Problem source

A men's endurance competition has an unlimited number of rounds. In each round, a competitor has, independently, a probability $p$ of making it through the round; otherwise, he fails the round. Once a competitor fails a round, he drops out of the competition; before he drops out, he takes part in every round. The grand prize is awarded to any competitor who makes it through a round which all the other remaining competitors fail; if all the remaining competitors fail at the same round the grand prize is not awarded. If the competition begins with three competitors, find the probability that:
\begin{questionparts}
\item all three drop out in the same round;
\item two of them drop out in round $r$ (with $r \ge 2$) and the third in an earlier round;
\item the grand prize is awarded.
\end{questionparts}

Solution source

\begin{questionparts}
\item This is the same as the sum of the probability that they all drop out in the $k$th round for all values of $k$, ie

\begin{align*}
\mathbb{P}(\text{all drop in same round}) &= \sum_{k=0}^\infty \mathbb{P}(\text{all drop out in the }k+1\text{th round}) \\
&= \sum_{k=0}^{\infty}(p^k(1-p))^3 \\
&= (1-p)^3 \sum_{k=0}^{\infty}p^{3k} \\
&= \frac{(1-p)^3}{1-p^3} \\
&= \frac{1+3p(1-p)(p-(1-p))-p^3}{1-p^3} \\
&= \frac{1-p^3-3p(1-p)(1-2p)}{1-p^3}
\end{align*}

\item There are $3$ ways to choose the person who drops out earlier, and then they can drop out in round $0, 1, \cdots r-1$

\begin{align*}
\mathbb{P}(\text{exactly two drop out in round }r\text{ and one before}) &= 3\sum_{k=0}^{r-2} (p^{r-1}(1-p))^2p^k(1-p) \\
&= 3p^{2r-2}(1-p)^3 \sum_{k=0}^{r-2}p^k \\
&=  3p^{2r-2}(1-p)^3 \frac{1-p^{r-1}}{1-p} \\
&= 3p^{2r-2}(1-p)^2(1-p^{r-1})
\end{align*}

\item The probability exactly $2$ finish after the third

\begin{align*}
\mathbb{P}(\text{exactly two drop out after third}) &= \sum_{r=2}^{\infty}\mathbb{P}(\text{exactly two drop out in round }r\text{ and one before}) \\
&= \sum_{r=2}^{\infty}3p^{2r-2}(1-p)^2(1-p^{r-1}) \\
&= 3(1-p)^2p^{-2}\sum_{r=2}^{\infty}(p^{2r}-p^{3r-1}) \\
&= 3(1-p)^2p^{-2} \left( \frac{p^4}{1-p^2} - \frac{p^5}{1-p^3} \right) \\
&= \frac{3(1-p)^2(p^2(1-p^3)-p^3(1-p^2))}{(1-p^2)(1-p^3)}\\
&= \frac{3(1-p)^3p^2}{(1-p^2)(1-p^3)}\\
\end{align*}

Therefore the probability the grand prize is not awarded is

\begin{align*}
P &= 1 - \frac{(1-p)^3}{1-p^3} - \frac{3(1-p)^3p^2}{(1-p^2)(1-p^3)} \\
&= \frac{(1-p^3)(1-p^2) - (1-p)^3(1-p^2)-3(1-p)^3p^2}{(1-p^2)(1-p^3)} \\
&= \frac{(1-p^3)(1-p^2) - (1-p)^3(1+2p^2)}{(1-p^2)(1-p^3)} \\
\end{align*}
\end{questionparts}