Editing Arithmetic coding (section)

===Equal probabilities===
In the simplest case, the probability of each symbol occurring is equal. For example, consider a set of three symbols, A, B, and C, each equally likely to occur. Encoding the symbols one by one would require 2 bits per symbol, which is wasteful: one of the bit variations is never used. That is to say, symbols A, B and C might be encoded respectively as 00, 01 and 10, with 11 unused.

A more efficient solution is to represent a sequence of these three symbols as a rational number in base 3 where each digit represents a symbol. For example, the sequence "ABBCAB" could become 0.011201<sub>3</sub>, in arithmetic coding as a value in the interval &#91;0,&nbsp;1). The next step is to encode this [[Ternary numeral system|ternary]] number using a fixed-point binary number of sufficient precision to recover it, such as 0.0010110001<sub>2</sub> – this is only 10 bits; 2 bits are saved in comparison with naïve block encoding. This is feasible for long sequences because there are efficient, in-place algorithms for converting the base of arbitrarily precise numbers.

To decode the value, knowing the original string had length&nbsp;6, one can simply convert back to base&nbsp;3, round to 6 digits, and recover the string.