# How many bits are required to encode information in probability set G = [0.001, 0.002, 0.003, 0.994]?

I am currently working on data compression and thought it would be a good time to read up on the basics of information theory to better understand data compression and its algorithms.

As I understand, given a set of data, we can compute the minimum no. of bits on average required to encode the data my multiplying the rounded-up entropy value with length of the set. Following is the formula to compute the entropy of a set:

$$H ( X ) = −\sum i = 1 N p ( x_i ) \text{log}_2 ⁡ ( p ( x_i ) )$$

So, for a set $$G = \{A,B,B,C,C,C,D,D,...D\}$$ of length 1000, the probability set would be $$\{0.001, 0.002, 0.003, 0.994\}$$ and the Entropy would be 0.06. Rounding this up would give 1. This means 1 * 1000 = 1000 bits would be required to encode this set.

This would entail I use only 1 bit per symbol to encode this whole set. I am unable to understand how can I use just 1 bit per symbol when there are 4 unique symbols in the set. Won't I require 2 bits per symbol at least? $$G = \{00, 01, 01, 10, 10, 10, 11, 11, ..., 11\}$$.

But this would lead to a usage of 2000 bits in total betraying the value computed using entropy. What am I missing here?