The Baron and Sir R-----'s latest wager comprised of first placing a draught piece upon the fifth lowest of a column of twelve squares and subsequently moving it up or down by one space depending upon the outcome of a coin toss until such time as it should escape, either by moving above the topmost or below the bottommost square. In the former outcome the Baron should have had a prize of three coins and in the latter Sir R----- should have had two.

Since the draught has an even chance of retreating or advancing after each toss of the coin, the probability \(p_n\) that Sir R----- should ultimately win the game if the draught is upon the \(n^\mathrm{th}\) square is

Specifically, figuring the odds of Sir R----- taking the game from the sixth rank yields

We can reckon the expected number of coin tosses \(t_n\) until the game concludes from the \(n^\mathrm{th}\) rank by noting that it is one more than the average expected number from those immediately before and after it

Since the draught has an even chance of retreating or advancing after each toss of the coin, the probability \(p_n\) that Sir R----- should ultimately win the game if the draught is upon the \(n^\mathrm{th}\) square is

\[
p_n = \tfrac12 \times p_{n-1} + \tfrac12 \times p_{n+1}
\]

with the first and last squares instead giving him the chances
\[
\begin{align*}
p_1 &= \tfrac12 \times 1 + \tfrac12 \times p_2\\
p_{12} &= \tfrac12 \times p_{11} + \tfrac12 \times 0
\end{align*}
\]

Noting that the Baron has the same chance of winning when the draught is \(n\) squares from the last rank as does Sir R----- when it is \(n\) squares from the first, we also have the symmetry
\[
p_{13-n} = 1 - p_n
\]

We may consequently iteratively express the odds of Sir R----- emerging victorious from each of the first five ranks in terms of that of the sixth, which we may ultimately find by forming simultaneous equations from the derived odds at the first and those stated above.Specifically, figuring the odds of Sir R----- taking the game from the sixth rank yields

\[
\begin{align*}
p_6 &= \tfrac12 \times p_5 + \tfrac12 \times p_7\\
2 \times p_6 &= p_5 + p_7 = p_5 + \left(1-p_6\right)\\
p_5 &= 3 \times p_6 - 1
\end{align*}
\]

From the odds at the fifth we can resolve those at the fourth with
\[
\begin{align*}
2 \times p_5 &= p_4 + p_6\\
p_4 &= 2 \times p_5 - p_6 = 2 \times \left(3 \times p_6 - 1\right) - p_6 = 5 \times p_6 - 2
\end{align*}
\]

Likewise, we can find the odds at the third with
\[
p_3 = 2 \times p_4 - p_5 = 2 \times \left(5 \times p_6 - 2\right) - \left(3 \times p_6 - 1\right) = 7 \times p_6 - 3
\]

and those at the second
\[
p_2 = 2 \times p_3 - p_4 = 2 \times \left(7 \times p_6 - 3\right) - \left(5 \times p_6 - 2\right) = 9 \times p_6 - 4
\]

and, finally, those at the first
\[
p_1 = 2 \times p_2 - p_3 = 2 \times \left(9 \times p_6 - 4\right) - \left(7 \times p_6 - 3\right) = 11 \times p_6 - 5
\]

Now, from our initial statement of the probabilities, we also know that
\[
2 \times p_1 = 1 + p_2 = 1 + \left(9 \times p_6 - 4\right) = 9 \times p_6 - 3
\]

and so
\[
\begin{align*}
22 \times p_6 - 10 &= 9 \times p_6 - 3\\
13 \times p_6 &= 7\\
p_6 &= \frac{7}{13}
\end{align*}
\]

Putting this into our formula for the chance of Sir R-----'s victory from the fifth rank yields
\[
p_5 = 3 \times p_6 - 1 = \frac{21}{13} - \frac{13}{13} = \frac{8}{13}
\]

so that his expected prize is
\[
\frac{8}{13} \times 2 - \frac{5}{13} \times 3 = \frac{16}{13} - \frac{15}{13} = \frac{1}{13}
\]

and I should have advised him to accept the Baron's wager, although only if he had sufficient patience since the game would likely take some time!We can reckon the expected number of coin tosses \(t_n\) until the game concludes from the \(n^\mathrm{th}\) rank by noting that it is one more than the average expected number from those immediately before and after it

\[
\begin{align*}
t_n &= 1 + \tfrac12 \times \left(t_{n-1} + t_{n+1}\right)\\
t_{n-1} &= 2 \times t_n - t_{n+1} - 2
\end{align*}
\]

and, similarly, that from the first rank is
\[
t_1 = 1 + \tfrac12 \times \left(0 + t_2\right)
\]

The symmetry of the game means that we should expect it to conclude after the same number of moves from equal distances to the first and last ranks
\[
t_{13-n} = t_n
\]

yielding
\[
\begin{align*}
t_5 &= 2 \times t_6 - t_7 - 2 = 2 \times t_6 - t_6 - 2 = t_6 - 2\\
t_4 &= 2 \times t_5 - t_6 - 2 = 2 \times \left(t_6 - 2\right) - t_6 - 2 = t_6 - 6\\
t_3 &= 2 \times t_4 - t_5 - 2 = 2 \times \left(t_6 - 6\right) - \left(t_6 - 2\right) - 2 = t_6 - 12\\
t_2 &= 2 \times t_3 - t_4 - 2 = 2 \times \left(t_6 - 12\right) - \left(t_6 - 4\right) - 2 = t_6 - 22\\
t_1 &= 2 \times t_2 - t_3 - 2 = 2 \times \left(t_6 - 22\right) - \left(t_6 - 12\right) - 2 = t_6 - 34
\end{align*}
\]

From the expected number of moves from the first rank we have
\[
2 \times t_1 = 2 + t_2
\]

so that
\[
\begin{align*}
2 \times t_6 - 68 &= t_6 - 20\\
t_6 &= 48
\end{align*}
\]

and
\[
t_5 = t_6 - 2 = 46
\]

I explained my line of reasoning to the Baron, but I fear that he may not have entirely understood it!
\(\Box\)

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