Question

Three persons $A,B,C$ in order 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 chances

• A. $18/37,10/37,9/37$
• B. $16/37,12/37,9/37$
• C. $20/37,10/37,7/37$
• D. none of these

Answer 1

The correct option is B $16/37,12/37,9/37$

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

A win a chance if he cuts spade at 1st , 4th, 7th turns i.e. A will get second chance if A,B,C all fails to cut a spade once then A cuts spade at 4th turn

A will third chance if A,B,C all fails to cut a spade twice then A cuts a spade at 7th 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}$