In the Baron's most recent game Sir R----- was to place a total of between four and eight coins upon four spaces after which the Baron would give him two coins and do likewise. A four sided die was then to be cast to determine who should have won the wager, with he having the most coins upon the space thus indicated taking all of the coins upon the table. In the event of a draw, each should have had his own coins back from it.

To figure the fairness of this wager we must determine Sir R-----'s most effective strategy for placing his coins. It most certainly could not have been to place just four since the Baron could then have placed one more than he upon each space to ensure a prize of four coins for his cost of two.

If Sir R----- were to have placed \(n\) coins in total and three or more coins upon any particular space then the Baron could have placed the same number of coins as Sir R----- upon each of his spaces, removed all of them from the most fortified and add one or more to each of the others for a three in four chance of victory and an expected prize of

Now if Sir R----- were to have placed no more than two coins upon any space, but less than two upon one or more of them then the Baron could have again mirrored Sir R-----'s play and then empty a space containing two coins and have had sufficient remaining to have added one to his remaining spaces to the same effect.

Sir R-----'s most effective strategy should therefore have been to have placed two coins upon each space, which the Baron should have countered by placing zero, two, three and three upon his for expected winnings of

To figure the fairness of this wager we must determine Sir R-----'s most effective strategy for placing his coins. It most certainly could not have been to place just four since the Baron could then have placed one more than he upon each space to ensure a prize of four coins for his cost of two.

If Sir R----- were to have placed \(n\) coins in total and three or more coins upon any particular space then the Baron could have placed the same number of coins as Sir R----- upon each of his spaces, removed all of them from the most fortified and add one or more to each of the others for a three in four chance of victory and an expected prize of

\[
\tfrac{1}{4} \times -n + \tfrac{3}{4} \times n = \tfrac{1}{2}n
\]

Given that we have already determined that \(n\) must be greater than four this represents a profit for the Baron over and above the two coins that he should have surrendered and Sir R----- should consequently not have placed more than two coins upon any of his spaces.Now if Sir R----- were to have placed no more than two coins upon any space, but less than two upon one or more of them then the Baron could have again mirrored Sir R-----'s play and then empty a space containing two coins and have had sufficient remaining to have added one to his remaining spaces to the same effect.

Sir R-----'s most effective strategy should therefore have been to have placed two coins upon each space, which the Baron should have countered by placing zero, two, three and three upon his for expected winnings of

\[
\tfrac{1}{4} \times -8 + \tfrac{1}{4} \times 0 + \tfrac{1}{2} \times 8 = -2 + 0 + 4 = 2
\]

This being exactly the sum that the Baron should have given Sir R----- before placing his coins, the game is perfectly fair and I should have advised Sir R----- to have no concern in taking this wager provided that he gave due thought to his strategy.
\(\Box\)

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