On One Against Many

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Recall that the Baron proposed a pair of dice contests in which Sir R-----, were he to best the Baron's score, stood to win a bounty of thirteen coins.
Upon paying his stake Sir R----- was to cast his die but, if unhappy with its outcome, could pay a further coin to cast it again. Likewise, if he were not satisfied with the second cast, he could elect to cast a third time for a further two coins. He could continue in this fashion for as long as he pleased with the cost rising by one coin for each additional cast of his die. The Baron was to have but a single cast of his die, with Sir R----- to determine whether after or before his own play according to his stake; seven coins for the former and eight for the latter.

In order to figure the fairness of these wagers it is first necessary to recognise that there will inevitably come a point beyond which it is not in Sir R-----'s interest to draw out his play, for the cost of doing so rises inexorably whilst the prize, should he best the Baron, rests in constancy. Specifically, Sir R----- should refrain from casting his die if he should expect on the average to be further out of pocket in doing so. If we figure when this will be we can work backwards from his final cast including the advantage, if any, he might derive from further casts as we proceed. I said as much to the Baron, but I do not believe that I had his full attention.
Now, Sir R----- will be most compelled to recast his die if his will assuredly lose the wager if he does not. Indeed, he should only choose not if his expected winnings should be less than the cost of casting. Fortunately for our reckoning we need not consider further casts after this since, given that their cost shall grow ever the greater, their outcomes for Sir R----- shall grow ever the worser.

In the first game, the probability that Sir R----- will best the Baron if he casts a one is zero. If he should cast a two, he has one chance in six of winning. If a three, then two chances in six, etcetera. The probability that he should win on his final cast is therefore simply
\[ \frac{1}{6} \times 0 + \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{2}{6} + \frac{1}{6} \times \frac{3}{6} + \frac{1}{6} \times \frac{4}{6} + \frac{1}{6} \times \frac{5}{6} = \frac{1+2+3+4+5}{36} = \frac{5}{12} \]
His expected winnings from that cast should consequently be
\[ \frac{5}{12} \times 13 = \frac{65}{12} = 5 \frac{5}{12} \]
and he should in no circumstance elect to proceed if the cost of doing so exceeds five coins, which it shall do after his sixth cast.
From this it is readily apparent that Sir R----- should expect a prize of five twelfths of a coin for his sixth cast of the die.

Now, if Sir R----- were to cast a one upon his fifth roll, then his expected profit from paying for a sixth exceeds his certain profit of zero and he should elect to roll again. In contrast, if he were to roll a two, then he should have one chance in six of taking the prize of thirteen coins, for an expected windfall of thirteen upon six parts of a coin, or in other words two and one sixth coins. Consequently, Sir R----- should stick with his fifth die if it shows two or greater for an expected prize of
\[ \begin{align*} &\frac{1}{6} \times \frac{5}{12} + \frac{1}{6} \times \frac{1}{6} \times 13 + \frac{1}{6} \times \frac{2}{6} \times 13 + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13\\ &\quad= \frac{5}{72} + \frac{13}{36} + \frac{26}{36} + \frac{39}{36} + \frac{52}{36} + \frac{65}{36}\\ &\quad= \frac{5+26+52+78+104+130}{72}\\ &\quad= \frac{395}{72} = 5\frac{35}{72} \end{align*} \]
Finally, we must subtract the cost of four coins to yield Sir R-----'s net gain from a fifth cast of one seventy second part of a coin shy of one and a half coins.

To figure the value of this wager to Sir R----- we must simply continue in this fashion, replacing expected winnings smaller than can be got from continuing play, until we reach the first cast of the die.

For the fourth cast we consequently have a net average take of
\[ \begin{align*} &\frac{1}{6} \times \frac{107}{72} + \frac{1}{6} \times \frac{1}{6} \times 13 + \frac{1}{6} \times \frac{2}{6} \times 13 + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13 - 3\\ &\quad= \frac{107}{432} + \frac{13}{36} + \frac{26}{36} + \frac{39}{36} + \frac{52}{36} + \frac{65}{36} - 3\\ &\quad= \frac{107}{432} + \frac{195}{36} - 3\\ &\quad= \frac{1,151}{432} = 2\frac{287}{432} \end{align*} \]
Note that since
\[ \frac{13}{6} = 2\frac{1}{6} < 2\frac{287}{432} \]
Sir R----- should pay for a fourth cast if his third is less than three, yielding a payoff for the latter of
\[ \begin{align*} &\frac{2}{6} \times \frac{1,151}{432} + \frac{1}{6} \times \frac{2}{6} \times 13 + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13 - 2\\ &\quad= \frac{2,302}{2,592} + \frac{182}{36} - 2\\ &\quad= \frac{5,111}{1,296} = 3\frac{1,223}{1,296} \end{align*} \]
Similarly, for the second cast we have
\[ \begin{align*} &\frac{2}{6} \times \frac{5,111}{1,296} + \frac{1}{6} \times \frac{2}{6} \times 13 + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13 - 1\\ &\quad= \frac{10,222}{7,776} + \frac{182}{36} - 1\\ &\quad= \frac{20,879}{3,888} = 5\frac{1,439}{3,888} \end{align*} \]
and, finally, for the first
\[ \begin{align*} &\frac{3}{6} \times \frac{20,879}{3,888} + \frac{1}{6} \times \frac{3}{6} \times 13 + \frac{1}{6} \times \frac{4}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} \times 13\\ &\quad= \frac{62,637}{23,328} + \frac{156}{36}\\ &\quad= \frac{54,575}{7,776} = 7\frac{143}{7,776} \end{align*} \]
Sir R----- will clearly earn, on average, a bounty greater than his stake and I would therefore in good conscience have recommended that he take the first of the Baron's wagers.

In the second of the Baron's wagers he was to cast his die before Sir R----- began his play. Sir R-----'s strategy should therefore have been informed by the Baron's score.
For each of the Baron's possible scores we can figure the final cast that Sir R----- should countenance in exactly the same fashion as we did for the first game. For example, had the Baron cast a six, Sir R----- should have immediately cut his losses since he should have surely lost the wager no matter how many casts he took. If the Baron had instead cast a five, then on each cast Sir R-----'s should have had one chance in six of besting him. On his final cast he should therefore have expected a prize of
\[ \frac{1}{6} \times 13 = 2\frac{1}{6} \]
and, given that his third cast would have cost him two coins, he should have in no event elected to continue beyond a third cast, at which his expected net gain would have been one sixth part of a coin.
Should he not best the Baron on his second cast he should most certainly cast again and, given that the second cast costs one coin, his expected winnings for it are
\[ \frac{1}{6} \times 13 + \frac{5}{6} \times \frac{1}{6} - 1 = \frac{47}{36} = 1\frac{11}{36} \]
Similarly, for his first cast he should expect
\[ \frac{1}{6} \times 13 + \frac{5}{6} \times \frac{47}{36} = \frac{703}{216} = 3\frac{55}{216} \]
Now, if the Baron should have cast a four then Sir R----- would have had two chances in six of winning on each cast and would have consequently had, for his last roll, an average prize of
\[ \frac{2}{6} \times 13 = \frac{26}{6} = 4\frac{1}{3} \]
and should therefore not have cast his die more than five times.
Working backwards from his fifth cast we have
\[ \begin{align*} 5:\;&4\frac{1}{3}-4 = \frac{1}{3}\\ 4:\;&\frac{2}{6} \times 13 + \frac{4}{6} \times \frac{1}{3} - 3 = 1\frac{5}{9}\\ 3:\;&\frac{2}{6} \times 13 + \frac{4}{6} \times 1\frac{5}{9} - 2 = 3\frac{10}{27}\\ 2:\;&\frac{2}{6} \times 13 + \frac{4}{6} \times 3\frac{10}{27} - 1 = 5\frac{47}{81}\\ 1:\;&\frac{2}{6} \times 13 + \frac{4}{6} \times 5\frac{47}{81} = 8\frac{13}{243} \end{align*} \]
Next, we must consider the case of the Baron casting a three, for which Sir R----'s final expected winnings will be
\[ \frac{3}{6} \times 13 = \frac{39}{6} = 6\frac{1}{2} \]
and so his last cast should be his seventh. Working retrograde once again we have
\[ \begin{align*} 7:\;&6\frac{1}{2}-6 = \frac{1}{2}\\ 6:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times \frac{1}{2} - 5 = 1\frac{3}{4}\\ 5:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 1\frac{3}{4} - 4 = 3\frac{3}{8}\\ 4:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 3\frac{3}{8} - 3 = 5\frac{3}{16}\\ 3:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 5\frac{3}{16} - 2 = 7\frac{3}{32}\\ 2:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 7\frac{3}{32} - 1 = 9\frac{3}{64}\\ 1:\;&\frac{3}{6} \times 13 + \frac{3}{6} \times 9\frac{3}{64} = 11\frac{3}{128} \end{align*} \]
In the same fashion we can figure that should the Baron cast a two Sir R----- should throw his die no more than nine times and that his expected winnings should be
\[ \begin{align*} 9:\;&\frac{4}{6} \times 13 - 8 = \frac{2}{3}\\ 8:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times \frac{2}{3} - 7 = 1\frac{8}{9}\\ 7:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 1\frac{8}{9} - 6 = 3\frac{8}{27}\\ 6:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 3\frac{8}{27} - 5 = 4\frac{62}{81}\\ 5:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 4\frac{62}{81} - 4 = 6\frac{62}{243}\\ 4:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 6\frac{62}{243} - 3 = 7\frac{548}{729}\\ 3:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 7\frac{548}{729} - 2 = 9\frac{548}{2,187}\\ 2:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 9\frac{548}{2,187} - 1 = 10\frac{4,922}{6,561}\\ 1:\;&\frac{4}{6} \times 13 + \frac{2}{6} \times 10\frac{4,922}{6,561} = 12\frac{4,922}{19,683} \end{align*} \]
Finally and, given the growing complexity of the figures thus far, most dauntingly, we must reckon the value of the wager should the Baron have cast a one.
\[ \begin{align*} 11:\;&\frac{5}{6} \times 13 - 10 = \frac{5}{6}\\ 10:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times \frac{5}{6} - 9 = 1\frac{35}{36}\\ 9:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 1\frac{35}{36} - 8 = 3\frac{35}{216}\\ 8:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 3\frac{35}{216} - 7 = 4\frac{467}{1,296}\\ 7:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 4\frac{467}{1,296} - 6 = 5\frac{4,355}{7,776}\\ 6:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 5\frac{4,355}{7,776} - 5 = 6\frac{35,459}{46,656}\\ 5:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 6\frac{35,459}{46,656} - 4 = 7\frac{268,739}{279,936}\\ 4:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 7\frac{268,739}{279,936} - 3 = 9\frac{268,739}{1,679,616}\\ 3:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 9\frac{268,739}{1,679,616} - 2 = 10\frac{3,627,971}{10,077,696}\\ 2:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 10\frac{3,627,971}{10,077,696} - 1 = 11\frac{33,861,059}{60,466,176}\\ 1:\;&\frac{5}{6} \times 13 + \frac{1}{6} \times 11\frac{33,861,059}{60,466,176} = 12\frac{275,725,763}{362,797,056} \end{align*} \]
We have one last and most tedious calculation to perform; to figure Sir R-----'s expected bounty we must average these results
\[ \frac{1}{6} \times \left(0 + 3\frac{55}{216} + 8\frac{13}{243} + 11\frac{3}{128} + 12\frac{4,922}{19,683} + 12\frac{275,725,763}{362,797,056}\right) \]
That figuring the result of this average was a Herculean feat of arithmetic hardly bears mention! Have I not erred then Sir R----- should expect from this wager a bounty of
\[ 7\frac{1,937,927,123}{2,176,782,336} \]
and I could not therefore have recommended it to him for a stake of eight coins.

That the reckoning of the fairness of the Baron's second wager was substantially more difficult than that of his first is all the more curious if we consider a slight change to these games. Specifically, let us imagine that the Baron had shaken his die in a cup and upended it upon the table before Sir R----- began his play.
If he revealed his score after Sir R----- was done then the reckoning of the game would be identical to that of the first of the Baron's wagers; if before he began, then identical to that of the second.
The Baron's score in both games is determined at the outset; the only difference being when Sir R----- learns of it. That the earlier informed Sir R----- is of the state of play, the harder it is for him to determine the consequences of entering into the Baron's wager has caused the more philosophically minded of my fellow students no small consternation.
For my part, I am content; it is so and that is reason enough!
\(\Box\)


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

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