Three persons A,B,C in order cut a pack of cards replacing them after each cut. The person who first cuts a club shall win a prize. The probabilities of their winning are

$\mathrm{p}=\frac{13}{52}=\frac{1}{4}\text{ and }\mathrm{q}=1-\mathrm{p}=1-\frac{1}{4}=\frac{3}{4}$
$\therefore$Probability of A’s winning $=\frac{\mathrm{P}}{1-{\mathrm{q}}^3}=\frac{\frac{1}{4}}{1-{\left(\frac{3}{4}\right)}^3}=\frac{16}{64-27}=\frac{16}{37}$
Probability of B’ s winning $=\frac{\mathrm{qp}}{1-{\mathrm{q}}^3}=\frac{\left(\frac{3}{4}\right)\left(\frac{1}{4}\right)}{-{\left(\frac{3}{4}\right)}^3}=\frac{12}{64-27}=\frac{12}{37}$
Probability of C’s winning $=\frac{{\mathrm{q}}^2\mathrm{p}}{1-{\mathrm{q}}^2}=\frac{{\left(\frac{3}{4}\right)}^2\left(\frac{1}{4}\right)}{1-{\left(\frac{3}{4}\right)}^3}=\frac{9}{64-27}=\frac{9}{37}$