Question

$A,B,C,D$ cut a pack of $52$ cards successively in the order given. If the person who cuts a spade first receives Rs.$350$, what are their respective expectations?

• A. $A=Rs.128$, $B=Rs.90$, $C=Rs.78$, $D=Rs.54$
• B. $A=Rs.128$, $B=Rs.96$, $C=Rs.72$, $D=Rs.54$
• C. $A=Rs.120$, $B=Rs.96$, $C=Rs.80$, $D=Rs.54$
• D. $A=Rs.124$, $B=Rs.96$, $C=Rs.76$, $D=Rs.54$

Answer 1

The correct option is B $A=Rs.128$, $B=Rs.96$, $C=Rs.72$, $D=Rs.54$

Let $E$ be the event of any one cutting a spade in one cut, and let $S$ be the sample space then
$n\left(E\right){=}^{13}{C}_1$ and $n\left(S\right){=}^{52}{C}_1$

$\therefore P\left(E\right)=p=\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^{13}{C}_1}{{}^{52}{C}_1}=\frac{13}{52}=\frac{1}{4}$

$\Rightarrow P\left(E\right)=p=\frac{1}{4}$
$\therefore p\left(\overline{E}\right)=q=1-p=\frac{3}{4}$
The probability of $A$ winning (when $A$ starts the game)
$=p+\left(qqqq\right)p+{\left(qqqq\right)}^{2}p+...\infty =p+q^{4}p+q^{8}p+...\infty$

$=\frac{p}{1-q^4}=\frac{\frac{1}{4}}{1-{\left(\frac{3}{4}\right)}^4}=\frac{64}{175}$
Therefore expectation of $A$ $=Rs350=Rs350\times \frac{64}{175}=Rs128$
The probability of $B$ winning $=qp+qqqq\left(qp\right)+{\left(qqqq\right)}^2\left(pq\right)+...\infty$
$=\frac{qp}{1-q^4}=\frac{\frac{3}{4}\times \frac{1}{4}}{{1-\left(\frac{3}{4}\right)}^4}=\frac{48}{175}$
Therefore Expectation of $B$ $=Rs350\times =Rs350\times \frac{48}{175}=Rs96$
The probability of $C$ winning $=qqp+qqqq\left(qqp\right)+{\left(qqqq\right)}^2\left(qqp\right)+...\infty$

$=\frac{qqp}{1-qqqq}=\frac{\frac{3}{4}\times \frac{3}{4}\times \frac{1}{4}}{1-{\left(\frac{3}{4}\right)}^4}=\frac{36}{175}$
Therefore expectation of $C$ $=Rs72$
Expectation of $D$ $=Rs350-$ (Sum of the expecatation of $A,B,C$)
$=Rs350-(Rs128+Rs96+Rs72)=Rs54$