A, B, C, D cut a pack of 52 well shuffled playing cards successively in the same order. If the person who cuts a spade first, wins the game and the game continues until this happens, then the probability that A wins the game is

$\begin{array}{r}n\left(E\right){=}^{13}{C}_1\\ n\left(S\right){=}^{52}{C}_1\end{array}$

$\text{ Now as we know probability of an event to occur is the ratio of the number of favorable outcome to the total number of outcomes}$

$\begin{array}{c}P\left(E\right)=p=\frac{n\left(E\right)}{n\left(S\right)}=\frac{{ }^{13}{C}_1}{{ }^{52}{C}_1}=\frac{13}{52}=\frac{1}{4}\\ P\left(E\right)=p=\frac{1}{4}\end{array}$

$\text{ Now let }q\text{ be the probability of event if card drawn from the pack is not a spade, hence we can write }$

$\begin{array}{c}P\left(\overline{E}\right)=q=1-p=1-\frac{1}{4}=\frac{3}{4}\\ P\left(\overline{E}\right)=q=\frac{3}{4}\end{array}$

$\text{ Now the probability of A winning the game when A starts the game will be }$

$=p+\left(qqqq\right)p+(qqqq{)}^{2}p+\dots \mathrm{\infty }=p+q^{4}p+q^{8}p+\dots \mathrm{\infty }=\frac{p}{1-q^4}$

$\text{ By substituting the values we get }$

$=\frac{p}{1-q^4}=\frac{\frac{1}{4}}{1-{\left(\frac{3}{4}\right)}^4}=\frac{64}{175}$