Giving 1 apple to 1 of 3 children fairly with a coinflip I wish to give an apple to 1 of 3 children fairly using a coin flip game. Each child calls heads or tails, and I flip the coin once for each child. If exactly one child calls correctly, that child gets the apple. If there is no one who calls correctly the game repeats. If two children call correctly, the game repeats between the 2 children until only of them calls correctly. 
Is this game fair; i.e., does each child have the same probability of winning? I am assuming yes intuitively.
What about a game in which I only flip the coin once, and each child calls. Is this game fair? I am assuming yes intuitively.
What is the expected number of coinflips in my original game? I recursively got 6:
Let $N_3, N_2$ be the number of flips for three and two children respectively.  Then for three children:
$$P(0\space correct\space calls) = 1/8 $$
$$P(1\space correct\space call) = 3/8 $$
$$P(2\space correct\space calls) = 3/8 $$
$$P(3\space correct\space calls) = 1/8 $$
Hence: $E(N_3) = \frac{3}{8} \cdot3 + \frac{2}{8}\cdot (3 + E(N_3)) + \frac{3}{8}\cdot(3 + E(N_2))$
For two children:
$$P(0\space correct\space calls) = 1/4$$
$$P(1\space correct\space call) = 2/4$$
$$P(2\space correct\space calls) = 1/4$$
Hence: $E(N_2) = \frac{1}{2}\cdot 2 + \frac{1}{2}\cdot (2 + E(N_2))$
Thus: $E(N_2) = 4$ and $E(N_3) = 6$
For the curious, I am trying to see if this "tournament game" is "isomorphic" to randomly assigning "T" to a child, then doing 3 coinflips until a permutation of {T,H,H} is achieved and hence the assigned child gets the apple, as described by Tim Crack in Heard On the Street (17e). That "assignment game" expects 8 coinflips whereas I am getting 6 in my "tournament game." I am probably misinterpreting his description of "tournament game" or incorrectly calculating 6.
 A: Your method is fair by symmetry, and it has the nice property that its fairness doesn't depend on the coin being fair (though the expected number of flips does).
MJD's proposals in the comments also have this property and have the same expected number of flips.
If you want to minimize the expected number of flips and you don't mind relying on the coin being fair, you can assign one of the four patterns for two successive flips to each of the children and start again if the fourth pattern occurs. The expected number $x$ of flips then satisfies $x=2+\frac14x$, with solution $x=\frac83$, a significant improvement over $6$.
Barrycarter's suggestion of generating binary digits until you know which third of the unit interval the resulting number falls into is slightly less efficient than that. Since $\frac13=0.\overline{01}_2$ and $\frac23=0.\overline{10}_2$, you know which third the number will fall into when a digit repeats. This occurs with probability $\frac12$ for each digit except the first, so the expected number of digits you need to generate is $3$.
