# 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: $$p_1 = \frac{1}{N_0}, \\ p_2 = \frac{2(N_0-1)}{N_0^2} .$$

• Beautifully presented, both visually as well as textually. – Evan Aad Oct 10 '16 at 12:21

(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 $${k\over 2N-1}\ .$$

• Well it's hardly getting simpler than that! Thanks for the solution – glS Oct 11 '16 at 9:06

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 $$1-\frac {2N_0-(k+1)}{2N_0-1}=\frac k{2N_0-1}$$

• Your solution is very elegant, but I don't get the derivation. The desired probability is that of drawing a single tea bag, any single tea bag. You, on the other hand, seem to be calculating the probability of drawing a specific tea bag. Then shouldn't the final probability be multiplied by $2N_0$? – Evan Aad Oct 10 '16 at 12:29
• The answer doesn't depend on what $i$ is, I just introduced it for specificity. Whichever bag you pick, your only question concerns the survival of its mate. – lulu Oct 10 '16 at 12:34
• By way of analogy: suppose I ask "what's the probability of drawing a pair if I select two cards from a deck?" To do that, I select the first (don't care what it is) and then remark that $\frac 3{51}$ choices for the second give me a pair. This works because the answer is independent of which I select first. it would be different if, say, I include the two jokers. Now I have to worry that I chose a joker first. That's because I broke the symmetry...jokers are different from the others. Similarly, in your question, the key point is that every tea bag is the same. – lulu Oct 10 '16 at 13:10
• Put in other way: $P(S_k) = \sum_i P(S_k \wedge t_k=i) = \sum_i P(S_k \mid t_k=i) P(t_k=i) = P(S_k \mid t_k=i)$ where the last equality results because $P(t_k=i)$ is a constant. Here $S_k$ is the event of interest (single bag in extraction $k$) and $t_k$ is the label of the bag. What is computed in lulu's answer. is $P(S_k \mid t_k=i)$ – leonbloy Oct 10 '16 at 22:49

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 $s$ $$\substack{\text{Probability}\\\text{of drawing s}\\\text{on the k+1th draw}\\\text{while its pair, t}\\\text{is still attached}} = \underset{\substack{\text{avoid s,t}\\\text{on 1st draw}}}{\underbrace{\frac{2N-2}{2N}}}\underset{\substack{\text{avoid s,t}\\\text{on 2nd draw}}}{\underbrace{\frac{2N-3}{2N-1}}}\cdots\underset{\substack{\text{avoid s,t}\\\text{on k-th draw}}}{\underbrace{\frac{2N-(k+1)}{2N-(k-1)}}}\underset{\substack{\text{draw s}\\\text{on k+1-th draw}}}{\underbrace{\frac{1}{2N-k}}}$$

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+1th 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 $$p_k = 1-\frac{2N-(k+1)}{2N-1} = \frac{k}{2N-1}$$

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.

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/collections/puer-teas

• There's yet another tea bag trick that I know, but it demands a degree of flexibility and is not altogether appropriate for this forum. – Evan Aad Oct 10 '16 at 21:50
• @EvanAad well, yes. I first heard of that in a movie, imdb.com/title/tt0126604 This is also why Bill Maher refers to the Tea Party as the Teabaggers. – Will Jagy Oct 10 '16 at 21:55
• mmmh... probably I'm missing something but.. what? – glS Oct 11 '16 at 9:07
• @glS My answer has nothing to do with mathematics. I was just charmed by the topic, a very familiar one. Usually my answers of this sort are deleted within a day, but perhaps nobody has thought to flag this one. For the second part, look up teabagging online, it really is not appropriate to give any detail here; since I just looked it up, apparently I was not entirely correct in what I thought was meant by the term. – Will Jagy Oct 11 '16 at 14:35