Question

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

• A. expectations of $A$ and $B$ are Rs.$6$ and Rs.$5$ respectively.
• B. expectations of $A$ and $B$ are Rs.$5$ and Rs.$6$ respectively.
• C. expectations of $A$ and $B$ are Rs.$7$ and Rs.$4$ respectively.
• D. expectations of $A$ and $B$ are Rs.$4$ and Rs.$7$ respectively.

Answer 1

The correct option is A expectations of $A$ and $B$ are Rs.$6$ and Rs.$5$ respectively.

In his first throw $A^{\prime }s$ chance is $\frac{1}{6}$, in his second its is $\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$,
because each players must have failed once before.
$A$ can have a second throw, in his third throw his chance is ${\left(\frac{5}{6}\right)}^4\times \frac{1}{6}$ because each players must have failed twice, and so on.
Thus $A^{\prime }s$ chance is the sum of the infinite series
$\frac{1}{6}(1+{\left(\frac{5}{6}\right)}^2+{\left(\frac{5}{6}\right)}^4+...)$
Similarly $B^{\prime }s$ chance is the sum of the infinite series $\frac{5}{6}.\frac{1}{6}(1+{\left(\frac{5}{6}\right)}^2+{\left(\frac{5}{6}\right)}^4)$
Therefore $A^{\prime }s$ chance is to $B^{\prime }s$ as $6$ is to $5$;
their respective chance are therefore $\frac{6}{11}$ and $\frac{5}{11}$
and their expectation are $rs6$ and $rs5$ respectively.