Question

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

• A. $\frac{16}{37},\frac{12}{37},\frac{9}{37}$
• B. $\frac{16}{21},\frac{4}{21},\frac{1}{21}$
• C. $\frac{9}{37},\frac{12}{37},\frac{16}{37}$
• D. $\frac{1}{21},\frac{4}{21},\frac{16}{21}$

Answer 1

The correct option is A

$\frac{16}{37},\frac{12}{37},\frac{9}{37}$

Let p be the chance of cutting a spade and q the chance of not cutting a spade from a pack of 52 cards.
Then $p=\frac{{}^{13}{C}_1}{{}^{52}{C}_1}=\frac{1}{4}q=1-\frac{1}{4}=\frac{3}{4}.$
Now A will win a prize if he cuts spade at 1st, 4th, 7th, 10th turns, etc. Note that A will get a second chance if A,B,C all fail to cut a spade once and then A cuts a spade at the 4th turn. Similarly he will cut a spade at the 7th turn when A,B,C fail to cut spade twice etc.
Hence A's chance of winning the prize
$=p+q^{3}p+q^{6}p+q^{6}p+q^{9}p+\cdots =\frac{p}{1-q^3}=\frac{\frac{1}{4}}{1-(\frac{3}{4}{)}^3}=\frac{16}{37}.$

Similarly B' s chance $=(qp+q^{4}p+q^{7}p+\cdots )$
$=q(p+q^{3}p+q^{6}p+\cdots )$
$\frac{3}{4}.\frac{16}{37}=\frac{12}{37}$
and C's chance = $\frac{3}{4}$ of B's chance$=\frac{3}{4}.\frac{12}{37}=\frac{9}{37}$