On A Day At The Races

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Most recently the Baron challenged Sir R----- to a race of knights around the perimeter of a chessboard, with the Baron starting upon the lower right hand square and Sir R----- upon the lower left. The chase proceeded anticlockwise with the Baron moving four squares at each turn and Sir R----- by the roll of a die. Costing Sir R----- one cent to play, his goal was to catch or overtake the Baron before he reached the first rank for which he would receive a prize of forty one cents for each square that the Baron still had to traverse before reaching it.

In reckoning Sir R-----'s expected winnings our first observation is that on the \(n^\mathrm{th}\) turn the Baron's knight would be at
\[ b_n = 4 \times n + 7 \]
and consequently returned to the first rank for
\[ \begin{align*} 4 \times n + 7 &\geqslant 28\\ n &\geqslant 5\tfrac14 \end{align*} \]
The greatest ground that Sir R----- can cover is six squares per turn and so cannot catch the Baron until
\[ \begin{align*} 6 \times n &\geqslant 4 \times n + 7\\ n &\geqslant 3\tfrac12 \end{align*} \]
Sir R----- will have caught the Baron with a roll of five or six if he were one square behind him and with a roll of six if he were two, so that
\[ \Pr\left(r_n \geqslant b_n\right) = \Pr\left(r_{n-1}=b_{n-1}-1\right) \times \tfrac13 + \Pr\left(r_{n-1}=b_{n-1}-2\right) \times \tfrac16 \]
The probability that Sir R----- will have reached the \(k^\mathrm{th}\) square upon the third turn is given by
\[ \Pr\left(r_3=k\right) = \frac{N_3(k)}{6^3} \]
where \(N_3(k)\) is the number of ways that three numbers between one and six can sum to \(k\).
\[ \begin{align*} N_3(3) &= N\left(\{1,1,1\}\right) = 1\\ N_3(4) &= N\left(\{1,1,2\}\right) = 3\\ N_3(5) &= N\left(\{1,1,3\}\right)+N\left(\{1,2,2\}\right) = 3+3 = 6\\ N_3(6) &= N\left(\{1,1,4\}\right)+N\left(\{1,2,3\}\right)+N\left(\{2,2,2\}\right) = 3+6+1 = 10\\ N_3(7) &= N\left(\{1,1,5\}\right)+N\left(\{1,2,4\}\right)+N\left(\{1,3,3\}\right)+N\left(\{2,2,3\}\right) = 3+6+3+3 = 15\\ N_3(8) &= N\left(\{1,1,6\}\right)+N\left(\{1,2,5\}\right)+N\left(\{1,3,4\}\right)+N\left(\{2,2,4\}\right)+N\left(\{2,3,3\}\right)\\ &= 3+6+6+3+3 = 21\\ N_3(9) &= N\left(\{1,2,6\}\right)+N\left(\{1,3,5\}\right)+N\left(\{1,4,4\}\right)+N\left(\{2,2,5\}\right)+N\left(\{2,3,4\}\right)+N\left(\{3,3,3\}\right)\\ &= 6+6+3+3+6+1 = 25\\ N_3(10) &= N\left(\{1,3,6\}\right)+N\left(\{1,4,5\}\right)+N\left(\{2,2,6\}\right)+N\left(\{2,3,5\}\right)+N\left(\{2,4,4\}\right)+N\left(\{3,3,4\}\right)\\ &= 6+6+3+6+3+3 = 27 \end{align*} \]
We can recover the figures for sums above ten by exploiting the symmetry
\[ N_3\left(21-k\right) = N_3\left(k\right) \]
yielding the probabilities
\[ \begin{align*} \Pr\left(r_3=3\right) &= \Pr\left(r_3=18\right) = \tfrac{1}{216}\\ \Pr\left(r_3=4\right) &= \Pr\left(r_3=17\right) = \tfrac{1}{72}\\ \Pr\left(r_3=5\right) &= \Pr\left(r_3=16\right) = \tfrac{1}{36}\\ \Pr\left(r_3=6\right) &= \Pr\left(r_3=15\right) = \tfrac{5}{108}\\ \Pr\left(r_3=7\right) &= \Pr\left(r_3=14\right) = \tfrac{5}{72}\\ \Pr\left(r_3=8\right) &= \Pr\left(r_3=13\right) = \tfrac{7}{72}\\ \Pr\left(r_3=9\right) &= \Pr\left(r_3=12\right) = \tfrac{25}{216}\\ \Pr\left(r_3=10\right) &= \Pr\left(r_3=11\right) = \tfrac{1}{8} \end{align*} \]
The probability that Sir R----- will win upon the fourth turn is consequently
\[ \Pr\left(r_4 \geqslant b_4\right) = \Pr\left(r_3=18\right) \times \tfrac13 + \Pr\left(r_3=17\right) \times \tfrac16 = \tfrac{1}{216} \times \tfrac13 + \tfrac{1}{72} \times \tfrac16 = \tfrac{5}{1,296} \]
and that he will do so upon the fifth
\[ \Pr\left(r_5 \geqslant b_5\right) = \Pr\left(r_4=22\right) \times \tfrac13 + \Pr\left(r_4=21\right) \times \tfrac16 \]
To calculate the probabilities that Sir R-----'s knight will be upon the twenty second or twenty first squares at the fourth turn we must use the convolution
\[ \Pr\left(r_4 = k \right) = \sum_{i=1}^6 \Pr\left(r_3 = k-i \right) \times \tfrac16 \]
where \(\sum\) is the summation sign. This is because some of the values of \(k-i\) would imply that Sir R----- had already caught the Baron. Specifically \(r_3\) cannot be greater than eighteen, yielding
\[ \begin{align*} \Pr\left(r_4=22\right) &= \Pr\left(r_3=21\right) \times \tfrac16 + \Pr\left(r_3=20\right) \times \tfrac16 + \Pr\left(r_3=19\right) \times \tfrac16\\ &+ \Pr\left(r_3=18\right) \times \tfrac16 + \Pr\left(r_3=17\right) \times \tfrac16 + \Pr\left(r_3=16\right) \times \tfrac16\\ &= 0 \times \tfrac16 + 0 \times \tfrac16 + 0 \times \tfrac16 + \tfrac{1}{216} \times \tfrac16 + \tfrac{1}{72} \times \tfrac16 + \tfrac{1}{36} \times \tfrac16 = \tfrac{5}{648} \end{align*} \]
and
\[ \begin{align*} \Pr\left(r_4=21\right) &= \Pr\left(r_3=20\right) \times \tfrac16 + \Pr\left(r_3=19\right) \times \tfrac16 + \Pr\left(r_3=18\right) \times \tfrac16\\ &+ \Pr\left(r_3=17\right) \times \tfrac16 + \Pr\left(r_3=16\right) \times \tfrac16 + \Pr\left(r_3=15\right) \times \tfrac16\\ &= 0 \times \tfrac16 + 0 \times \tfrac16 + \tfrac{1}{216} \times \tfrac16 + \tfrac{1}{72} \times \tfrac16 + \tfrac{1}{36} \times \tfrac16 + \tfrac{5}{108} \times \tfrac16 = \tfrac{5}{324} \end{align*} \]
so that
\[ \Pr\left(r_5 \geqslant b_5\right) = \tfrac{5}{648} \times \tfrac13 + \tfrac{5}{324} \times \tfrac16 = \tfrac{5}{972} \]
The expected number of squares that the Baron will have left to traverse is therefore
\[ \Pr\left(r_4 \geqslant b_4\right) \times \left(28-b_4\right) + \Pr\left(r_5 \geqslant b_5\right) \times \left(28-b_5\right) = \tfrac{5}{1,296} \times 5 + \tfrac{5}{972} \times 1 = \tfrac{95}{3,888} \]
and so Sir R-----'s expected winnings are
\[ \tfrac{95}{3,888} \times 41 - 1 = \tfrac{3,895-3,888}{3,888} = \tfrac{7}{3,888} \]
and I should have advised him to take on the Baron!
\(\Box\)


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

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