A 3-way coin toss

How do you produce a random number 1-3 with a (fair) coin?
Answer:
TH = 1
TT = 2
HT = 3
HH = repeat

Here's a slicker way of saying it:
Toss until you get a tail.
"first occurance of T" has a 2/3 chance of being on an odd,
a 1/3 chance of being on even,
  (b/c odd goes first)
so in case it's odd, toss again to divide.

BTW, how long does this take?
Expected is:
E = 2 + 1/4 E,
so 3/4 E = 2, E = 8/3
This is optimal.

Funny that -less- information (a 3-ary bit)
takes -more- tosses
(b/c it doesn't fit as well)

you can obviously do this for any kind of division;
[by binary expansion]
there are questions of efficiency,
and it's basically "finding diadically measurable sets
with given measures"
EG, if you toss a coin twice and toss out the first result,
you're dividing the interval as
0101
instead of
0011
...which requires finer divisions
(hence more tosses)