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The Poisson distribution

  1. Mar 31, 2003 #1
    If you have a lottery (Megamillions) and you sell 20,000,000 tickets, the probability of them all losing is given by:
    (135,145,919/135,145,920)^20,000,000 = 0.862448363

    A close approximation is given by:

    e^-(20,000,000/135,145,920) = 0.8624413

    I just learned this from a book. That's pretty cool!
  2. jcsd
  3. Mar 31, 2003 #2


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    I suggest that you go back and read that book again!

    Did it say anything about where it got the figures
    "135,145,919" and "135,145,920". For that matter, did it say anything about the legality of a lottery in which there is an 86% chance that EVERYONE loses?
  4. Mar 31, 2003 #3


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    I bet people would still play... ;)

    - Warren
  5. Mar 31, 2003 #4
    This is because ln(1+x) ~ x for small x.
    => ln[(k/(k+1))^n] = n ln [k/(k+1)] = n ln [1 - 1/(k+1)]
    ~ -n/(k+1)
    => (k/(k+1))^n ~ e^[-n/(k+1)].

    However I'm puzzled because it doesn't say how many tickets there are in total. Don't we have to know this to figure out the 135,145,919?
    Last edited: Mar 31, 2003
  6. Mar 31, 2003 #5


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    Staff: Mentor

    Ever hear of PowerBall?
  7. Mar 31, 2003 #6
    At face value this doesn't seem like a good approximation for some similar applications, like 1/(1-ln(1+x)) where x=1.
  8. Mar 31, 2003 #7
    Loren Booda, by 'small x' I meant |x| << 1. So it's not valid for x=1, of course.
  9. Mar 31, 2003 #8
    I assume there are 135,145,920 possible numbers/choices, with one chosen as the winner.

    The rules say there is a power-number 1-52, and 5 distinct numbers 1-52 which can be in any order. So that's 52 * 52! / (5!*47!) , which is right.
  10. Mar 31, 2003 #9
    mea culpa, arcnets
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