Three players A, B, C in this order, cut a pack of cards, and the whole pack is reshuffled after each cut. If the winner is one who first draws a diamond then C’s chance of winning is:

9/37

Probability (C will win)

$= P(\bar A)P(\bar B)P(C) + P(\bar A)P(\bar B)P(\bar C)P(\bar A)P(\bar B)P(C) + .....∞$

$=\frac{3}{4}×\frac{3}{4}×\frac{1}{4}+(\frac{3}{4}×\frac{3}{4}×\frac{3}{4})×\frac{3}{4}×\frac{3}{4}×\frac{1}{4}+.....∞$

It is infinite G.P. Using the formula for sum of infinite G.P., we get:

$=[\frac{3}{4}×\frac{3}{4}×\frac{1}{4}][1+(\frac{3}{4})^3+(\frac{3}{4})^6....]$

$P(C\,wins)=\frac{\frac{9}{64}}{1-(\frac{3}{4})^3}=\frac{9}{37}$