- #26

- 15,415

- 686

The solution R=1 is not just a mathematical artifact. The expression pRI'm not sure I understand mathematically why you can claim that R=1 is not the correct solution.

^{n}-R+1-p=0 has either two or zero positive solutions when p>0. Since R=1 is always a solution, this expression must have two positive solutions when p>0. One cannot reject the solution R=1 out-of-hand because this is the correct solution when p<1/n. In the case p<1/n, the other solution to pR

^{n}-R+1-p=0 is greater than one, which obviously is incorrect. When p=1/n, the solution R=1 is a double root. When p>1/n, the other solution is between 0 and 1.

Note that in the extreme case of p=1, the two solutions are R=0 and R=1. The R=0 solution is obviously correct in this extreme case. If the coin always comes up heads, the probability of ruin is 0, not 1.

This plus continuity suggests a way around the dilemma: The correct solution is the lesser of the two solutions to pR

^{n}-R+1-p=0.

This is consistent with my unwieldy sums. Those sums can be expressed in closed form for small n:

- n=2: ##R=\frac{1-|2p-1|}{2p}##, or

[tex]R=\begin{cases}

1 & p\le \frac 1 2 \\[4pt]

\frac{1-p} p & p > \frac 1 2

\end{cases}[/tex] - n=3: ##R = \frac{2\sin(\frac 1 3 \sin^{-1}(\frac 3 2 (1-p)\sqrt{3p}))}{\sqrt{3p}}##, or

[tex]R=\begin{cases}

1 & p\le \frac 1 3 \\[4pt]

\frac{\sqrt{(4-3p)p}-p} {2p} & p > \frac 1 3

\end{cases}[/tex]