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    Probability to win




    Hello,

    I would like to calculate the probability to win 3 times in a lottery considering the following:
    1 - the chance to win 1 time is 1 in 3.268.760
    2 - it is not consecutive, so there is for exemple 1 year between one prize and another.

    Could someone help me?

    Thank you for attention.

    Regards,

    Marcus

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    Re: Probability to win

    I am not clear which probability you want.

    Scenario A: The lottery runs once a week. A person plays one week and wins. S/he does not play for a few weeks, then plays again and wins again. Once more, s/he does not play for a few weeks, then plays again and wins again. So the person wins three times and plays only three times.

    Scenario B: The lottery runs once a week. A person plays every week. After winning the first time, the person continues to play each week, and wins again after not winning several times, and then winning for the third time after again not winning several times. So the person wins three times but plays more than three times.

    Which of these scenarios do you mean? Or do you mean something else altogether?

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    Re: Probability to win

    Hello Con-Tester, thanks for your reply. The lottery runs three times in a week and I was thinking in scenario B, a person plays every week on the three games.

    Regards,

    Marcus

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    Re: Probability to win

    Okay, I’m still confused which scenario you actually mean because you seem to be saying that the person plays three times and wins three times, which is scenario A, not B.

    There appears to be a language problem.

    So tell me this: How many times does the person play to win three times? Does s/he play three times only, or more than three times?

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    Re: Probability to win

    Sorry about the confusion, the person played 1.212 times.

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    Re: Probability to win

    Okay, so its a binomial problem with p = 1/3 268 760, n = 1 212 and k = 3. Therefore, the answer is

    \Pr \left( X = 3 \right) = {{1\ 212}\choose 3}\cdot{ \left( \frac{1}{3\ 268\ 760} \right) }^3\cdot{ \left( 1-\frac{1}{3\ 268\ 760} \right) }^{1\ 212-3}

    = {{1\ 212}\choose 3}\cdot{ \left( \frac{1}{3\ 268\ 760} \right) }^3\cdot{ \left( \frac{3\ 268\ 759}{3\ 268\ 760} \right) }^{1\ 209}

    The answer is very small at ≈ 8.472e-12.

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    Re: Probability to win

    Thank you very much.

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    Re: Probability to win

    Hello Con-Tester, Im trying the following formula in Excel and Im getting another result, could you please tell me if Im doing something wrong?

    =(1212/3)*(1/3268760)^3*(1-1/3268760)^(1212-3) -> 1,156E-17

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    Re: Probability to win

    Er, the first expression, “{{1\ 212}\choose 3}”, is the mathematical notation for “1 212 choose 3,” as in “combinations and permutations” (See here).

    That is, the expression quantifies how many different combinations of 3 can be made from a total of 1 212 options. The Excel function “COMBIN()” must be used for this, and so the Excel expression you want to enter is this:
    =COMBIN(1212,3)*(1/3268760)^3*(1-1/3268760)^(1212-3)

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    Re: Probability to win

    Thank you again. This solve my problem.

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    Re: Probability to win

    Quote Originally Posted by Con-Tester View Post
    =COMBIN(1212,3)*(1/3268760)^3*(1-1/3268760)^(1212-3)
    If you're using binomial pdf values then binom.dist would be easier to use if you need to stay in excel

    Code: 
    =BINOM.DIST(3, 1212, 1/3268760, FALSE)
    I don't have emotions and sometimes that makes me very sad.

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    Re: Probability to win


    Sure, but doing so would hide the mathematical principles and thereby hobble understanding, even while the answers are the same. Thanks for pointing it out, though.

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