The tea bag problem: probability of extracting a single bag of tea Suppose you have a bunch of tea bags in a box, initially in pairs, like these:

Let us suppose the box initially contains only joined pairs of tea bags, say $N_0$ of them (thus making for a total of $2N_0$ tea bags).
Every time you want to make yourself a tea, you put a hand in the box and randomly extract a tea bag.
Sometimes you will find yourself with a joined pair, in which case you split it, take one for your tea, and put the other back into the box.
If you instead extract a single tea bag (which was already split before), you just take it.
Now if you ever happened to be in a similar situation, you will probably have noticed that after a while you will almost always extract single tea bags and seldolmly find doubles (which is not surprising of course).
The question is, what exactly is the probability $p_k$ of extracting a single tea bag, after $k$ tea bags have already been picked?
Suppose for this problem that each time there is an equal probability of extracting any of the tea bags, regardless of them being joined with another or not, so that after the first step (in which we necessarily extract and split a double) the probability of extracting a single bag is $p_1=\frac{1}{2N_0-1}$.
It is relatively easy, just by computing the values of $p_k$ for the first $k$s, to see that the answer to the problem is quite nice:
$$p_k = \frac{k}{2N_0 -1}.$$
How can we prove this?

An interesting variation of the problem is asking what happens if we instead consider the picking of a pair as a single event (instead that as two, as in the above considered case).
With this assumption the previous formula does not hold, as computing the first values of $p_k$ shows:
$$ p_1 = \frac{1}{N_0}, \\ p_2 = \frac{2(N_0-1)}{N_0^2} .$$
 A: Say you select tea bag $i$.  Your question, then, is what is the probability that teabag $\hat i$ has previously been selected. (where $\forall i$, teabag $i$ was initially paired with $\hat i$).  thus we are asked to compute the probability that a given teabag survived the selection process conditioned on the fact that we know some other specified teabag did survive.  the conditioning just requires us to start with a sample of $2N_0-1$ and then the probability that a given teabag survived $k$ selections is $$\frac {2N_0-2}{2N_0-1}\times \frac {2N_0-3}{2N_0-2}\times \cdots\times \frac {2N_0-(k+1)}{2N_0-k}=\frac {2N_0-(k+1)}{2N_0-1}$$
The answer you seek is the compliment of this, hence $$1-\frac {2N_0-(k+1)}{2N_0-1}=\frac k{2N_0-1}$$
A: lulu wrote an excellent answer. Nevertheless I would like to offer a small variation on lulu's derivation that I find easier to understand.
For every tea-bag $s$
$$
\substack{\text{Probability}\\\text{of drawing $s$}\\\text{on the $k+1$th draw}\\\text{while its pair, $t$}\\\text{is still attached}} = \underset{\substack{\text{avoid $s,t$}\\\text{on 1st draw}}}{\underbrace{\frac{2N-2}{2N}}}\underset{\substack{\text{avoid $s,t$}\\\text{on 2nd draw}}}{\underbrace{\frac{2N-3}{2N-1}}}\cdots\underset{\substack{\text{avoid $s,t$}\\\text{on $k$-th draw}}}{\underbrace{\frac{2N-(k+1)}{2N-(k-1)}}}\underset{\substack{\text{draw $s$}\\\text{on $k+1$-th draw}}}{\underbrace{\frac{1}{2N-k}}}
$$
Repeat for every tea bag and add up to obtain
$$
\begin{align}
\substack{\text{Probability}\\\text{of drawing a tea bag}\\\text{on the $k+1$th draw}\\\text{while its pair}\\\text{is still attached}} &= 2N\ \frac{2N-2}{2N}\frac{2N-3}{2N-1}\cdots\frac{2N-(k+1)}{2N-(k-1)}\frac{1}{2N-k} \\
&= \frac{2N-2}{2N-1}\frac{2N-3}{2N-2}\cdots\frac{2N-(k+1)}{2N-k} \\
&= \frac{2N-(k+1)}{2N-1} 
\end{align}
$$
Now take the complement to obtain
$$
p_k = 1-\frac{2N-(k+1)}{2N-1} = \frac{k}{2N-1}
$$
A: (This is a condensed version of @lulu 's answer.)
If in drawing${}_{k+1}\>$ I pick a certain bag $b$ then any other  bag, in particular the partner of $b$,  is among the $k$ previously drawn bags with probability
$${k\over 2N-1}\ .$$
A: I found a simple way to deal with this. I get Barry's Tea, which does come in paired bags. Furthermore, each bag is $3$ grams of tea, where the vast majority of teabags one sees are $2$ grams each. I get the Gold Blend and the Irish Breakfast. 
I take both teabags and put them in the mug and then pour boiling water on them. And wait five minutes. More flavor. Oh, I do separate the two teabags, because I later pull them out with tweezers and squeeze them with teabag tongs to get most of the liquid out. That would be clumsy with joined bags.  
http://www.barrystea.ie/ 
http://www.englishteastore.com/tea-bag-squeezer.html 
Afterthought: Barry's is the first tea that I have regularly purchased that comes with no string. It is entirely possible that teabags with strings are typically two grams and teabags without them are typically three grams. Oh, and I sometimes get a very nice pu-erh, loose leaves, from a nearby shop called Far Leaves.  
http://www.farleaves.com/ 
http://www.farleaves.com/collections/puer-teas
