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Math Help - Finding expected value

  1. #1
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    Finding expected value

    A casino patron will continue to make $5 bets on red in roulette until she has won 4 of these bets.
    On each bet, she will either win $5 with probability \frac {18}{38} or lose $5 with probability \frac {20}{38}.
    If W is her final winnings and X is the number of bets she makes, then, since she would have won 4 bets and lost (X-4) bets, it follows that
    W=20-5(X-4)=40-5X
    Hence E[W]=40-5E[X]=40-5[4/(\frac {9}{19})]=-\frac{20}{9}
    but why is E[X]=4/(\frac{9}{19})?
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  2. #2
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    Quote Originally Posted by noob mathematician View Post
    A casino patron will continue to make $5 bets on red in roulette until she has won 4 of these bets.
    On each bet, she will either win $5 with probability \frac {18}{38} or lose $5 with probability \frac {20}{38}.
    If W is her final winnings and X is the number of bets she makes, then, since she would have won 4 bets and lost (X-4) bets, it follows that
    W=20-5(X-4)=40-5X
    Hence E[W]=40-5E[X]=40-5[4/(\frac {9}{19})]=-\frac{20}{9}
    but why is E[X]=4/(\frac{9}{19})?
    The number of failures is k = X - 4.

    k follows a negative binomial distribution: Negative binomial distribution - Wikipedia, the free encyclopedia

    p = 18/38 = 9/19 and the number of successes is r = 4.

    E(k) = r \cdot \left( \frac{1 - p}{p} \right).

    Therefore E(X) = E(k + 4) = E(k) + 4 = 4 \cdot \left( \frac{1 - \frac{9}{19}}{\frac{9}{19}} \right) + 4 = \, ....
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