Certainty that one has found all of the socks in a pile Suppose that I have a pile of $n$ socks, and, of these, $2k$ are "mine." Each of the socks that is mine has a mate (so that there are $k$ pairs of my socks) I know $n$, but not $k$. Assume that all possible values of $k$ are equally likely.
Now, I draw socks from the pile, uniformly and at random. After I draw a sock, I keep it if it is mine, and discard it otherwise.
If, after I have drawn $m$ of my socks from the pile, I have drawn only pairs of my socks (that is, it I have drawn one of my socks, I have also drawn it's mate), how confident am I that I have drawn all $2k$ of my socks from the pile?

Suppose after $m$ draws, you might have exactly $j$ pairs of your socks and no singletons. Then, given $K=k,m,n$ and provided that  $j \le k$, $2j \le m$, and $2k \le n$, the probability of this event is 
$$\Pr(J=j \text{ pairs and no singletons}| K=k,m,n) = {k \choose j}{n-2k \choose m-2j}\big/ {n \choose m}$$
and so the probability that $J=j$ given $K=j,m,n$ is  
$$\Pr(J=j \text{ pairs and no singletons}| K=j,m,n) = {n-2j \choose m-2j}\big/ {n \choose m}.$$
You have suggested a flat prior for $K$, so the posterior probability given you have only have pairs is 
$$\Pr(K=j| J=j \text{ pairs and no singletons}, m,n) = \frac{\displaystyle{n-2j \choose m-2j}}{\displaystyle\sum_{k=j}^{\lfloor n/2 \rfloor} {k \choose j}{n-2k \choose m-2j}}.$$
If you do not know $j$ and you are just observing the that the number of your odd socks is $0$ after $m$ draws from the pile then the calculation becomes a little more complicated with extra sums
$$\Pr(K=J| \text{no singletons}, m,n) = \frac{\displaystyle \sum_{j=0}^{\lfloor m/2 \rfloor} {n-2j \choose m-2j}}{\displaystyle\sum_{j=0}^{\lfloor m/2 \rfloor}\displaystyle\sum_{k=j}^{\lfloor n/2 \rfloor} {k \choose j}{n-2k \choose m-2j}}.$$ 
If your prior for $K$ was not flat, it could easily be introduced into the results above to give
$${k \choose j}{n-2k \choose m-2j} \Pr(K=k) \big/ {n \choose m}\, , \, \, {n-2j \choose m-2j} \Pr(K=j) \big/ {n \choose m} \, ,$$
$$\frac{\displaystyle  {n-2j \choose m-2j}\Pr (K=j)}{\displaystyle\sum_{k=j}^{\lfloor n/2 \rfloor} {k \choose j}{n-2k \choose m-2j}\Pr (K=k)} \text{ and }  \frac{\displaystyle \sum_{j=0}^{\lfloor m/2 \rfloor} {n-2j \choose m-2j}\Pr (K=j)}{\displaystyle\sum_{j=0}^{\lfloor m/2 \rfloor}\displaystyle\sum_{k=j}^{\lfloor n/2 \rfloor} {k \choose j}{n-2k \choose m-2j}\Pr (K=k)}$$ respectively.
