<s>[INST] Is it possible to 'split' coin flipping 3 ways? When flipping a coin to make important decisions in life you can flip once to choose between 2 possible outcomes. (Heads I eat cake, Tails I eat chocolate!)
You can also flip twice to choose between 4 outcomes. (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads)
Can you use a coin to choose evenly between three possible choices? If so how?
(Ignoring slight abnormalities in weight)
 [/INST] Here's my method. Like voltrevo's method the expected number of tosses is also only $2\frac{2}{3}$. 
In order to explain this easily, lets say there are three people named p1, p2 and p3. If the coin comes up heads on the first flip then p3 looses and p1 and p2 flip to decide who wins. However, if it comes up tails on the first flip then you keep flipping until you get either two tails in a row or two heads. If it's two tails then p3 wins, and if it's two heads then p1 and p2 flip to decide who wins again. 
To see why this works, lets first find the probability that p3 loses. This can happen in the following ways where something like TTH represents tails followed by tails followed by heads. The probability of each case is also shown next to the case: 
H         = $\frac{1}{2^1}$
THH       = $\frac{1}{2^3}$
THTHH     = $\frac{1}{2^5}$
THTHTHH   = $\frac{1}{2^7}$
THTHTHTHH = $\frac{1}{2^9}$
...
In each new case we see that two new flips must be added, and so the probability of any one of these cases happening (ie the probability that p3 looses) can be given by the following geometric sum: 
$$\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{2n+1}$$
Solving this sum: 
$$\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{2n+1}=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{2n}=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n=\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}=\boxed{\frac{2}{3}}$$
From this we see that the probability of p3 losing is $\frac{2}{3}$ which means he has a $\frac{1}{3}$ chance of winning, and there is a $\frac{2}{3}$ chance of either p1 or p2 winning. So by having p1 and p2 flip an additional time whenever p3 loses they split this $\frac{2}{3}$ chance among each other giving each person a $\frac{1}{3}$ chance of winning.
Finally, we can make a probability model for the number of tosses in a given game. Here is a table which represents the number of tosses (x), the cases which yielded them and the probability of obtaining them (p(x)). 
$$\boxed{
\begin{matrix}
x & 1 & 2 & 3 & 4 & 5 & 6 & ...\\ 
Case & N/A & H,TT & N/A  & THH, THTT & N/A & THTHH,THTHTT & ... \\ 
p(x) & 0 & \left ( \frac{1}{2^1}+\frac{1}{2^2} \right ) & 0 & \left ( \frac{1}{2^3}+\frac{1}{2^4} \right ) & 0 &\left ( \frac{1}{2^5}+\frac{1}{2^6} \right ) & ...
\end{matrix}
}$$
You can see that all of the even numbers have two cases and these two cases correspond to P3 loosing and P3 winning respectively. Analyzing the pattern we can make a general formula for the probability of the game ending in x flips where x is an even integer: 
$$p(x)=\left ( \frac{1}{2^{x-1}}+\frac{1}{2^x} \right )$$
Using this formula, we can find the expected number of tosses.
$$\mathbb{E}(x)=\sum_{x=0}^{\infty}\left(2x\right)\left(\frac{1}{2^{\left(2x-1\right)}}+\frac{1}{2^{2x}}\right)=\sum_{n=0}^{100}\frac{n}{4^{n-1}}+\sum_{n=0}^{100}\frac{n}{2^{2n-1}}$$
To find the value of these sum, we will first consider the general case of the geometric series and take the derivative of both sides. 
$$\frac{d}{dx}\sum_{n=0}^{\infty}x^n=\frac{d}{dx}\frac{1}{1-x}$$
$$\boxed{\sum_{n=0}^{\infty}nx^{\left(n-1\right)}=\frac{1}{\left(1-x\right)^2}}$$
You can then see that the above two sums are just special cases of this formula and so we get that: 
$$\mathbb{E}(x)=\sum_{n=0}^{100}\frac{n}{4^{n-1}}+\sum_{n=0}^{100}\frac{n}{2^{2n-1}} = \frac{16}{9}+\frac{8}{9}=\boxed{2\frac{2}{3}}$$
So, in summary, this method should take less than 3 flips on average to evenly split a coin toss among three people. 
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