# Entropy of a loaded dice throw

1. Nov 8, 2014

### ZetaOfThree

1. The problem statement, all variables and given/known data
A 6 sided dice is loaded such that 6 occurs twice as often as 1. What is the total probability of rolling a 6 if the Shannon entropy is a maximum?

2. Relevant equations
Shannon Entropy:
$$S=-\sum_i p_i \ln{p_i}$$
where $p_i$ is the probability that we roll $i$.

3. The attempt at a solution
We know that $\sum_i p_i = 1$ and we are given that $p_1=p_6/2$. So $$p_2+p_3+p_4+p_5+\frac{3}{2} p_6 =1 \Rightarrow p_5 = 1-p_2-p_3-p_4-\frac{3}{2} p_6$$ There we can write the Shannon entropy as $$S=-\left( \frac{p_6}{2} \ln{\frac{p_6}{2}} + p_2 \ln{p_2}+p_3 \ln{p_3}+p_4 \ln{p_4}+(1-p_2-p_3-p_4-\frac{3}{2} p_6) \ln(1-p_2-p_3-p_4-\frac{3}{2} p_6) + p_6 \ln{p_6}\right)$$
$$=-\left( \frac{3 p_6}{2} \ln{p_6} + p_2 \ln{p_2}+p_3 \ln{p_3}+p_4 \ln{p_4}+(1-p_2-p_3-p_4-\frac{3}{2} p_6) \ln(1-p_2-p_3-p_4-\frac{3}{2} p_6) - \frac{p_6}{2} \ln{2}\right)$$
To find an extremum, we partial differentiate and set it equal to zero:
$$\frac{\partial S}{\partial p_2} = - \left(\ln{p_2} +1 - \ln(1-p_2-p_3-p_4-\frac{3}{2} p_6) -1 \right) = -\left(\ln{p_2} - \ln(1-p_2-p_3-p_4-\frac{3}{2} p_6) \right) =0$$
$$\Rightarrow \ln p_2 = \ln(1-p_2-p_3-p_4-\frac{3}{2} p_6)$$
Differentiating with respect to $p_3$ and $p_4$, we find the same condition, so we conclude that $p_2=p_3=p_4=p_5$. We now write the Shannon entropy as $$S=- \left(\frac{3 p_6}{2} \ln{p_6}+4p_2 \ln{p_2} - \frac{p_6}{2} \ln{2}\right)$$ So $$\frac{\partial S}{\partial p_6}= -\left(\frac{3}{2} \ln{p_6}+\frac{3}{2} - \frac{1}{2} \ln{2}\right) = 0$$
Therefore we find that $p_6 = \frac{2^{1/3}}{e}$. But this is wrong, because if we plug in the probabilities into the Shannon entropy formula, we do not get a maximum. For example, we get a higher Shannon entropy if we plug in $p_1=p_2=p_3=p_4=p_5=1/7$ and $p_6=2/7$. Where did I go wrong? Maybe I found a minimum or something? If so, how do I get the maximum?
Thanks for any help.

2. Nov 9, 2014

### BvU

It looks as if you put $\displaystyle {d p_2\over d p_6}=0$. But isn't there a constraint that $4p_2+{3\over 2}p_6 = 1 \ \Rightarrow \ 4d p_2 + {3\over 2}dp_6 = 0$ ?

3. Nov 9, 2014

### ZetaOfThree

What step are you referring to? I don't think I used this.
Yes, you can use that condition to find $p_2$ after $p_6$ is found. So in the case above, we have $p_1 = \frac{1}{2^{2/3}e}$, $p_2=p_3=p_4=p_5= \frac{1}{8} \left(2- \frac{3 \sqrt [3] {2}}{e} \right)$ and $p_6 = \frac{\sqrt[3]{2}}{e}$.

4. Nov 9, 2014

### Staff: Mentor

The partial derivative gives you an optimum if p6 is independent of the other probabilities. It is not. You should get the correct result if you express p2 as function of p6 and then calculate the partial derivative.

5. Nov 9, 2014

### Dick

Given that you know $p_2=p_3=p_4=p_5$, and given the two other constraints you could write the entropy as a function of a single variable like $p_6$ and maximize that as a single variable problem. That should be straightforward.

6. Nov 9, 2014

### ZetaOfThree

Sounds about right. I did that, and got an answer I'm pretty sure is correct. Thank you all for the help!