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)
Best Answer
If you throw your coin $n$ times you have $2^n$ outcomes, the probability of each of which is $\frac{1}{2^n}$. The larger $n$ is, the better you can divide $2^n$ into three approximately equal parts:
Just define $a_n=[2^n/3]$ and $b_n=[2\cdot 2^n/3]$, where $[\cdot]$ denotes rounding off (or on). Since $\frac{a_n}{2^n}\to\frac{1}{3}$ and $\frac{b_n}{2^n}\to\frac{2}{3}$ as $n\to\infty$, each of the three outcomes
"the number of Heads is between $0$ and $a_n$",
"the number of Heads is between $a_n$ and $b_n$", and
"the number of Heads is between $b_n$ and $2^n$"
has approximately the probability $\frac{1}{3}$.
Alternatively, you could apply your procedure to get four outcomes with the same probability (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads) to your problem in the following way:
Associate the three outcomes Heads-Heads, Tails-Tails, Heads-Tails with your three possible choices. In the case that Tails-Heads occurs, just repeat the experiment.
Sooner or later you will find an outcome different from Tails-Heads.
Indeed, by symmetry, the probability for first Heads-Heads, first Tails-Tails, or first Heads-Tails is $\frac{1}{3}$, respectively.
(Alternatively, you could of course throw a die and select your first choice if the outcome is 1 or 2, select your second choice if the outcome is 3 or 4, and select your third choice if the outcome is 5 or 6.)