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Thread: 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 $\displaystyle \frac{1}{5}.$ Bob throws second and hits with probability b. What value of b makes each boy have probability $\displaystyle \frac{1}{2}$ of winning?

    What I've got:

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

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

    Then I set them equal and got $\displaystyle 16b^2 -12b + 1 = 0$
    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 $\displaystyle \frac{1}{5}.$ Bob throws second and hits with probability b. What value of b makes each boy have probability $\displaystyle \frac{1}{2}$ of winning?

    What I've got:

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

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

    Then I set them equal and got $\displaystyle 16b^2 -12b + 1 = 0$
    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 $\displaystyle \tfrac{1}{5}.$
    Bob throws second and hits with probability $\displaystyle b.$
    What value of $\displaystyle b$ makes each boy have probability $\displaystyle \tfrac{1}{2}$ of winning?

    What I've got:

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

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

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

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


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

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

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