Question

A, B, C in order cut a pack of cards, replacing them after each cut, on condition that the first who cuts a spade shall win a prize: find their respective chances.

Answer 1

Let p be the chance of cutting a spade
Let q be the chance of not cutting a spade from pack of $52$ cards
Total no. of spade$=13$
$p=\frac{13}{52}=\frac{1}{4}$
$q=\frac{39}{52}=\frac{3}{4}$
A will win if the cuts spade $1^{th},4^{th},7^{th},{10}^{th}$ turns
His chance$=\frac{1}{4}+{\left(\frac{3}{4}\right)}^3\left(\frac{1}{4}\right)+{\left(\frac{3}{4}\right)}^6\frac{1}{4}.....$
A's winning$=\frac{\frac{1}{4}}{1-{\left(\frac{3}{4}\right)}^3}=\frac{1}{4}\times \frac{64}{37}$
$\Rightarrow \frac{16}{37}$
Similarly B's winning
$=\frac{3}{4}\times \frac{1}{4}+{\left(\frac{3}{4}\right)}^4\left(\frac{1}{4}\right)+{\left(\frac{3}{4}\right)}^7\frac{1}{4}...$
$=\frac{\frac{3}{4}\times \frac{1}{4}}{1-{\left(\frac{3}{4}\right)}^3}$
$=\frac{3}{4}\times \frac{64}{37}$
$\Rightarrow \frac{12}{37}$
Similarly C's winning$=\frac{{\left(\frac{3}{4}\right)}^2\times \frac{1}{4}}{1-{\left(\frac{3}{4}\right)}^3}\Rightarrow \frac{9}{64}\times \frac{64}{37}$
$\Rightarrow \frac{9}{37}$