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

Let ‘E’ be the event of getting a spade, and ‘S’ be the sample space.
Then $\mathrm{n}\left(\mathrm{E}\right){=}^{13}{\mathrm{C}}_1\text{and}\mathrm{n}\left(\mathrm{S}\right){=}^{52}{\mathrm{C}}_1$
Now, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{{ }^{13}{C}_1}{{ }^{52}{C}_1}=\frac{13}{52}=\frac{1}{4}=p$
$\mathrm{q}=1-\mathrm{p}=1-\frac{1}{4}=\frac{3}{4}$
Now, probability of A wins the game is 
$\begin{array}{rl} & =p+q^{4}p+q^{8}p+\dots \mathrm{\infty }\\ & =\frac{p}{1-q^4}=\frac{1/4}{1-{\left(\frac{3}{4}\right)}^4}=\frac{64}{256-81}=\frac{64}{175}\end{array}$