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The infinite series \(S\) is given by \[ S = 1 + (1 + d)r + (1 + 2d)r^2 + \cdots + (1+nd)r^n +\cdots\; ,\] for \(\vert r \vert <1\,\). By considering \(S - rS\), or otherwise, prove that \[ S = \frac 1{1-r} + \frac {rd}{(1-r)^2} \,.\] Arthur and Boadicea shoot arrows at a target. The probability that an arrow shot by Arthur hits the target is \(a\); the probability that an arrow shot by Boadicea hits the target is \(b\). Each shot is independent of all others. Prove that the expected number of shots it takes Arthur to hit the target is \(1/a\). Arthur and Boadicea now have a contest. They take alternate shots, with Arthur going first. The winner is the one who hits the target first. The probability that Arthur wins the contest is \(\alpha\) and the probability that Boadicea wins is \(\beta\). Show that \[ \alpha = \frac a {1-a'b'}\,, \] where \(a' = 1-a\) and \(b'=1-b\), and find \(\beta\). Show that the expected number of shots in the contest is \(\displaystyle \frac \alpha a + \frac \beta b\,.\)
Solution: Notice that \begin{align*} && S - rS &= 1 + dr + dr^2 + \cdots \\ &&&= 1 + dr(1 + r+r^2+ \cdots) \\ &&&= 1 + \frac{rd}{1-r} \\ \Rightarrow && S &= \frac{1}{1-r} + \frac{rd}{(1-r)^2} \end{align*} The number of shots Arthur takes is \(\textrm{Geo}(a)\), so it's expectation is \(1/a\). The probability Arthur wins is: \begin{align*} \alpha &= a + a'b'a + (a'b')^2a + \cdots \\ &= a(1+a'b' + \cdots) \\ &= \frac{a}{1-a'b'} \\ \\ \beta &= a'b + a'b'a'b + \cdots \\ &= a'b(1+b'a' + (b'a')^2 + \cdots ) \\ &= \frac{a'b}{1-a'b'} \end{align*} The expected number of shots in the contest is: \begin{align*} E &= a + 2a'b + 3a'b'a + 4a'b'a'b + \cdots \\ &= a(1 + 3a'b' + 5(a'b')^2 + \cdots) + 2a'b(1 + 2(a'b') + 3(a'b')^2 + \cdots) \\ &= a \left ( \frac{1}{1-a'b'} + \frac{2a'b'}{(1-a'b')^2} \right) + 2a'b \left ( \frac{1}{1-a'b'} + \frac{a'b'}{(1-a'b')^2}\right) \\ &= \frac{a}{1-a'b'} \left (1 + \frac{2a'b'}{(1-a'b')} \right) + 2\frac{a'b}{1-a'b'} \left ( 1 + \frac{a'b'}{(1-a'b')}\right) \\ &= \alpha \frac{1+a'b'}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ &= \alpha \frac{1+1-a-b+ab}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ \end{align*}