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Lottery probability

  1. Apr 21, 2012 #1
    thanks for helping :)
     
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 21, 2012 #2
    Think about this as follows: how many ways are there to choose 2 numbers from 18? (no replacement, order doesn't matter)
     
  4. Apr 21, 2012 #3
    i was thinking about this:

    numerator: 18C2, 18C2, 19C2 --- but i'm not sure if you ADD or MULTIPLY these
    denominator: 55C6
     
  5. Apr 21, 2012 #4
    [tex]
    \frac{C^{2}_{18} \, C^{2}_{18} \, C^{2}_{19}}{C^{6}_{55}}
    [/tex]

    Wolfram Alpha

    This looks terribly high?!
     
  6. Apr 21, 2012 #5
    i was thinking of that.
    if you add it on the other hand, its too low

    i'm not quite sure with my solution
    :(
     
    Last edited: Apr 21, 2012
  7. Apr 21, 2012 #6
    No, but this not a probability ot win! This is just a fraction of the total number of outcomes that satisfy your condition. It doesn't mean that if you fill A ticket satisfying this condition that you have this probability to win.
     
  8. Apr 21, 2012 #7
    so you think the solution is right? :)
     
  9. Apr 21, 2012 #8
    Yes. The solution is correct.
     
  10. Apr 21, 2012 #9
    thanks! :)
     
  11. Apr 21, 2012 #10
    The intuition for why you have to multiply the combinations is that for every possible combination, say, in [1,18] you can have 18C2 of combinations in [19,36] and 19C2 in [37,55].
     
  12. Apr 21, 2012 #11
    now i understand. thanks! :)
     
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