So the problem is in Tanis & Hogg, Probability and Statistical inference section 1.5 Independence of Events.
An Urn contains five balls, one marked WIN and four marked LOSE. You and another player take turns selecting a ball at random from the urn, one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done with replacement.
I can't see why the probability isn't $\displaystyle\frac{1}{5}$. The book states it is
$\displaystyle\sum_{k=0}^\infty(\displaystyle\frac{1}{5}\times\displaystyle\frac{4}{5}2k)$
Thanks!
Best Answer
First lets find the probability that you win on your $k^{th}$ draw. Let us denote it by $p_k$. Then the probability you win is given as $\displaystyle \sum_{k=1}^{\infty} p_k$.
The probability you win on your $k^{th}$ draw can be computed as follows. If you have to win on your $k^{th}$ draw, then in each of your previous $k-1$ draws, you should have chosen one of the balls marked LOSE else you would have won before your $k^{th}$ draw. Your opponent must have also chosen one of the balls marked LOSE in each of his $k-1$ draws else he would have won before you. In each draw, the probability that a person chooses one of the balls marked LOSE is $\dfrac45$ and the probability that a person chooses the ball marked WIN is $\dfrac15$. Hence, \begin{align} p_k & = \underbrace{\left(\dfrac45 \times \dfrac45 \times \cdots \times \dfrac45 \right)}_{\substack{(k-1) \text{ times}\\ \text{You lost your first $k-1$ draws}}} \times \underbrace{\left(\dfrac45 \times \dfrac45 \times \cdots \times \dfrac45 \right)}_{\substack{(k-1) \text{ times}\\ \text{Opponent lost his first $k-1$ draws}}} \times \underbrace{\dfrac15}_{\text{You chose the right ball on your $k^{th}$ draw}}\\ & = \left(\dfrac45 \right)^{2k-2} \times \dfrac15 \end{align} Hence, the probability you win is $$\sum_{k=1}^{\infty} p_k = \sum_{k=1}^{\infty} \left(\dfrac45 \right)^{2k-2} \times \dfrac15 = \dfrac15 \left( \sum_{k=1}^{\infty} \left(\dfrac{16}{25} \right)^{k-1}\right) = \dfrac15 \left( \dfrac1{1-\dfrac{16}{25}} \right) = \dfrac59$$