# On Fruitful Opals

Recall that the Baron’s game consisted of guessing under which of a pair of cups was to be found a token for a stake of four cents and a prize, if correct, of one. Upon success, Sir R----- could have elected to play again with three cups for the same stake and double the prize. Success at this and subsequent rounds gave him the opportunity to play another round for the same stake again with one more cup than the previous round and a prize equal to that of the previous round multiplied by its number of cups.
In other words, if Sir R----- had earned the right to play the $$n^\mathrm{th}$$ round, he would have had to pay a stake of four cents to do so and must have guessed beneath which of $$n+1$$ cups was the token for a prize of
$1 \times 2 \times 3 \times \dots \times n$
cents.
At first glance the game seems to be heavily biased in the Baron’s favour. Indeed, Sir R----- could only have come out ahead if he won four games in a row since
$1 + 2 + 2 \times 3 = 9 < 3 \times 4 = 12$
and
$1 + 2 + 2 \times 3 + 2 \times 3 \times 4 = 33 > 4 \times 4 = 16$
The chance of his having done so was a paltry
$\tfrac12 \times \tfrac13 \times \tfrac14 \times \tfrac15= \tfrac1{120}$
strongly suggesting that he should not have taken up the Baron’s challenge.
Appearances, however, can be deceptive. We can reckon Sir R-----'s expected winnings, in cents, with the never-ending formula
$-4 + \tfrac12\times\big(1 - 4 + \tfrac13 \times \big(2 - 4 + \tfrac14 \times \big(2 \times 3 - 4 + \tfrac15 \times \dots$
Rearranging this yields
$-4 - \tfrac12 \times 4 - \tfrac12 \times \tfrac13 \times 4 - \tfrac12 \times \tfrac13 \times \tfrac14 \times 4 - \dots\\ + \tfrac12 + \tfrac12 \times \tfrac13 \times 2 + \tfrac12 \times \tfrac13 \times \tfrac14 \times 2 \times 3 + \dots$
and thus
$-4 \times \left(1 + \tfrac12 + \tfrac{1}{2 \times 3} + \tfrac{1}{2 \times 3 \times 4} + \dots \right) + \left(\tfrac12 + \tfrac13 + \tfrac14 + \dots\right)$
Using an exclamation mark to denote the product of one and every integer greater than zero and less than or equal to that that immediate precedes it, known as the factorial, and a capital sigma to represent the sum of the expression that follows it with the variable term replaced one after the other by each of the integers between those below and above it, we have
\begin{align*} -4 \times \sum_{i=1}^\infty\frac{1}{i!} + \sum_{i=2}^\infty\frac{1}{i} &= -4 \times \left(\left(\sum_{i=0}^\infty\frac{1}{i!}\right)-1\right) + \left(\left(\sum_{i=1}^\infty\frac1i\right)-1\right)\\ &= 3 - 4 \times \sum_{i=0}^\infty\frac{1}{i!} + \sum_{i=1}^\infty\frac1i \end{align*}
The first sum is known as the exponential series and is equal to the mathematical constant $$e$$, or approximately 2.7183. The second sum is known as the harmonic series and is infinite in value! I expressed as much to the Baron, but I am not entirely sure he grasped my point.
This fact that the sum of these ever diminishing terms grows without limit seems unlikely, but it is relatively easy to demonstrate by grouping them together.
\begin{align*} & 1 + \tfrac12 + \tfrac13 + \tfrac14 + \tfrac15 + \tfrac16 + \tfrac17 + \tfrac18 + \tfrac19 + \tfrac1{10} + \tfrac1{11} + \tfrac1{12} + \tfrac1{13} + \tfrac1{14} + \tfrac1{15} + \tfrac1{16} + \dots\\ =& 1 + \tfrac12 + \left(\tfrac13 + \tfrac14\right) + \left(\tfrac15 + \tfrac16 + \tfrac17 + \tfrac18\right) + \left(\tfrac19 + \tfrac1{10} + \tfrac1{11} + \tfrac1{12} + \tfrac1{13} + \tfrac1{14} + \tfrac1{15} + \tfrac1{16}\right) + \dots\\ >& 1 + \tfrac12 + \left(\tfrac14 + \tfrac14\right) + \left(\tfrac18 + \tfrac18 + \tfrac18 + \tfrac18\right) + \left(\tfrac1{16} + \tfrac1{16} + \tfrac1{16} + \tfrac1{16} + \tfrac1{16} + \tfrac1{16} + \tfrac1{16} + \tfrac1{16}\right) + \dots\\ =& 1 + \tfrac12 + \tfrac12 + \tfrac12 + \tfrac12 + \dots \end{align*}
Figure 1: Bounds For The Harmonic Series
The question that remains is how long must Sir R----- have been willing to play in order to exploit this advantage.
Unfortunately, I know of no tidy formula with which to express the sum of the first $$n$$ terms of the harmonic series. We can, however, figure bounds upon it using the integral calculus.

The sum of the first $$n$$ terms of this series is trivially equal to the area under the curve that takes the value of one between zero and one, of one half between one and two, of one third between two and three and so on and so forth up to $$\frac1n$$ between $$n-1$$ and $$n$$. Now this curve is everywhere above that of $$\frac{1}{x+1}$$ and nowhere above that of the lesser of one and $$\frac1x$$, as sketched in figure 1.
Using the integral calculus to calculate the areas under these two bounding curves we can conclude that
$\int_0^n \frac{1}{x+1} \mathrm{d}x < \sum_{i=1}^n \frac1i \leqslant 1 + \int_1^n \frac1x \mathrm{d}x$
and hence, by the rules of integration
$\ln\left(n+1\right) < \sum_{i=1}^n \frac1i \leqslant 1+ \ln n$
If we treat the first $$n$$ terms of the exponential series as if it were equal to its entirety, a reasonable supposition if $$n$$ is not small, Sir R-----'s expected winnings if he elects to never play more than $$n$$ games, $$\mathrm{E}_n$$ cents, would have been approximately bounded by
$3 - 4 \times e + \ln\left(n+1\right) < \mathrm{E}_n \leqslant 3 - 4 \times e + \left(1+ \ln n\right)$
Now the number of games that Sir R----- must have been prepared to play if he wished to come out ahead on average must satisfy
$3 - 4 \times e + \ln\left(n+1\right) < 0 \leqslant 3 - 4 \times e + \left(1+ \ln n\right)$
and therefore, for large $$n$$, have approximate bounds of
\begin{align*} 4 \times e - 4 \leqslant &\ln n < 4 \times e - 3\\ 6.8731 \leqslant &\ln n < 7.8731\\ e^{6.8731} \leqslant &\;\;n\;\; < e^{7.8731} \end{align*}
or, in more familiar notation, roughly
$965 < n < 2,626$
I should consequently have advised Sir R----- to take up the Baron’s challenge, but only if he had a taste for a very large prize at very long odds and plenty of time on his hands!
$$\Box$$

Based upon an article I wrote for ACCU's CVu magazine.