled 1, 2, 3,... and the teachers A,B,C,D,.. then the committee 1,A,2,3,4, corresponding to selecting 1 and A as your guaranteed teacher & student and 2, 3, 4 as the 3 chosen from the 31 people left, is counted again as 2,A,1,3,4, where 2 and A are your guaranteed student and teacher and 1,3,4 are from the 31 remaining. (note: the order I've used for the committee is supposed to reflect the reasoning behind your answer)

    You could try to correct this overcounting or you could take a different approach. You have 4 acceptable possibilities to make #of students+#of teachers=5, namely 1+4, 2+3, 3+2, and 4+1. Find the number of committees in each case and add.
     
  11. Feb 28, 2005 #10

    Alkatran

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    Has anyone every calculated the average amount won per lottery ticket? (Including all those little sub prizes). I'd estimate it at max a dollar per ticket, min 10 cents per ticket, probably from 20 to 70.
     
  12. Mar 1, 2005 #11

    jcsd

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    It's actyally a nice little maths project for someone to do at school, I guess that like the UK lottery the expected payout is actually dependent on how many people play the lottery and things like 'roll-overs'.
     
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