Question

Three persons $A,B,C$ take turns in order to cut a pack of cards , replacing them after each cut, on the condition that the first who cuts a card of spade shall win a prize. Find their respective probabilities of winning.

• A. $\frac{16}{37},\frac{12}{37},\frac{9}{37}$
• B. $\frac{16}{37},\frac{9}{37},\frac{12}{37}$
• C. $\frac{12}{37},\frac{9}{37},\frac{16}{37}$
• D. $\frac{17}{37},\frac{11}{37},\frac{9}{37}$

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 $q$ be the chance of not cutting a spade from pack of $52$ cards.

Then $p=\frac{{}^{13}{C}_1}{52C_1}=\frac{1}{4}$ and $q=1-p=\frac{3}{4}$

$A$ wins if he cuts spade at 1st , $4^{th}$, $7^{th}$ turns i.e. $A$ will win in second chance if $A,B,C$ all fail to cut a spade once then $A$ cuts spade at $4^{th}$ turn

$A$ will in third chance if $A,B,C$ all fail to cut a spade twice then $A$ cuts a spade at $7^{th}$ turn

Hence, A's chance of winning the prize $=p+q^{3}p+q^{6}p+...$$=\frac{p}{1-q3}$

$=\frac{\left(\frac{1}{4}\right)}{1-{\left(\frac{3}{4}\right)}^3}=\frac{16}{37}$

Similarly, $B$ has the second, sixth, cuts hence his probability of success is

$=qp+q^{4}p+q^{7}p+...$$=qp(1+q^3+q^6+...)$

$=\frac{qp}{1-q^3}=\frac{\frac{1}{4}\times \frac{3}{4}}{1-{\left(\frac{3}{4}\right)}^3}=\frac{12}{37}$

And for $C$ $=q^2+q^{5}p+q^{8}p+...$$=q^{2}p(1+q^3+q^6+...)$$=\frac{q^{2}p}{(1-q^3)}$

$=\frac{{\left(\frac{3}{4}\right)}^2\left(\frac{1}{4}\right)}{(1-{\left(\frac{3}{4}\right)}^3)}=\frac{9}{37}$