The Baron's most recent wager set Sir R----- the task of placing tokens upon spaces numbered from zero to nine according to the outcome of a twenty sided die upon which was inscribed two of each of those numbers. At a cost of one coin per roll of the die, Sir R-----'s goal was to place a token upon every space for which he should receive twenty nine coins and twenty nine cents from the Baron.

To figure the fairness of this wager we must first note that if an experiment has a probability of success of \(p\) then the expected number \(n\) made before success is achieved is given by

To figure the fairness of this wager we must first note that if an experiment has a probability of success of \(p\) then the expected number \(n\) made before success is achieved is given by

\[
\mathrm{E}[n] = \frac1p
\]

If \(k\) of the ten places contain tokens then the chance of a cast of the die adding another is
\[
p = \frac{10-k}{10}
\]

and the expected number of rolls to fill all ten is consequently
\[
\begin{align*}
\mathrm{E} &= \frac{10}{10} + \frac{10}{9} + \frac{10}{8} + \frac{10}{7} + \frac{10}{6} + \frac{10}{5} + \frac{10}{4} + \frac{10}{3} + \frac{10}{2} + \frac{10}{1}\\
&= \frac{190}{90} + \frac{10}{8} + \frac{10}{7} + \frac{10}{6} + \frac{10}{5} + \frac{10}{4} + \frac{10}{3} + \frac{30}{2}\\
&= \frac{2,420}{720} + \frac{10}{7} + \frac{10}{6} + \frac{10}{5} + \frac{10}{4} + \frac{110}{6}\\
&= \frac{24,140}{5,040} + \frac{10}{6} + \frac{10}{5} + \frac{500}{24}\\
&= \frac{195,240}{30,240} + \frac{2,740}{120}\\
&= \frac{106,286,400}{3,628,800}
= \frac{7,381}{252}
\end{align*}
\]

Sir R-----'s expected winnings were therefore
\[
\begin{align*}
\frac{2,929}{100} - \frac{7,381}{252}
= \frac{2,929 \times 252 - 7,381 \times 100}{25,200}
= \frac{738,108 - 738,100}{25,200}
= \frac{8}{25,200}
= \frac{1}{3,150}
\end{align*}
\]

and I should have had no reservations in recommending that he accept the challenge!
\(\Box\)

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