Question

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

Answer 1

In his first throw A's chance is $\frac{1}{6};$ in his second throw it is $\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$, because each player 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 player must have failed twice; and so on.
Thus A'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's chance is the sum of the infinite series
$\frac{5}{6}\cdot \frac{1}{6}\{1+{\left(\frac{5}{6}\right)}^2+{\left(\frac{5}{6}\right)}^4+......\};$
$\therefore$ A's chance is to B's as $6$ is to $5$; their respective chances are therefore $\frac{6}{11}$ and $\frac{5}{11}$, and their expectations are pounds $6$ and pounds $5$ respectively.