Once again, you are a frog in a pond. But this time, the pond has infinitely many lily pads, which are numbered “1,” “2,” “3,” etc. As before, you are currently on pad $2$, and your goal is to make it to pad $1$, which you would happily stay on forever.
Whenever you are on pad $k$, you will hop to pad $k−1$ with probability $1/k$, and you will hop to pad $k+1$ with probability $(k−1)/k.$
Now, what is the probability that you will ultimately make it to pad $1$?
Taking off where we landed from the Classic Fiddler, let's expand the $4$ lily pad Markov chain to an $n$ state Markov chain and then see what happens when $n \to \infty.$ Skipping right to the canonical form of the transition matrix we get \begin{align*}\hat{T}_n &= \begin{pmatrix}
0 & \frac{1}{n-1} & 0 & 0 & \cdots & 0 & 0 & 0 & 0 & \frac{n-2}{n-1}\\
\frac{n-3}{n-2} & 0 & \frac{1}{n-2} & 0 & \cdots & 0 & 0 & 0 & 0 & 0 \\
0 & \frac{n-4}{n-3} & 0 & \frac{1}{n-3} & \cdots & 0 & 0 & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & \frac{2}{3} & 0 & \frac{1}{3} & 0 & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 & 1
\end{pmatrix} \\ &= \begin{pmatrix} Q & R \\ 0 & I \end{pmatrix}.\end{align*} Again, we have the fundamental matrix $N = (I - Q)^{-1}$ and $P = NR,$ where here can focus of wanting $P_{n-2,1}.$ Instead of trying to code up an $(n-2) \times (n-2)$ matrix inversion, let's instead rewrite the matrix equation as a tridiagonal system of equations $$(I-Q) x = \begin{pmatrix}
1 & -\frac{1}{n-1} & 0 & 0 & \cdots & 0 & 0 & 0 \\
-\frac{n-3}{n-2} & 1 & -\frac{1}{n-2} & 0 & \cdots & 0 & 0 & 0 \\
0 & -\frac{n-4}{n-3} & 1 & -\frac{1}{n-3} & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & -\frac{2}{3} & 1 & -\frac{1}{3} & \\
0 & 0 & 0 & 0 & \cdots & 0 & -\frac{1}{2} & 1 \\ \end{pmatrix} x = \frac{1}{2} e_{n-2},$$ and use the Thomas algorithm to find the probability of ending up on pad $1$ forever in the case of $n$ lily pads as $p_{1,n} = x_{n-2} = P_{n-2,1}.$ In particular, the main diagonal entries are $b_i = 1, i = 1, 2, \dots, n-2.$ The upper diagonal is $c_i = -\frac{1}{n-i}, i = 1, 2, \dots, n-3$ and lower diagonal are $a_i = -\frac{n-i-1}{n-i}, i = 2, 3, \dots, n-2.$ The Thomas algorithm consists of two sweeps, first $c^\prime\in \mathbb{R}^{n-3}$ and $d^\prime \in \mathbb{R}^{n-2}$ and then another that starts with solving for $x_{n-2}$ and then sweeps backward from $i = n-1$ to $i=1.$ In particular, we define $c^\prime_1 = c_1 = -\frac{1}{n-1}$ and $$c^\prime_i = \frac{c_i}{1 - a_ic^\prime_{i-1}}, i = 2, \dots, n-3$$ and $d^\prime_1 = d_1,$ and $$d^\prime_i = \frac{d_i - a_id^\prime_{i-1}}{1 - a_ic^\prime_{i-1}}, i = 2, 3, \dots, n-2,$$ and finally $x_{n-2} = d^\prime_{n-2}$ and $$x_i = d^\prime_i - c^\prime_i x_{i+1}, i = n-1, n-2, \dots, 1.$$ Additionally, since $d_i = 0, i = 1, \dots, n-3,$ we have $d^\prime_i = 0, i = 1, \dots, n-3,$ with $$p_{1,n} = P_{n-2,1} = x_{n-2} = d^\prime_{n-2} = \frac{\frac{1}{2}}{1 - \left(-\frac{1}{2}\right) c^\prime_{n-3}} = \frac{1}{2 + c^\prime_{n-3}},$$ and since we only care about $x_{n-2},$ we don't have to do any of the backward sweep.
So we can simply code up one recursive function that sweeps through the $c^\prime_i$ variables for any $i = 1, \dots, n-3,$ for instance the following Python script:
def prob_zero(N):
c = -1.0 / float(N - 1.)
for i in range(2, N-2):
c = (-1.0 / float(N - i)) / (1. - (-1. + 1.0 / float(N - i)) * c)
return (0.5 / (1 + 0.5 * c))
We can just get the values of the probability of absorption at pad $1$ for many, many values of $n$ and emprically see that as $n$ gets large we get close (fairly quickly to $\lim_{n \to \infty} p_{1,n} = 63.21205588285577....\%,$ which is disturbingly close to $1 - e^{-1}.$ So much so, that if we wanted we could likely call it quits here and claim that the probability of ending up forever on pad $1$ in the inifinite lily pad case is $p_{1,\infty} = 1 - e^{-1}.$
| # of Lily Pads |
Absorption Prob. |
| 4 | 60% |
| 5 | 62.5% |
| 6 | 63.07692307692307% |
| 7 | 63.19018404907976% |
| 8 | 63.20899335717935% |
| 9 | 63.21167883211679% |
| 10 | 63.21201448891889% |
| 11 | 63.21205178374104% |
| 12 | 63.21205551321910% |
| 13 | 63.21205585226252% |
| 14 | 63.21205588051614% |
| 15 | 63.21205588268949% |
| 16 | 63.21205588284473% |
| 17 | 63.21205588285508% |
| 18 | 63.21205588285572% |
| 19 | 63.21205588285577% |
| 20 | 63.21205588285577% |
However, perhaps we can do better than just our anecdotal, experimental evidence. Let's start at $$p_{1,n} = \frac{1}{2+c^\prime_{n-3}(n)}.$$ Using the recursive formula, we get $$c^\prime_{n-3}(n) = \frac{c_{n-3}}{1 - a_{n-3} c^\prime_{n-4}(n)} = \frac{-\frac{1}{3}}{1 + \frac{2}{3} c^\prime_{n-4}(n)} = -\frac{1}{3 + 2 c^\prime_{n-4}(n)}.$$ Going a step further we get $$c^\prime_{n-4}(n) = \frac{ -\frac{1}{4} }{1 - \frac{3}{4} c^\prime_{n-5}(n)} = -\frac{1}{4 + 3c^\prime_{n-5}(n)},$$ which plugging back into the previous equation gives $$c^\prime_{n-3}(n) = - \frac{1}{3 + 2 \frac{ -1 }{4 + 3 c^\prime_{n-5}(n)} } = - \frac{4 + 3 c^\prime_{n-5}(n)}{10 + 9 c^\prime_{n-5}(n)}.$$ We see that we can define $c^\prime_{n-3}(n)$ as a rational function of $c^\prime_{n-k}(n),$ for each $k = 4, \dots, n-1.$ In particular, let's say that $$c^\prime_{n-3}(n) = -\frac{ \alpha_k + (\alpha_k - 1) c^\prime_{n-k}(n) }{\beta_k + (\beta_k - 1) c^\prime_{n-k}(n)},$$ where $\alpha_4 = 1$ and $\beta_4 = 3.$ From the recursion formula for $c^\prime$ we get that $$c^\prime_{n-3}(n) = - \frac{\alpha_k + (\alpha_k - 1) \frac{ - 1 }{k + (k-1) c^\prime_{n-k-1}(n)} }{ \beta_k + (\beta_k - 1) \frac{ -1}{ k + (k-1) c^\prime_{n-k-1}(n)}} = -\frac{ k \alpha_k - (\alpha_k - 1) + (k-1) \alpha_k c^\prime_{n-k-1}(n)} { k \beta_k - (\beta_k - 1) + (k-1) \beta_k c^\prime_{n-k-1}(n)},$$ so we can derive the recursion formulae for $\alpha_k$ and $\beta_k$ as \begin{align*}\alpha_{k+1} &= (k-1) \alpha_k + 1 \\ \beta_{k+1} &= (k-1) \beta_k + 1,\end{align*} for all $k \geq 4,$ with initial conditions $\alpha_4 = 1$ and $\beta_4 = 3.$
So, trying to organize the situation a bit, we notice that since $c_1^\prime(n) = -\frac{1}{n-1}$ we would then have to stop the recursion, we get $$c^\prime_{n-3}(n) = - \frac{ \alpha_{n-1} - \frac{\alpha_{n-1} - 1}{n-1} }{ \beta_{n-1} - \frac{\beta_{n-1} - 1}{n-1} } = \frac{ \alpha_{n-1} (n-2) + 1 }{\beta_{n-1} (n-2) + 1},$$ where $\alpha_{n-1}$ and $\beta_{n-1}$ can be read from the recursion formulae above. Plugging this back into the absorption probability, we see that \begin{align*}p_{1,n} &= \frac{1}{2 + c_{n-3}^\prime(n)} \\ &= \frac{1}{2 - \frac{ \alpha_{n-1} (n-2) + 1}{ \beta_{n-1} (n-2) + 1} } \\ & = \frac{ \beta_{n-1} (n-2) + 1 }{ (2 \beta_{n-1} - \alpha_{n-1}) (n-2) + 1 } \\ &= \frac{ \beta_{n-1} + O (n^{-1}) }{ (2 \beta_{n-1} - \alpha_{n-1}) + O(n^{-1}) }.\end{align*} Now, let's turn our attention to the growth of $\alpha_{n-1}$ and $\beta_{n-1}.$
Given the recursion formulae and initial conditions, we can show that in fact $$\alpha_k = \lfloor (e-2) (k-2)! \rfloor$$ and $$\beta_k = \lfloor (e-1) (k-2)! \rfloor$$ for all $k \geq 4.$ Clearly, the initial conditions hold since $\lfloor (e-2) \cdot (4-2)! \rfloor = \lfloor 1.436\dots \rfloor = 1$ and $\lfloor (e-1) \cdot (4-2)! \rfloor = \lfloor 3.436\dots \rfloor = 3.$ Assume that for some $j \geq 4$ that \begin{align*}\alpha_j &= \lfloor (e-2) (j-2)! \rfloor \\ \beta_j &= \lfloor (e-1) (j-2)! \rfloor.\end{align*} From the recursion formula, we get \begin{align*} \alpha_{j+1} &= (j-1) \alpha_j + 1 = (j-1) \lfloor (e-2) (j-2)! \rfloor + 1 \\ \beta_{j+1} &= (j-1) \beta_j + 1 = (j-1) \lfloor (e-1) (j-2)! \rfloor + 1.\end{align*} Now at this point we will rely on the following lemma (that can be proven using an application of the Taylor remainder theorem) that states that for any $\ell = 2, \dots, \infty,$ $$\sum_{k = \ell}^\infty \frac{1}{k!} \lt \frac{1}{(\ell - 1)!}.$$ So using this lemma we see that \begin{align*}\lfloor e (j-1)! \rfloor &= \left\lfloor \sum_{k=0}^{j-1} \frac{(j-1)!}{k!} + (j-1)! \sum_{k=j}^\infty \frac{1}{k!} \right\rfloor \\ &= (j-1)! \sum_{k=0}^{j-1} \frac{1}{k!} \\ &= (j-1)! \sum_{k=0}^{j-2} \frac{1}{k!} + 1 \\ &= (j-1) \left( (j-2)! \sum_{k=0}^{j-2} \frac{1}{k!} \right) + 1 \\ &= (j-1) \lfloor e (j-2)! \rfloor + 1.\end{align*} So plugging back into the recursion formulae above, since $j!$ and $2 \cdot j!$ are always integers for any $j,$ we confirm the induction step, since \begin{align*} \alpha_{j+1} &= (j-1) \lfloor (e-2) (j-2)! \rfloor + 1 = \lfloor (e-2) (j-1) ! \rfloor \\ \beta_{j+1} &= (j-1) \lfloor (e-1) (j-2)! \rfloor + 1 = \lfloor (e-1) (j-1)! \rfloor,\end{align*} thus proving the formulae $$ \alpha_k = \lfloor (e-2) (k-2)! \rfloor, \beta_k = \lfloor (e-1) (k-2)! \rfloor, \,\, \forall k \geq 4.$$
Finally, returning our attention to the formula for the absorption probability we get \begin{align*}p_{1,n} &= \frac{ \beta_{n-1} + O (n^{-1}) }{ (2 \beta_{n-1} - \alpha_{n-1}) + O(n^{-1})}\\ &= \frac{ \lfloor (e-1) (n-2)! \rfloor + O(n^{-1}) }{ ( 2 \lfloor (e-1) (n-2)! \rfloor - \lfloor (e-2) (n-2)! \rfloor + O(n^{-1}) } \\ &= \frac{ (n-2)! \sum_{k=1}^{n-2} \frac{1}{k!} + O(n^{-1}) }{ (n-2)! \left( 2 \sum_{k=1}^{n-2} \frac{1}{k!} - \sum_{k=2}^{n-2} \frac{1}{k!} \right) + O(n^{-1})} \\ &= \frac{ \sum_{k=1}^{n-2} \frac{1}{k!} + O\left( \frac{1}{ (n-1)!}\right) }{ \left(2 \sum_{k=1}^{n-2} \frac{1}{k!} - \sum_{k=1}^{n-2} \frac{1}{k!} \right) + O \left(\frac{1}{(n-1)!}\right)},\end{align*} so we have that the probability of ending up on pad $1$ in the infinite lily pad problem is \begin{align*}p_{1,\infty} = \lim_{n \to \infty} p_{1,n} &= \lim_{n\to \infty} \frac{ \sum_{k=1}^{n-2} \frac{1}{k!} + O\left( \frac{1}{ (n-1)!}\right)}{ \left(2 \sum_{k=1}^{n-2} \frac{1}{k!} - \sum_{k=1}^{n-2} \frac{1}{k!} \right) + O\left( \frac{1}{ (n-1)!}\right)}\\ &= \frac{\sum_{k=1}^\infty \frac{1}{k!}}{ 2 \sum_{k=1}^\infty \frac{1}{k!} - \sum_{k=2}^\infty \frac{1}{k!}} \\ &= \frac{e-1}{2(e-1) - (e-2)} = 1 - e^{-1},\end{align*} as experimentally derived.
