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Math Help - Winning at darts

  1. #1
    Super Member Anonymous1's Avatar
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    Winning at darts

    Two boys take turns throwing darts at a target. Al throws first and hits with probability \frac{1}{5}. Bob throws second and hits with probability b. What value of b makes each boy have probability \frac{1}{2} of winning?

    What I've got:

    P(Al Wins) = \frac{1}{5} \sum_{i=0}^{\infty} (\frac{4}{5} (1-b))^i = \frac{1+4b}{5}

    P(Bob Wins) = b \sum_{i=0}^{\infty} (\frac{4}{5})^{i+1} (1-b)^i = \frac{4b}{1+4b}.

    Then I set them equal and got 16b^2 -12b + 1 = 0
    Which makes no sense. What am I doing wrong?
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  2. #2
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    Quote Originally Posted by Anonymous1 View Post
    Two boys take turns throwing darts at a target. Al throws first and hits with probability \frac{1}{5}. Bob throws second and hits with probability b. What value of b makes each boy have probability \frac{1}{2} of winning?

    What I've got:

    P(Al Wins) = \frac{1}{5} \sum_{i=0}^{\infty} (\frac{4}{5} (1-b))^i = \frac{1+4b}{5} Mr F says: No. This is equal to {\color{red} \frac{1}{1 + 4b}}. Recheck your calculation.

    P(Bob Wins) = b \sum_{i=0}^{\infty} (\frac{4}{5})^{i+1} (1-b)^i = \frac{4b}{1+4b}.

    Then I set them equal and got 16b^2 -12b + 1 = 0
    Which makes no sense. What am I doing wrong?
    ..
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  3. #3
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    Hello, Anonymous1!

    Mr f is right . . . Your algebra is off.


    Two boys take turns throwing darts at a target.
    Al throws first and hits with probability \tfrac{1}{5}.
    Bob throws second and hits with probability b.
    What value of b makes each boy have probability \tfrac{1}{2} of winning?

    What I've got:

    P(\text{Al wins}) \;=\; \underbrace{\frac{1}{5} \sum_{n=0}^{\infty} \bigg[\tfrac{4}{5} (1-b)\bigg]^n}_{\text{This is correct}} \;\;=\;\; \underbrace{\frac{1+4b}{5}}_{\text{This is not}}

    The infinite series has: first term a = 1\:\text{ and }\:\text{common ratio }r = \tfrac{4}{5}(1-b)

    . . Its sum is: . \frac{a}{1-r} \;=\;\frac{1}{1 - \frac{4}{5}(1-b)} \;=\;\frac{1}{1-\frac{4}{5} + \frac{4}{5}b} \;=\; \frac{1}{\frac{1}{5} + \frac{4}{5}b} \;=\;\frac{5}{1 + 4b}

    . . Hence: . P(\text{Al wins}) \:=\;\frac{1}{5}\cdot\frac{5}{1+4b} \;=\;\frac{1}{1+4b}


    Then: . P(\text{Al wins}) \:=\:\frac{1}{2} \quad\Rightarrow\quad \frac{1}{1+4b} \:=\:\frac{1}{2}

    . . From which we get: . b \,=\,\frac{1}{4}

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