Student-b

Dear Sir/Madam,

In this letter, I derive and tabulate the maximum entropy values for the probabilities of each side of biased $n$-sided dice, for $n=3,4,6,8,10,12,\mathrm{15<span\; style="font-size:\; revert;\; font-family:\; Lato,\; sans-serif;">,\; and </span>}$$20$. These probabilities for each of the n options (sides), are those which have the least input information beyond what we know, which is nothing more than the bias or average score on the $n$-sided die. I generalise the “Brandeis dice” problem from E T Jaynes’ 1963 lectures, to an $n$-sided die, from the 6-sided case. To calculate these probabilities, I obtain the solution of an $n+1$-order polynomial equation, derived using a power series identity, for the value of the Lagrange multiplier, $\lambda$. The resulting maximally-equivocated prior probabilities at the 5th, 17th, 34th, 50th (fair), 66th, 83rd, and 95th percentiles of the range from 1 up to $\mathrm{n }$will aid in decision-making, where the options are the conditions we cannot influence, but across which we may have a non-linear payoff.

We use the standard variational principle in order to maximise the entropy in the system.

$$\sum _{i=1}^n{p}_i{f}_k(x_i)=F_k$$

$$\sum _{i=1}^n{p}_i=1$$

where the index $k$ is not summed over in the first equation, and where the $p_i$ are the probabilities of the $n$ options, e.g. sides of an n-dice. $F_k$ are the numbers given in the problem statement (constraints or biases), and $f_k(x_i)$ are functions of the Lagrange multipliers ${\lambda }_i$. The second equation is just the probability axiom requiring the probabilities to sum to one. This set of constraints is solved by using Lagrange multipliers. The formal solution is

$$p_i=\frac{1}{Z}\exp [-{\lambda }_1{f}_1(x_1)-...-{\lambda }_m{f}_m(x_i)]$$

where $$Z({\lambda }_1,...,{\lambda }_m)=\sum _{i=1}^n\exp [-{\lambda }_1{f}_1(x_1)-...-{\lambda }_m{f}_m(x_i)]$$is the partition function and $-{\lambda }_k$ are the set of multipliers, of which for a solution to the problem there need to be fewer than n, though in our current problem as we shall see, there is only one. The constraints are satisfied if:

$$F_k=-\frac{\partial }{\partial {\lambda }_k}{\log }_{e}Z$$

for k ranging from 1 to m.

Our measure of entropy is given by $S=-\sum _{i=1}^n{p}_i{\log }_e{p}_i$ and in terms of our constraints, i.e. the data, this function is:

$$S(F_1,...,F_m)={\log }_{e}Z+\sum _{k=1}^m{\lambda }_k{F}_k$$

The solution for the maximum of S is:

$${\lambda }_k=\frac{\partial S}{\partial F_k}$$

For k in same range up to m. For our set of n-sided dice, $m=1$ and so I can simplify $F_k$to F. The $f_k(x_i)$ are simply the set of $i$ the values on the n sides of our die.

For the problem at hand of the biased die, I introduce the quantity q which I define as the tested, trusted average score on the given n-sided die in hand. That is, I set $F=q$ here, our bias constraint number, which can range from the lowest die value 1 through to the highest value, n.

$$q=q_0:=\frac{1}{2}(n+1)$$

i.e. 3.5 on a 6-sided die, then the die is fair, otherwise, it has a bias and therefore an additional constraint. I assume this is all I know and believe about the die, other than the number of sides, n.

We see that ${\lambda }_k$ becomes just $\lambda$ and the equation for $F_k$ reduces to

$$F=-\frac{\partial }{\partial \lambda }{\log }_{e}Z$$

and the equation for $S_k$ reduces to

$$S(F)={\log }_{e}Z+\lambda F$$

and its solution is

$$\lambda =\frac{\partial S}{\partial F}$$

After a little algebra, I found that the partition function Z is given by

$$Z=\frac{x(x^n-1)}{(x-1)}$$

and after some further algebra, I found that in order to determine the value of $x$, where $x=e^{-\lambda }$, corresponding to the maximum entropy (least input information) set of probabilities, we must find the positive, real root of the following equation, which is not unity:

$$(n-q)x^{n+1}+(q-(n+1))x^{n-1}+qx+(1-q)=0$$

By inspection, this equation is always satisfied by the real solution x = 1, which corresponds to the fair or unbiased die, with all probabilities equal to 1/n for n sides. We need the other real root, and we obtain this by simple numerical calculation. From the solution $x=x_q$ for the given value of bias q, the set of probabilities corresponding to maximum entropy for each side of the relevant n-sided die are easily generated.

The following tables may be of use in decisionmaking in business and other contexts, especially where the agent (the organisation or individual making a decision) has a non-linear desirability or utility function over the outcomes (i.e. the values of the discrete set of possible options), does not have perfect intuition and does not wish to put any more information into the decision that is not within the agent’s state of knowledge.

I present tables for n = 3, 4, 6, 8, 10, 12, 15, and 20 here, each at 7 bias values of q for each n, corresponding to percentages of the range from 1 to n of 5%, 17, 34, 50, 66, 83 and 95%. There is transformation group symmetry in this problem. If $i$ represents the side with $i$ spots up, then when we reflect from $i\to n+1-i$ and transform $x\to \frac{1}{x}$ we obtain the same probability, e.g. the probability of a 1 on a six sided die at 5th percentile bias is the same as a 6 at 95th percentile bias. This is why in our tables we can observe the corresponding symmetry in the values of the probabilities and in the entropy, which is maximal of all biases when there is no bias and thus no constraint. Readers may wish arbitrarily to adjust any of the probabilities in the tables in the appendix and recalculate the entropy $S=-\sum _{i=1}^n{p}_i{\log }_e{p}_i$, which will be lower than the maximum entropy value in the table.

APPENDIX Student-b Maximum Entropy Probability Tables

APPENDIX Student-b Maximum Entropy Probability Tables (ctd)

APPENDIX Student-b Maximum Entropy Probability Tables (ctd)

APPENDIX Student-b Maximum Entropy Probability Tables (ctd)