# The Entropy of the phrase “Eile Mit Weile”

I want to calculate the Entropy of the phrase "Eile mit Weile". I found the probability of each letter as the following

$$P(e)=\frac{4}{12}$$ $$P(i)=\frac{3}{12}$$ $$P(l)=\frac{2}{12}$$ $$P(m)=\frac{1}{12}$$ $$P(t)=\frac{1}{12}$$ $$P(w)=\frac{1}{12}$$

Then I used the formula for entropy $$H(p_1,...,p_k)=-\sum_{i=1}^{k}p_i\log_2(p_i)$$ $$H(p_1,...,p_k)=-(\frac{4}{12}\log_2(\frac{4}{12})+\frac{3}{12}\log_2(\frac{3}{12})+\frac{2}{12}\log_2(\frac{2}{12})+\frac{1}{12}\log_2(\frac{1}{12})+\frac{1}{12}\log_2(\frac{1}{12})+\frac{1}{12}\log_2(\frac{1}{12}))≈2.355388542 \ Bits$$ But the correct answer is $1.63263$ Bits. I've made a mistake somewhere but can't figure out what I did wrong.

• Keep in mind that this depends on many assumptions you implicitly took. For instance, you disregard spaces, and the case of each letter. You also assume that each letter is independent, and is generated according to the frequencies of this particular sentence. With the same method, the entropy of "A AaA aa" would be zero, which could be absurd in several scenarios. – chi Apr 2 '18 at 16:22

• $1$ nat = $\frac{1}{In{2}}$bits and I want to find out what $2.35538854$bits is equal to, I'd just do cross multiplication but I come to the answer $28.2646625$nats - which is wrong – Ski Mask Apr 2 '18 at 13:01