Lowest score never made In cricket, a player's score can reach into the hundreds, and the lowest score never made was 229 (as of 2004).  Players have made higher scores, and if anyone made 229 exactly, the new lowest-never-made would be 238.
Assume that a player's score is a Poisson process with parameter $\lambda$.  What proportion of scores will ever be the lowest-never-made?  Or, what is the limiting probability that $N$ is ever the lowest score never made?
This is just idle wondering while cricket takes a break for rain.
 A: Let $X_n$ be the $n$'th score ever recorded (in chronological order).  I'm assuming all $X_n$ are independent Poisson($\lambda$) random variables. 
Let $p(x) = P(X_n = x) = e^{-\lambda} \lambda^x/x!$. For nonnegative integers $s$ let $T(s)$ be the first $n$ for which
$X_n = s$  (with probability $1$ there will be some).  Then $s$ is at some time the lowest score never made if and only if $T(s) > T(t)$ for all positive integers $t < s$.  For any permutation $\pi$ of $\{0,1,\ldots,s\}$, let
$M(\pi)$ be the probability that $T(\pi_0) < T(\pi_1) < \ldots < T(\pi_s)$.
It is easily seen that 
$$M(\pi) = \frac{p(\pi_0)}{p(\pi_0) + \ldots + p(\pi_s)} \frac{p(\pi_1)}{p(\pi_1)+ \ldots + p(\pi_s)} \ldots \frac{p(\pi_{s-1})}{p(\pi_{s-1}) + p(\pi_s)} $$ 
Then the probability $P(A_s)$ that $s$ is at some time the lowest score ever made
is the sum of $M(\pi)$ for all permutations of $\{0,1,\ldots,s\}$ that end
in $s$.  So according to my calculations (with Maple's help)
$$\eqalign{P(A_0) &= M([0]) = 1\cr
P(A_1) &= M([01]) = \dfrac{p(0) }{(p(0)+p(1))} = \dfrac{\lambda}{\lambda+1}\cr
P(A_2) &=  M([012]) + M([102]) = {\frac {8 \left({\lambda}^{2}+\lambda+1\right)}{ \left( {\lambda}^{2}+2\,\lambda+2
 \right)  \left( \lambda+2 \right)  \left( {\lambda}^{2}+2 \right) }}
\cr
P(A_3) &= M([0123]) + M([0213]) + M([1023]) + M([1203]) + M([2013]) + M([2103])\cr
&= {\frac { 648\left( {\lambda}^{5}+3\,{\lambda}^{4}+6\,{\lambda}^{3}+
15\,{\lambda}^{2}+18\,\lambda+18 \right)  \left( {\lambda}^{6}+3\,{
\lambda}^{5}+9\,{\lambda}^{4}+15\,{\lambda}^{3}+12\,{\lambda}^{2}+18\,
\lambda+12 \right) }{ \left( {\lambda}^{3}+3\,{\lambda}^{2}+6\,\lambda
+6 \right)  \left( {\lambda}^{2}+3\,\lambda+6 \right)  \left( \lambda+
3 \right)  \left( {\lambda}^{2}+6 \right)  \left( {\lambda}^{3}+3\,{
\lambda}^{2}+6 \right)  \left( {\lambda}^{3}+6 \right)  \left( {
\lambda}^{3}+6\,\lambda+6 \right) }}
\cr}
$$
etc.
A: Well, I think I can make a worthy crack at this, but I can't really solve the problem.
The probability that the lowest never score after $n$ rounds is $0$ is the probability that nobody ever scores $0$. The probability that the lowest never score after $n$ rounds is $1$ is the probability that someone has scored $0$, and nobody has ever scored $1$. We needn't bother with the other scores. The probability that the lowest never score after $n$ rounds is $s$ is the probability that someone has scored $0, 1, 2, ..., s-1$, and that nobody has scored $s$. The other scores are irrelevant.
We'll let $L_n$ be the random variable denoting the lowest never score after $n$ matches, $S_n(i)$ denote the event that at least someone has scored exactly $i$ runs in the past $n$ matches, $A^c$ denote the set complement of $A$, and $p_\lambda(i)$ be the probability that a $Po(\lambda)$ random variable takes the value $i$. Trivially,
$$
\mathbb{P}(S_n(i)) = (1- p_\lambda(i))^n
$$
Now,
$$
\mathbb{P}(L_n = i) = \mathbb{P}[S_n(0) \cap S_n(1) \cap ... \cap S_n(i-1) \cap S_n(i)^c] 
$$
The probability that the lowest never score is $i$ is the probability that all scores up to $i$ have been attained and $i$ has not been attained. Further,
$$
\mathbb{P}(L_n = i) = \mathbb{P}[S_n(0)]\mathbb{P}[ S_n(1) \cap ... \cap S_n(i-1) \cap S_n(i)^c \;\;|\;\; S_n(0)]
$$
Given $S_n(0)$, we know that at least one poor sod in the entire history of cricket must have scored $0$. The effect of this conditioning does nothing but fix exactly one of our iid random variables to $0$. The probability that at least one person has scored $i\neq 0$ runs in $n$ matches given that at least one has scored $0$ is exactly the probability that at least one person has scored $i$ runs in $n-1$ matches, so...
$$
\mathbb{P}(L_n = i) = \mathbb{P}[S_n(0)]\mathbb{P}[ S_{n-1}(1) \cap ... \cap S_{n-1}(i-1) \cap S_{n-1}(i)^c]
$$
We can repeat this process a few times, and find that
$$
\mathbb{P}(L_n = i) = \mathbb{P}[S_n(0)]\mathbb{P}[S_{n-1}(1)]...\mathbb{P}[S_{n-i+1}(i-1)] \mathbb{P}[S_{n-i}(i)^c]
$$
Each of these factors is known from above, and making those substitutions yields the following minor mess:
$$
\mathbb{P}(L_n = i) = (1-p_\lambda(0))^n...(1-p_\lambda(i-1))^{n-i+1} (1-(1-p_\lambda(i))^{n-i})
$$
So, from here, we have a value for the probability that the lowest score never made from a set of $n$ matches is $i$. The behaviour as $n$ and $i$ vary is something of a mystery to me. Finding the probability that a particular score is ever the lowest never score would involve accounting for some form of ordering in these random variables, which I'm not sure how to deal with elegantly.
