CAT 2009 | Question: 20

Question

$\text{A}$ and $\text{B}$ throw with one dice for a stake of $\text{Rs.}\;11$ which is to be won by the player who first throw $6$. If $\text{A}$ has the first throw, what are their respective expectations.

1. $\text{Rs.}\; 7, \text{Rs.}\; 4$
2. $\text{Rs.}\; 6, \text{Rs.}\; 5$
3. $\text{Rs.}\; 4, \text{Rs.}\; 7$
4. $\text{Rs.}\; 5, \text{Rs.}\; 6$

Comments

B.

Answer 1

If A throw first then A's chances of winning =$\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+...$

$=1/6\left(1+\left(\frac{5}{6}\right)^2+(\frac{5}{6}.\frac{5}{6})^2+ \dots \right)$

$=1/6\left(\frac{1}{1-(5/6)^2}\right)=6/11$

And B's chances of winning = $\frac{5}{6}.\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+\dots$

$=\frac{5}{6}.\frac{1}{6}\left(1+\frac{5}{6}.\frac{5}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}+\dots \right)$

$=5/11$
 

Expected return for A $= \text{Probability of A winning} \times \text{Amount of money on offer} \\=\frac{6}{11} \times 11 = 6.$

Expected return for B $= \text{Probability of B winning} \times \text{Amount of money on offer} \\=\frac{5}{11} \times 11 = 5.$

So, ans B)