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Math Help - Statistics - Expected win/loss for lottery question

  1. #1
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    Statistics - Expected win/loss for lottery question

    Hello! I need help on this question. I've been thinking about it for ages and I have absolutely no idea where to start.

    The local lottery sets up a game wherein a player has a chance of collecting $750. The player must choose a 3-digit number and if it matches, the player receives $750.
    It costs $1 to play the game.
    What is the expected profit (or loss) for each player?

    I'm not sure if there was a thread on this before. I googled but the link was not valid anymore.

    Please help! What theory/formula do I use here?

    Thank you.
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  2. #2
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    Hello,

    Let X denote the earning for each game.
    X=-1 if the number doesn't match.
    X=750 if the number matches.

    The value you're looking for is E(X)=\sum_x x P(X=x), with x taking values from \{-1;750\}

    Now let's calculate the probabilities.



    Choosing a 3 digit number is equivalent to choosing randomly three times a number between 0 and 9 (except for the first digit).

    YZT

    Y \in \{1;2; \dots ; 9\} -> probability to get the matching one : \frac{1}{9}

    Z \in \{0;1;2; \dots ;9 \} -> probability to get the matching one : \frac{1}{10}

    T \in \{0;1;2; \dots ;9 \} -> probability to get the matching one : \frac{1}{10}


    Another way to do it, more direct, is to know that there are 999-100+1=900 number between 100 and 999.


    Hence, the probability of getting the right number is \frac{1}{900}
    The probability of getting a wrong number is \frac{899}{900}


    ----> E(X)=-1 \times \frac{899}{900}+750 \times \frac{1}{900} \approx -0.1656
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  3. #3
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    Quote Originally Posted by defygravityy View Post
    Hello! I need help on this question. I've been thinking about it for ages and I have absolutely no idea where to start.

    The local lottery sets up a game wherein a player has a chance of collecting $750. The player must choose a 3-digit number and if it matches, the player receives $750.
    It costs $1 to play the game.
    What is the expected profit (or loss) for each player?

    I'm not sure if there was a thread on this before. I googled but the link was not valid anymore.

    Please help! What theory/formula do I use here?

    Thank you.
    Excluding 0 but including numbers like 001, 021 etc, there are 999 different three digit numbers. So your probability of winning $750 is 1/999 and of losing $1 (that is, winning $-1) is 998/999.

    Expected profit ($) = (750)(1/999) + (-1)(998/999) = -248/999, which represents a loss.

    If you're not allowed to include numbers like 001, 021 etc, then there are (9)(10)(10) = 900 different three digit numbers. The above logic stays the same .....
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