Recall that in the Baron's latest game, for a cost of one coin, Sir R----- could roll two dice and have from the Baron coins equal in number to the best of them. In return, the Baron would have from Sir R----- coins equal to the roll of a solitary die.

The key to figuring the fairness of this wager lies in the working out of the expected score of the best of two dice since the expected score of a single die is trivially

To work out the expected best score of two dice we must first find the probability of it being equal to one, two, three and so on and so forth. The trick to doing so is to first find the probability of it being less than or equal to each score. For a single die, this is simply

Since the outcomes of two dice are wholly unrelated we can simply multiply the probabilities that each are less than or equal to any given score to recover the probability that they both are

Now, this must equal the probability that the greater score is less than or equal to \(n\), since if it did not then one of them would neccessarily be greater than \(n\), so we have

Expanding this out, we have

The key to figuring the fairness of this wager lies in the working out of the expected score of the best of two dice since the expected score of a single die is trivially

\[
\tfrac{1}{6} \times 1 +
\tfrac{1}{6} \times 2 +
\tfrac{1}{6} \times 3 +
\tfrac{1}{6} \times 4 +
\tfrac{1}{6} \times 5 +
\tfrac{1}{6} \times 6\\
= \tfrac{1}{6} \times \left(1+2+3+4+5+6\right)\\
= \tfrac{1}{6} \times 21\\
= 3 \tfrac{1}{2}
\]

which most students of wager will have long since committed to memory.To work out the expected best score of two dice we must first find the probability of it being equal to one, two, three and so on and so forth. The trick to doing so is to first find the probability of it being less than or equal to each score. For a single die, this is simply

\[
\Pr(\mathrm{d}6 \leqslant n) = \tfrac{1}{6}n
\]

where \(\mathrm{d}6\) stands for result of a roll of a single die.Since the outcomes of two dice are wholly unrelated we can simply multiply the probabilities that each are less than or equal to any given score to recover the probability that they both are

\[
\Pr(\mathrm{d}6_1 \leqslant n \; \mathrm{and} \; \mathrm{d}6_2 \leqslant n) = \tfrac{1}{6}n \times \tfrac{1}{6}n = \tfrac{1}{36}n^2
\]

in which the subscripts distinguish between the dice.Now, this must equal the probability that the greater score is less than or equal to \(n\), since if it did not then one of them would neccessarily be greater than \(n\), so we have

\[
\Pr(\max\left(\mathrm{d}6_1, \mathrm{d}6_2\right) \leqslant n) = \tfrac{1}{36}n^2
\]

Note that this is an example of the general rule that governs the probabilities of the maximum of a set of unrelated random quantities. Specifically if we have a set of such quantities
\[
X_1, X_2, X_3, \dots, X_n
\]

with cumulative distribution functions determining the probability that they will be equal to or less than any given value
\[
F_1, F_2, F_3, \dots, F_n
\]

then the maximum observed value has the cumulative distribution function
\[
F_1 \times F_2 \times F_3 \times \dots \times F_n
\]

Armed with this function we can often derive many other properties of the maximimum value. For example, returning to Sir R-----'s pair of dice, given that the probability that the greater score is equal to \(n\) is simply the probability that it is less than or equal to \(n\) minus the probability that it is less than or equal to \(n-1\), we have
\[
\begin{align*}
\Pr(\max\left(\mathrm{d}6_1, \mathrm{d}6_2\right) = n) &= \Pr(\max\left(\mathrm{d}6_1, \mathrm{d}6_2\right) \leqslant n) - \Pr(\max\left(\mathrm{d}6_1, \mathrm{d}6_2\right) \leqslant n-1)\\
&= \tfrac{1}{36}n^2 - \tfrac{1}{36}(n-1)^2\\
&= \tfrac{1}{36}\left(n^2 - \left(n^2-2n+1\right)\right)\\
&= \tfrac{1}{36}(2n-1)
\end{align*}
\]

That the expected score is but the sum of each possible score multiplied by its probability leads us to conclude that it is given by
\[
\begin{align*}
\mathrm{E}\left[\max\left(\mathrm{d}6_1, \mathrm{d}6_2\right)\right]
&= \sum_{n=1}^6 \, n \times \tfrac{1}{36}(2n-1)\\
&= \tfrac{2}{36} \sum_{n=1}^6 \, n^2 - \tfrac{1}{36} \sum_{n=1}^6 \, n
\end{align*}
\]

in which \(\sum\) is the sign that my fellow students and I habitually use to indicate a sum.Expanding this out, we have

\[
\begin{align*}
\mathrm{E}\left[\max\left(\mathrm{d}6_1, \mathrm{d}6_2\right)\right]
&= \tfrac{2}{36} \left(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2\right) - \tfrac{1}{36} (1 + 2 + 3 + 4 + 5 + 6)\\
&= \tfrac{2}{36} \left(1 + 4 + 9 + 16 + 25 + 36\right) - \tfrac{1}{36} (1 + 2 + 3 + 4 + 5 + 6)\\
&= \tfrac{2}{36} \times 91 - \tfrac{1}{36} \times 21\\
&= \tfrac{1}{36} \times (182-21)\\
&= \tfrac{1}{36} \times 161
\end{align*}
\]

So we can figure Sir R-----'s expected winnings as the difference between his expected prize and the Baron's, less his cost of one coin
\[
\begin{align*}
\mathrm{E}\left[\max\left(\mathrm{d}6_1, \mathrm{d}6_2\right)\right] - \mathrm{E}[\mathrm{d}6] - 1
&= \tfrac{1}{36} \times 161 - \tfrac{1}{6} \times 21 - 1\\
&= \tfrac{1}{36} \times 161 - \tfrac{1}{36} \times 126 - \tfrac{1}{36} \times 36\\
&= \tfrac{1}{36} \times (161 - 126 - 36)\\
&= -\tfrac{1}{36}
\end{align*}
\]

Being a loss for Sir R-----, I should most certainly have advised him to decline the Baron's wager.
\(\Box\)

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