Probability. White and black sheeps Of these days I met an interesting theoretical probability problem.
There is a flock of sheep, which initially included one white and one black sheep. Every day, the following procedure: choose randomly one sheep from the flock, and remember its color. Then it returned to the flock and add there a new one black sheep.
If any of the selected days choosing one is a white sheep, and the previous day took place the same, it is decided to make it a barbecue. 
So the quiestion is, what is the probability of such an event with an unlimited repetition of the procedure?
Added: make it a barbecue means that when white sheep comes two days in a row, the experiment is over.
 A: (Edit: The first version of this answer, while correct, was incomplete.)
Denote by $b_n$ the probability that you reach a state with $n$ sheep whereby the last picked sheep was black, and by $w_n$ the probability that you reach a state with $n$ sheep whereby the last picked sheep was white. Then $b_2=1$, $w_2=0$, and we have the recursion
$$b_{n+1}={n-1\over n}(b_n+w_n),\quad w_{n+1}={1\over n}b_n\qquad(n\geq2)\ ,\tag{1}$$
in particular $b_3={1\over2}$. From $(1)$ we obtain by eliminating $w_n$ the recursion
$$b_{n+1}={n-1\over n}b_n+{1\over n}b_{n-1}\ .$$
The sequence $(c_n)_{n\geq2}$ defined by
$$b_n:={ c_n\over(n-1)!}\qquad(n\geq2)$$
then satisfies
$$c_2=1, \quad c_3=1,\qquad c_{n+1}=(n-1)(c_n+c_{n-1})\ ,\tag{2}$$
whence is a sequence of natural numbers. One computes $(c_n)_{n\geq2}=(1,1,4,15,76,455,\ldots)$. This sequence can be found in OEIS as A002467. It turns out that $c_n$, defined by $(2)$, can be combinatorially identified as the number of non-derangements (i.e. permutations having at least one fixed point) in the symmetric group ${\cal S}_{n-1}$. Now it is well known that for large $n$ the fraction ${1\over e}$ of all permutations in ${\cal S}_{n-1}$ are derangements. This implies that
$$\lim_{n\to\infty} b_n=\lim_{n\to\infty}{c_n\over(n-1)!}=1-{1\over e}\ .$$
It follows that with probability $1-{1\over e}$ the sheep picking will go on forever. From this we can conclude that with probability
$$p={1\over e}$$ we shall see a barbecue taking place.
A: Edit: This answer only gives an upper bound for the requested probability, through the inequality:
$$P\left(\bigcup_{t=1}^{+\infty}A_t\right)< \sum_{t=1}^{+\infty}P(A_t)$$
for events $A_t$ that are not disjoint.

There is always exactly $1$ white sheep in the flock (the initial one, since you only add black ships). Hence the probability of drawing the white sheep on day $t$ with $t \in \mathbb N_{>0}$ is equal to $\frac{1}{t+1}$ and the probability of drawing the white sheep on two consecutive days is equal (due to independency of the drawings) to $$\frac{1}{t+1}\cdot\frac{1}{t+2}$$ Now, this can happen on days $1$ and $2$ or days $2$ and $3$ or days $3$ and $4$ etc. Assume we never stop drawing and denote with $A_t$ the event that the white sheep is drawn on days $t$ and $t+1$. Of course, for $t\ge 1$ the events $A_t$ are not disjoint. Hence the probability $p$ to "make it a barbecue" is the probability of the union of $A_t$ and is (strictly) less than the sum of their probabilities $$p<\sum_{t=1}^{+\infty}\frac{1}{(t+1)(t+2)}$$ For any $n \in \mathbb N$ you have that \begin{align}\sum_{t=1}^{n}\frac{1}{(t+1)(t+2)}&=\sum_{t=1}^{n}\left(\frac{1}{t+1}-\frac1{t+2}\right)\\[0.1cm]&=\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\dots+\left(\frac1{n-1}-\frac1n\right)\\[0.1cm]&=\frac12+\left(\frac13-\frac13\right)+\left(\frac14-\frac14\right)+\dots+\left(\frac1{n-1}-\frac1{n-1}\right)-\frac1n\\[0.1cm]&=\frac12-\frac1N\tag{1}\end{align} Hence letting $n\to \infty$ you have that the probability $p$ to make it a barbecue is less than $$p<\sum_{t=1}^{+\infty}\frac{1}{(t+1)(t+2)}=\lim_{n\to +\infty}\sum_{n=1}^{n}\frac{1}{(t+1)(t+2)}\overset{(1)}=\lim_{n\to +\infty}\frac12-\frac1n=\frac12$$
A: The answer is the following addition which comes from considering every possible way of having 2 white sheeps in a row:
$\sum_{k=1}^{\infty} \frac{1}{k(k+1)(k+2)}$
Using the formula below, we can see that the summation is equal to $1/4$ (notice how fractions with same denominator cancel out).
$\frac{1}{x(x+1)(x+2)}=\frac{0.5}{x}-\frac{1}{x+1}+\frac{0.5}{x+2}$
