A deadly game of chance

The easiest way to solve this problem is through complements: find the probability that Daniel does not survive, then subtract that value from 1 to find the probability that he will survive.

Suppose the game ends after only one round. That means Aaron, Benjamin, and Charlie all chose cards that were not the ace of spades, then Daniel chose the ace of spades. This happens with probability $\left ( \dfrac{51}{52} \right )^3 \left ( \dfrac{1}{52} \right ).$ If the game ends after two rounds, then all four people chose non-ace of spades cards during the first round, then in the second round Daniel is the first to draw the ace of spades. This happens with probability $\left ( \dfrac{51}{52} \right )^7 \left ( \dfrac{1}{52} \right ).$ In general, if the game lasts $n$ rounds, the probability that Daniel loses is $\left ( \dfrac{51}{52} \right )^{4n - 1} \left ( \dfrac{1}{52} \right ),$ and the overall probability that he loses is

$\begin{aligned} \sum_{n = 1}^{\infty} \left ( \dfrac{51}{52} \right )^{4n - 1} \left ( \dfrac{1}{52} \right ) &= \dfrac{ ( \frac{51}{52} )^3 ( \frac{1}{52} )}{1 - ( \frac{51}{52} )^4} \\ &= \dfrac{51^3}{52^4 - 51^4}. \end{aligned}$

Therefore, the probability that Daniel will survive is $1 - \dfrac{51^3}{52^4 - 51^4} \approx \boxed{0.757}.$