On South Seas Roll 'Em


The Baron's latest game consisted of up to four turns throwing a pair of dice and cost nine coins to play. After each turn Sir R----- may have elected to stop playing and collect their sum as winnings.
On hearing these rules, it immediately occurred to me that he should only continue the game if he has thrown a sum less than that he might expect to win in future turns.
We can thus reckon the expected winnings before throwing the dice by considering the expected winnings after throwing them, conditional upon the cases of having done better and having done worse than expected in future turns.

I explained this insight to the Baron, but fear I may not have done so with sufficiently clarity since I was struck with the impression that he had not fully understood me. Hopefully I shall do better with this exposition!

Let us assume that the expected winnings when there are \(n\) turns left are equal to \(e_n\). The expected winnings when there are \(n+1\) turns left can then be figured as the sum of \(e_n\) times the probability that we roll less than or equal to \(e_n\) and the expected sum of a throw given that it is greater than \(e_n\) times the probability that we make such a throw.
We write this as
\[ e_{n+1} = p\left(x \leqslant e_n\right) \times e_n + p\left(x > e_n\right) \times \mathrm{E}\left[x | x > e_n\right] \]
where \(x\) represents possible sums of the pair of dice, \(p\) the probability of observing the event that follows it and \(\mathrm{E}\) the expected value of the symbol before the bar assuming it satisfies the condition after it.
We can figure this conditional expectation with the formula
\[ \mathrm{E}\left[x|x>e_n\right] = \frac{\sum_{y>e_n}y \times p(x=y)}{p\left(x>e_n\right)} \]
where the \(\sum\) stands for the sum of the terms that follow it for every value \(y\) greater than \(e_n\).

The expected sum of a roll of a pair of dice is seven, and so
\[ e_1 = 7 \]
We can therefore figure the expected winnings in a game of two turns with
\[ e_2 = p\left(x\leqslant7\right) \times 7 + p\left(x>7\right) \times \mathrm{E} \left[x|x>7\right] \]
Noting that the probabilities of rolling sums from two to twelve are equal to
\[ \begin{align*} p(x=2) &= \frac{1}{36}\\ p(x=3) &= \frac{2}{36}\\ p(x=4) &= \frac{3}{36}\\ p(x=5) &= \frac{4}{36}\\ p(x=6) &= \frac{5}{36}\\ p(x=7) &= \frac{6}{36}\\ p(x=8) &= \frac{5}{36}\\ p(x=9) &= \frac{4}{36}\\ p(x=10) &= \frac{3}{36}\\ p(x=11) &= \frac{2}{36}\\ p(x=12) &= \frac{1}{36} \end{align*} \]
we can rewrite this as
\[ \begin{align*} e_2 &= \left(\frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}\right) \times 7\\ &\quad\quad + \left(\frac{5}{36}+\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\right) \times \frac{\left(8 \times \frac{5}{36} + 9 \times \frac{4}{36} + 10 \times \frac{3}{36} + 11 \times \frac{2}{36} + 12 \times \frac{1}{36}\right)}{\left(\frac{5}{36}+\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\right)}\\ &= \frac{21}{36} \times 7 + \left(\frac{40}{36}+\frac{36}{36}+\frac{30}{36}+\frac{22}{36}+\frac{12}{36}\right)\\ &= \frac{147}{36}+\frac{140}{36} = \frac{287}{36} = 7\frac{35}{36} \end{align*} \]
We can then use this to reckon the expected winnings in a game of three turns
\[ \begin{align*} e_3 &= p\left(x \leqslant e_2\right) \times e_2 + p\left(x > e_2\right) \times \mathrm{E}\left[x | x > e_2\right]\\ &= p\left(x \leqslant 7\frac{35}{36}\right) \times 7\frac{35}{36} + p\left(x > 7\frac{35}{36}\right) \times \mathrm{E}\left[x | x > 7\frac{35}{36}\right]\\ &= \left(\frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}\right) \times \frac{287}{36}\\ &\quad\quad + \left(\frac{5}{36}+\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\right) \times \frac{\left(8 \times \frac{5}{36} + 9 \times \frac{4}{36} + 10 \times \frac{3}{36} + 11 \times \frac{2}{36} + 12 \times \frac{1}{36}\right)}{\left(\frac{5}{36}+\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\right)}\\ &= \frac{21}{36} \times \frac{287}{36} + \frac{140}{36} = \frac{6027}{1296} + \frac{5040}{1296} = \frac{11067}{1296} = 8\frac{233}{432} \end{align*} \]
and finally those in a game of four turns
\[ \begin{align*} e_4 &= p\left(x \leqslant e_3\right) \times e_3 + p\left(x > e_3\right) \times \mathrm{E}\left[x | x > e_3\right]\\ &= p\left(x \leqslant 8\frac{233}{432}\right) \times 8\frac{233}{432} + p\left(x > 8\frac{233}{432}\right) \times \mathrm{E}\left[x | x > 8\frac{233}{432}\right]\\ &= \left(\frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}+\frac{5}{36}\right) \times \frac{3689}{432}\\ &\quad\quad + \left(\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\right) \times \frac{\left(9 \times \frac{4}{36} + 10 \times \frac{3}{36} + 11 \times \frac{2}{36} + 12 \times \frac{1}{36}\right)}{\left(\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\right)}\\ &= \frac{26}{36} \times \frac{3689}{432} + \frac{100}{36} = \frac{95914}{15552} + \frac{43200}{15552} = \frac{139114}{15552} = 8\frac{7349}{7776} \end{align*} \]
This being 427 of 7776 parts of a coin less than the cost to play, the game is biased in favour of the Baron and I should consequently have advised Sir R----- to decline the wager.

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

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