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Math Help - About lottery

  1. #1
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    About lottery

    Hey, I was just wondering if there is way to tell how many times person needs to do lottery in order to have 50% chance to actually win the big prize. We use 39 numbers and must pick 7 of them. Chances to win are 1/15380397 and I calculated that if you try to do it for 15380397 weeks always using 1 row at time, chances are

    1 - (15380396/15380397)^15380937 = 0,63

    So how many weeks is reguired to have almost exact 0,5 chance to win? I think I might have asked my math teacher about this problem, but she said there is no exact way to calculate that with a pen and paper.
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  2. #2
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    e^(i*pi)'s Avatar
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    From what I understand of the question the chance will always be {49 \choose 7} as the events are independent.
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  3. #3
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    Yeah but I'm talking about chance to win. For example, if you roll dice for 2 times you have 30& chance to get at least 1 "6" even chance is always 1/6. In this case, I'm asking how many lottery tries is reguired to have 50% chance to win at least once.
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  4. #4
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    Quote Originally Posted by Crookshanks View Post
    Hey, I was just wondering if there is way to tell how many times person needs to do lottery in order to have 50% chance to actually win the big prize. We use 39 numbers and must pick 7 of them. Chances to win are 1/15380397 and I calculated that if you try to do it for 15380397 weeks always using 1 row at time, chances are

    1 - (15380396/15380397)^15380937 = 0,63

    So how many weeks is reguired to have almost exact 0,5 chance to win? I think I might have asked my math teacher about this problem, but she said there is no exact way to calculate that with a pen and paper.
    Let X be the random variable 'number of times you win'.

    X ~ Binomial(n = ?, p = 1/15,380,397).

    You require the smallest integer value of n such that \Pr(X \geq 1) \geq 0.5 \Rightarrow \Pr(X = 0) \leq 0.5. Which means finding the smallest integer value of n that solves:

    \left( 1 - \frac{1}{15,380,397} \right)^n \leq 0.5.

    This can be done by trial and error using a simple scientific calculator. Alternatively an an exact algebraic solution to \left(1 - \frac{1}{15,380,397} \right)^n = 0.5 can be easily found and then appropriate rounding done to get the required answer.

    It should not surprise anyone to find that the value of n is between 10 million and 11 million.
    Last edited by mr fantastic; July 16th 2010 at 03:25 PM.
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