The coin decides

A fair coin lands on heads or on tails, and each has a chance of one half. When you cannot choose between two things, tossing it is a good way to decide.

A fair coin: $P(H)=P(T)=\frac{1}{2}$. Two things, one toss.

$P(H)=P(T)=\frac{1}{2}$.

To choose among three things, toss twice. Heads-heads means the first, heads-tails the second, tails-heads the third. Tails-tails means: toss twice again. Each of the four pairs has a chance of one quarter.

Three things, two tosses: $HH\to 1$, $HT\to 2$, $TH\to 3$, and for $TT$ toss again. Each pair has chance $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$.

$HH\to 1,HT\to 2,TH\to 3,TT\to \circlearrowleft$ and $P(HH)=\frac{1}{4}$.

So a round ends with a choice three times out of four. The chance that the first thing is chosen in the end is one quarter divided by three quarters, which is one third, and the same for the second and the third: all three are treated fairly. On average you need four thirds of a round, which is eight thirds of a toss, about two point seven tosses.

Let $D$ be: the round decides. $P(D)=\frac{3}{4}$ and $P(1\mid D)=\frac{1/4}{3/4}=\frac{1}{3}$, and the same for $2$ and $3$. Rounds needed on average: $\frac{1}{3/4}=\frac{4}{3}$, so tosses: $2\times \frac{4}{3}=\frac{8}{3}\approx 2.7$.

Round decides $D$: $P(D)=\frac{3}{4}$ and $P(1\mid D)=\frac{1/4}{3/4}=\frac{1}{3}$. Tosses $E$: $E=2\cdot \frac{4}{3}=\frac{8}{3}\approx 2.7$.

The surprise: while the coin is in the air, you often find out which side you are hoping for. If so, you have already decided, and the toss was not needed.

While it spins, if you hope for one side, $h\in \{H,T\}$, then you have decided and the toss was not needed.

Hope $h$: $h\in \{H,T\}$, so decided.