Is the probability $\frac12$? I have $n$ players in total. Note that $n$ is even. We want to pick $\frac n2$ players uniformly at random. We have access to only one unbiased coin. We want to make this bisection in minimum expected number of coin tosses. What should I do?
My approach: I will keep tossing-if it is H I will add the player to team A else to team B. If there are already $\frac n2$ players in any team, I stop tossing and put the rest in other team. Leaving aside the problem of proving why this will give minimum number of tosses, I am not sure why the players have equal probability of getting to team A or B. It is clear for the first $\frac n2$ players but after that it gets a little messy.
Please don't say that it is symmetric so probability is half trivially.
 A: It can't be done with a finite number of tosses at all. If $n=4$, then there are $\binom42=6$ possible bisections, so you want the probability of each bisection to be $\frac16$, which is not a sum of powers of $\frac12$.
You could consider a protocol using an unbounded number of tosses, and then ask for a protocol that minimizes the expected number of tosses. Or a protocol that maximizes the probability of being done as soon as possible, or something like that.
A: Step 1: Set $k={n\choose n/2}$.
Step 2: Choose an integer between $1$ and $k$, uniformly at random with your coin.  This is a well-known problem, e.g. here.  If $k$ happens to be a power of $2$ it can be done in a finite number of flips; otherwise, either the number of flips is unbounded (but very unlikely to be large) or your distribution is not precisely uniform (but very close).
Step 3: Choose the subset of size $n/2$, based on the outcome of step 2, and a lexicographic ordering of the subsets.  For example, if $n=4$ and the players are $A,B,C,D$, then the first subset is $AB$, the second is $AC$, etc.
A: Let's work this out fully for $n=4.$
We have at least two flips and no more than three, because by the time $n-1$ players have been placed on teams one team will be full and the last player's team will already be determined.
Let $X_1,$ $X_2,$ and $X_3$ (if needed) 
be the outcomes of the flips in that order.
We'll first consider the first two flips, and for each of those cases consider the remaining flip if we need to.
Case $X_1 = H, X_2 = H.$ Then players $1$ and $2$ go on team $A$ and the other players go on team $B$; the outcome is $AABB,$ with probability $\frac14.$
Case $X_1 = T, X_2 = T.$ Then players $1$ and $2$ go on team $B$ and the other players go on team $A$; the outcome is $BBAA,$ with probability $\frac14.$
Case $X_1 = H, X_2 = T.$ Then players $1$ and $2$ go on teams $A$ and $B,$ respectively, and we have to flip again. This produces two sub-cases:


*

*$X_1 = H, X_2 = T, X_3=H.$ The outcome is $ABAB,$ with probability $\frac18.$

*$X_1 = H, X_2 = T, X_3=T.$ The outcome is $ABBA,$ with probability $\frac18.$


Case $X_1 = T, X_2 = H.$ Then players $1$ and $2$ go on teams $B$ and $A,$ respectively, and we have to flip again. This produces two sub-cases:


*

*$X_1 = T, X_2 = H, X_3=H.$ The outcome is $BAAB,$ with probability $\frac18.$

*$X_1 = T, X_2 = H, X_3=T.$ The outcome is $BABA,$ with probability $\frac18.$


Observe that for player number $k,$ for any $k$ such that $1 \leq k \leq n,$
for every sequence of flips that places player $k$ on team $A$ there is another sequence of flips of the same length (and the same probability) that places player $k$ on team $B.$
Therefore player $k$ has equal probability (specifically, $\frac12$) to be on either team.
This is true in general for any $n,$ not just for $n=4.$
But notice that there are two outcomes ($AABB$ and $BBAA$) that place players $1$ and $2$ on the same team, and the event consisting of these two outcomes has probability $\frac12$;
whereas the outcomes that place players $2$ and $3$ on the same team
($BAAB$ and $ABBA$) constitute an event whose probability is $\frac14.$
The criterion about each player having $\frac12$ probability to be on either team is only a small part of what we think of as the uniformity of random selection of players for the teams.
After all, we can achieve the $\frac12$ probability just by flipping a single coin and assigning the outcome $AABB$ to heads and $BBAA$ to tails.
That's why I would prefer a uniform distribution over all $\binom{n}{n/2}$ possible lists of assignments, as discussed in the other answers.
A: To make the numbers integers, let 2k = n. If you've assigned a players to team A and b players to team B, then you have k-a left for team A and k-b left for team B. The next player should therefore have probability (k-a)/(2k-a-b) probability to be assigned to team A. The simplest method is to find the smallest p such that 2p >= 2k-a-b, flip the coin p times, convert the result to binary, assign to team A if the result is <= k-a, assign to team B if the result is between k-a+1 and 2k-a-b, and repeat otherwise. From there, there's the the low hanging fruit of dividing out by the gcd of k-a and k-b. 
Further improvements, however, are more complicated. Suppose k=5. The first player can be assigned by a coin flip. The next should have a 4/9 chance for one team and 5/9 for the other. So you can flip the coin four times, convert to binary, and send them to one team for 1-4 and the other for 5-9. 10-16 will result in a repeat, so there are seven outcomes that are "wasted". But if you're unlucky enough to have to repeat, then there are 49 ways that can happen (7*7), and you can can split those 49 possibilities into five groups of 9 with 4 left over. So only those 4 are wasted. If you flip the coin four more times and get one of the 7 "wasted" possibilities, you can multiply that by the 4 wasted and get 28, which you can divide into three groups of 9 with 1 left over. And so on.
